Instructor's Solutions Manual for Optics, 5 Edition Eugene Hecht

SOLUTION MANUAL

1

Chapter 2 Solutions

Chapter 2 Solutions 2.1

2.2

 2 1  2  z 2  2 t 2   2( z   t ) z  2 2 z 2   2 ( z   t ) t  2  2 2 t 2 It’s a twice differentiable function of ( z   t ), where  is in the negative z direction.

 2 1  2  2 2 2  t y

 ( y, t )  ( y  4t )2  y

 2( y  4t )

 2 2 y 2

  8( y  4t ) t  2  32 t 2 Thus,   4,  2  16, and,  2 1  2  2  16 t 2 y 2

The velocity is   4 in the positive y direction.

2.3

Starting with:

 ( z, t ) 

A ( z   t )2  1

 2 1  2  z 2  2 t 2 (z   t )   2 A z [( z   t )2  1]  2( z   t )2   2 1 A    2  2 2 3 2 2  z  [( z   t )  1] [( z   t )  1]   4( z   t )2 ( z   t )2  1   2 A   2 3 2 3   [( z   t )  1] [( z   t )  1]  3( z   t )2  1  2A [( z   t )2  1]3

2

Chapter 2 Solutions  (z   t )  2 A t [( z   t )2  1]2  (z   t )  2   2 A   t  [( z   t )2  1]2  t 2  4 ( z   t )    2 A   (z   t ) 2 2 z t z [(   )  1] [(   t )2  1]3     [( z   t )2  1] 4 ( z   t )2   2 A    2 2 [( z   t )2  1]3   [( z   t )  1]  2 A 2

3( z   t )2  1 [( z   t )2  1]3

Thus since  2 1  2  z 2  2 t 2

The wave moves with velocity  in the positive z direction.

2.4

c   c 3  10 8 m /s  5.831  1014 Hz    5.145  10 7 m

2.5

Starting with:  ( y, t )  A exp[ a(by  ct )2 ]

 ( y, t )  A exp[ a(by  ct )2 ]  A exp[ a(by  ct )2 ]  2 Aa c  c    2  y  t  exp[ a(by  ct )2 ] t b  b b  2 4 Aa 2 c 2  c  2  4  y  b t  exp[ a(by  ct ) ] b b2  t 2  2

c  2 Aa     2  y  t  exp[ a(by  ct )2 ] b  y b  c   2 4 Aa 2   4  y  t  exp[ a(by  ct )2 ] 2 b  b  y 2

Thus  ( y, t )  A exp[ a(by  ct )2 ] is a solution of the wave equation with   c /b in the + y direction.

2.6

(0.003) (2.54  10 2 /580  10 9 )  number of waves  131, c   ,

  c /  3  10 8 /1010 ,   3 cm. Waves extend 3.9 m.

2.7

  c /  3  10 8 /5  1014  6  10 7 m  0.6  m.   3  10 8 /60  5  10 6 m  5  10 3 km.

2.8

    5  10 7  6  10 8  300 m/s.

3

Chapter 2 Solutions

2.9

The time between the crests is the period, so   1 / 2 s; hence   1/  2.0 Hz. As for the speed   L /t  4.5 m/1.5 s  3.0 m/s. We now know  ,  , and  and must determine . Thus,

  /  3.0 m/s/2.0 Hz  1.5 m.

2.10

 =  = 3.5  103 m/s =  (4.3 m);  = 0.81 kHz.

2.11

 =  = 1498 m/s = (440 HZ) ;  = 3.40 m.

2.12

 = (10 m)/2.0 s) = 5.0 m/s;  = / = (5.0 m/s)/(0.50 m) = 10 Hz.

2.13

    (/2 )  and so   (2 /).

2.14 q

 /2

 /4

0

 /4

/2

3/4

sin q

1

 2 /2

0

2 /2

1

2 /2

cos q

0

2 /2

1

2 /2

0

 2 /2

sin(q  /4)

 2 /2

1

 2 /2

0

2 /2

1

sin(q   /2)

0

 2 /2

1

 2 /2

0

2 /2

sin(q 3 /4)

2 /2

0

 2 /2

-1

 2 /2

0

sin(q   /2)

0

2 /2

1

2 /2

0

 2 /2

q



5/4

3/2

7/4

2

sin q

0

 2 /2

1

 2 /2

0

cos q

1

 2 /2

0

2 /2

1

sin(q   /4)

2 /2

0

 2 /2

1

 2 /2

sin(q   /2)

1

2 2

0

 2 /2

1

sin(q  3 /4)

2 /2

1

2 /2

0

 2 /2

sin(q   /2)

1

 2 /2

0

2 /2

1

/2

/ 4

0

/4

/2

3/4





/2

0

/2



3/2

2

cos(kx   /4)

 2 /2

 2 /2

2 2

2 2

 2 /2

 2 /2

2 2

cos(kx  3 /4)

2 /2

2 /2

 2 /2

 2 /2

2 /2

2 /2

 2 /2

t

 /2

 /4 /2

 /4 /2

 /2 





0 0

3 /4

 t  (2 / )t sin( t   /4)

3/2

 2 /2

 2 /2

2 2

2 2

 2 /2

 2 /2

 2 /2

sin( /4   t )

 2 /2

2 /2

2 /2

 2 /2

 2 /2

2 /2

2 /2

sin q leads sin(q  p/2).

2.15

x kx 

2 x



2.16

4 2.17

Chapter 2 Solutions Comparing y with Eq. (2.13) tells us that A  0.02 m. Moreover, 2 /  157 m1 and so   2 /(157ml)  0.0400 m. The relationship between frequency and wavelength is    , and so  = / = (1.2 m/s)/0.0400 m  30 Hz. The period is the inverse of the frequency, and therefore   l/  0.033 s.

2.18

(a)   (4.0  0.0) m  4.0 m (b)    , so  20 m/s    5.0 Hz  4.0 m (c)  ( x, t )  A sin( kx   t   ) From the figure, A = 0.020 m 2 2   0.5 m 1 ;   2  2 (5.0 Hz)  10 rad/s k  4.0 m

    x  10 t    0.020 cos  x  10 t  2 2 2 

 ( x, t )   0.020 m  sin  2.19

(a)   (30.0  0.0) cm  30.0 cm. (c)   , so   /  (100 cm/s)/(30.0 cm)  3.33 Hz

2.20

(a)   (0.20  0.00) s  0.20 s. (b)   1/  1/(0.20 s)  5.00 Hz. (c)   , so   /  (40.0 cm/s)/(5.00 s1)  8.00 cm.

2.21

  A sin 2 (k x  t), 1  4sin2(0.2x  3t). (a)   3, (b)   1/0.2, (c)   1/3, (d) A  4, (e)   15, (f) positive x   A sin(kx  t), 2  (1/2.5) sin(7x  3.5t). (a)   3.5/2, (b)   2/ 7, (c)   2/3.5, (d) A  1/2.5, (e)   1/2, (f) negative x

2.22

From of Eq. (2.26) (x, t)  A sin(kx  t) (a)   2 , so    /2  (20.0 rad/s)/2, (b) k  2/, so   2/k  2/(6.28 rad/m)  1.00 m, (c)   1/, so   1/  1/(10.0/ Hz)  0.10s, (d) From the form of , A  30.0 cm, (e)   /k  (20.0 rad/s)/(6.28 rad/m)  3.18 m/s, (f) Negative sign indicates motion in  x direction.

2.23

(a) 10, (b) 5.0 × 1014 Hz, (c)  



(e) 2.24

1



c





3.0  108 8  6.0  10 7 m, (d) 3.0 × 10 m/s, 5.0  1014

   2.0  10 15 s, (f)  y direction

 2 /x 2   k 2 and  2 /t 2   k 2 2 . Therefore

 2 /x 2  (1/ 2 ) 2 /t 2  (  k 2  k 2 )  0.

2.25

 2 /x 2   k 2 ;  2 /t 2   2 ;  2 / 2  (2 v ) 2 / 2  (2 / ) 2  k 2 ;

therefore,  2 /x 2  (1/ 2 ) 2 /t 2  (  k 2  k 2 )  0.

2.26

(x, t)  A cos(kx  t  (/2))  A{cos(kx  t) cos(/2)  sin(kx  t) sin(/2)}  A sin(kx  t)

2.27

y  A cos(kx  t  ), ay  2y. Simple harmonic motion since ay  y.

5

Chapter 2 Solutions

2.28

  2.2  1015 s; therefore   1/  4.5  1014 Hz;   ,   /  6.7  107 m and k  2/  9.4  106 m1. (x, t)  (103V/m) cos[9.4  106m1(x  3  108(m/s)t)]. It’s cosine because cos 0  1.

2.29

y(x, t)  C/[2  (x  t)2].

2.30

 (0, t)  A cos(kt  )  A cos(kt)  A cos(t), then  (0, /2)  A cos(/2)  A cos ()  A,  (0, 3/4)  A cos(3/4)  A cos(3/2)  0.

2.31

Since (y, t)  (y   t) A is only a function of (y   t), it does satisfy the conditions set down for a wave. Since  2 /y 2   2 /t 2  0, this function is a solution of the wave equation. However, (y, 0)  Ay is unbounded, so cannot represent a localized wave profile.

2.32

k  3  106 m1,   9  1014 Hz,   /k  3  108 m/s.

2.33

     /     (2.0 m/s)(1/4 s)  0.5 m 

z t    0.50 m 1/4 s 

 ( z, t )  (0.020 m)sin 2 

 1.5 m 2.2 s     0.50 m 1/4 s 

 ( z, t )  (0.020 m)sin 2 

(z, t)  (0.020 m) sin 2(3.0  8.8) (z, t)  (0.020 m) sin 2(11.8) (z, t)  (0.020 m) sin 23.6 (z, t)  (0.020 m) (0.9511) (z, t)  0.019 m 2.34

d /dt  ( /x )( dx /dt )  ( /y )( dy /dt ) and let y  t whereupon d /dt   /x( )   /t  0 and the desired result follows immediately.

2.35

d /dt  ( /x )( dx /dt )   /t  0  k ( dx /dt )  k and this is zero provided dx/dt   , as it should be. For the particular wave of d Problem 2.32,   /y (   )   /t   3  10 6 (   )   9  1014  0 dt and the speed is 3  108 m/s. 

2.36

a(bx + ct)2  ab2(x + ct/b)2  g(x +  t) and so  c/b and the wave travels in the negative x-direction. Using Eq. (2.34)   /t  x /   /x t  [ A(2a)(bx  ct ) c exp[ a(bx  ct )2 ]] / [ A(2a)(bx  ct )b exp[ a(bx  ct )2 ]]  c /b;  

the minus sign tells us that the motion is in the negative x-direction.

2.37

(z, 0)  A sin(kz  ); ( /12, 0)  A sin( /6  )  0.866; (/6, 0)  A sin(/3  )  1/2; (/4, 0)  A sin (/2  )  0.

A sin (/2  )  A(sin /2 cos   cos /2 sin )  A cos   0,    /2.

A sin(/3   /2)  A sin(5/6)  1/2; therefore A  1, hence  (z, 0)  sin (kz  /2). 

2.38

Both (a) and (b) are waves since they are twice differentiable functions of z   t and x   t, respectively. Thus for (a)   a2(z  bt /a)2 and the velocity is b/a in the positive z-direction. For (b)   a2(x  bt/a  c/a)2 and the velocity is b/a in the negative x-direction.

6

Chapter 2 Solutions

2.39

(a) (y, t)  exp[ (ay  bt)2], a traveling wave in the y direction, with speed   /k  b/a. (b) not a traveling wave. (c) traveling wave in the x direction,   a/b, (d) traveling wave in the x direction,   1.

2.40

(x, t)  5.0 exp [  a( x  b /at )2 ], the propagation direction is negative x; 2   b /a  0.6 m/s. (x, 0)  5.0 exp(25x ).

2.41

  /  0.300 m; 10.0 cm is a fraction of a wavelength viz.

(0.100 m)/(0.300 m)  1/3; hence 2 /3  2.09 rad. 

2.42

8  1   3  10  30° corresponds to /12 or     42nm.

14   12   6  10 

2.43

(x, t)  A sin 2(x/ ± t/),   60 sin 2(x/400  109  t/1.33  1015),   400 nm,   400  109/1.33  1015  3  108 m/s.   (1/1.33)  1015 Hz,   1.33  1015 s.

2.44

exp[ i ]exp[i  ]  (cos   i sin  )(cos   i sin  )  (cos  cos   sin  sin  )  i (sin  cos   cos  sin  )  cos(   )  i sin(   )  exp[i(   )]  *  A exp[it] A exp[–it]  A2;  *  A. In terms of Euler’s formula  *  A2(cost  i sin  t)(cos t  i sin t)  A2(cos2t  sin2 t)  A2. 





2.45 2.46









If z  x  iy, then z*  x  iy and z  z*  2yi. z1  x1  iy1 z2  x2  iy2 z1  z2  x1  x2  iy1  iy2 Re( z1  z2 )  x1  x2 Re( z1 )  Re( z2 )  x1  x 2

2.47

z1  x1  iy1 z2  x2  iy2 Re( z1 )  Re( z2 )  x1 x2 Re( z1  z2 )  Re( x1 x2  ix1 y2  ix2 y1  y1 y2 )  x1 x2  y1 y2

Thus Re( z1 )  Re( z2 )  Re( z1  z2 ).

2.48

  A exp i(kxx  kyy  kzz), kx  k, ky  k, kz  k,  k  [( k )2  ( k  )2  ( k )2 ]1/ 2  k ( 2   2   2 )1/ 2 .

2.49

Consider Eq. (2.64), with  2 /x 2   2 f ,  2 /y 2   2 f ,  2 /z 2   2 f ,  2 /t 2   2 f . Then  2  (1/ 2 ) 2 /t 2  ( 2   2   2  1) f   0 whenever

 2   2   2  1.

*************************<<INSERT MATTER OF 2.50 IS MISSING>>***********************

7

Chapter 2 Solutions

2.51

Consider the function: (z, t)  Aexp[(a2z2  b2t2  2abzt)].

Where A, a, and b are all constants. First factor the exponent: 2

(a2z2  b2t2  2abzt)  (az  bt)2 

1  b  z a t . a2   

2 1  b    z t .

2  a    a  This is a twice differentiable function of (z   t), where   b /a, and travels in the  z direction.

 ( z, t )  Aexp  

Thus,

2.52 2.53

  (h/m)  6.6  1034/6(1)  1.1  1034 m.

 k can be constructed by forming a unit vector in the proper direction and multiplying it by k. The unit vector is [(4  0)iˆ  (2  0) ˆj  (1  0)kˆ ]/ 4 2  2 2  12  (4iˆ  2 ˆj  kˆ )/ 21 and   k  k (4iˆ  2 ˆj  kˆ ) / 21. r  xiˆ  yjˆ  zkˆ, hence

 ( x, y, z, t )  A sin[(4 k / 21) x  (2 k / 21) y  (k / 21)z   t ].

2.54

  k  (1iˆ  0 ˆj  0 kˆ ), r  xˆi  yˆj  zkˆ, so,     A sin(k  r   t   )  A sin(kx   t   ) where k  2 / (could use cos instead of sin).             (r1 , t )   [r2  (r2  r1 ), t ]   (k  r1 , t )   [ k  r2  k  (r2  r1 ), t ]         (k  r2 , t )   (r2 , t ) since k  (r2  r1 )  0     A exp[i(k  r   t   )] 

2.55

2.56

 A exp[i(k x x  k y y  k z z   t   )] The wave equation is: 1  2  2  2 2 v t where, 2 2 2 2  2  2  2 x y z

 2  (k x 2  k y 2  kz 2 ) A exp[i(k x x  k y y  kz z   t   )]  2   2 A exp i  k x x  k y y  k z z   t     t 2 where k  k x 2  k y 2  kz 2 k 2  k x 2  k y 2  kz 2

then,  2  k 2 A exp[i(k x x  k y y  k z z   t   )]

This means that  is a solution of the wave equation if  2   2 /k 2     /k.

8

Chapter 2 Solutions

2.57

θ sin θ 2 sin θ 3 sin θ 2.58

2.59

 /2

 /4

0

1

1/ 2

0

 /4 1/ 2

 /2

3 /4



5 / 4

3/2

7/4

1

1/ 2

0

1/ 2

1

1/ 2

2 0

2

 2

0

2

2

2

0

 2

2

 2

0

3

3/ 2

0

3/ 2

3

3/ 2

0

3/ 2

3

3/ 2

0



/2

/4 1/ 2

sin 

1

sin(  /2)

0

1/ 2

sin   sin(   /2)

1

 2

0

/4

/2

3 /4



5 / 4

3/2

0

1/ 2

1

1/ 2

0

1/ 2

1

7/4 2 1/ 2 0

1/ 2

0

1/ 2

1

1/ 2

0

1/ 2

0

1

2

1

0

1

 2

3/4 3/2 0 0 0

2 1 1 0

1 1



Note that the amplitude of {sin()  sin(  /2)} is greater than 1, while the amplitude of {sin()  sin(  3/4) is less than 1. The phase difference is /8.

2.60

x kx cos kx cos (kx  ) cos kx  cos (kx  )

/2  1 1 0

/4 /2 0 0 0

0 0 1 1 0

 /4 /2

/2 

0 0 0

1 1 0





1 1

9

Chapter 3 Solutions

Chapter 3 Solutions

3.1

Compare E y  2 cos[2  1014 (t  x /c )   /2] to E y  A cos[2 (t  x / )   /2]. (a)   1014 Hz,   c, and

  c /  3  108 /1014  3  10 6 m, moves in positive x-direction, A  2 V/m,    /2 linearly polarized in the y-direction. (b) Bx  0, By  0, Bz  E y /c.

3.2

Ez  0, E y  E x  E0 sin (kz   t ) or cosine; Bz  0, By   Bx  E y /c, or if you like,

 E E  0 (iˆ  ˆj )sin(kz   t ), 2

 E B  0 ( ˆj  iˆ )sin(kz   t ). c 2

3.3

First, by the right-hand rule, the directions of the vectors are right. Then kE   B and so (2 / )E   B  2 B, hence E   B  cB.

3.4

E /x  kE0 sin(kx   t ); B /t   B0 sin(kx   t ); kE0   B0 ; E0  ( /k ) B0 and Eq. (2.33)  /k  c.

3.5

(a) The electric field oscillates along the line specified by the vector  6iˆ  3 5 ˆj. (b) To find E0 , dot E0 with itself and take the square root, thus 36  45 10 4 V/m  9  10 4 V/m. (c) From the exponential    k  r  ( 5 x  2 y)( /3)  10 7 , hence k  ( 5iˆ  2 ˆj )( /3)  10 7 and     because the phase is k  r   t rather than k  r   t the wave moves in the    direction of k . (d) k  k  (  10 7 )2 , k    10 7 m 1 and

  2 /k  200 nm. (e)   9.42  1015 rad/s and    /2  1.5  1015 Hz. (f)     3.00  108 m/s.

3.6

(a) The field is linearly polarized in the y-direction and varies sinusoidally from zero and z  0 to zero at z  z0 . (b) Using the wave equation  2 Ey

 2 Ey

 2 Ey

x

y

z

 2

 2

 2

1  Ey  0, c 2 t 2 2

 2  2 2  z  k  2  2  E0 sin cos(kx   t )  0 z0 z c 0  

and since this is true for all x, z, and t each term must equal zero and so k  ( /c) 1  (c / z0 )2 . (c) Moreover,    /k  c / 1  (c / z0 )2 .

3.7

 (10 v/m) B (cos 0.5  )ˆj c

3.8

(a) c   , so   c /  (3  108 m/s)/(550  10 9 m)  5.45  1014 Hz. (b)   2  2 (5.45  1014 Hz)  3.42  1015 rad/s; k  2 /  2 /(550  10 9 m)  1.14  10 7 m 1 . (c) E0  cE0 , so B0  E0 /c  (600 V/m) / (3  10 8 m/s)  2  10 6 V-s/m 2  2  10 6 T.

10

Chapter 3 Solutions

(d) E ( y, t )  E0 sin(ky   t   ); E (0, 0)  0  E0 sin( ),   0; B(y, t)  B0 sin(ky   t   ); B (0, 0) = 0  B0 sin( ),   0; E (y, t)  (600 V/m) sin((1.14  10 7 m 1 ) y  (3.42  1015 rad/s) t ); B (y, t) (2  10 6 T )sin((1.14  10 7 m 1 ) y  (3.42  1015 rad/s) t ).

3.9

3.10

 E B  (ˆi  ˆj) sin(kz   t   /6) c  E B(0, 0)  (ˆi  ˆj) sin( /6) c  E  (ˆi  ˆj)E0 sin(kz   t   /6)  E( /2, 0)  (ˆi  ˆj)E0 sin(   0   /6)  E( /2, 0)  (ˆi  ˆj)E0 sin( 5  /6)  E( /2, 0)  (ˆi  ˆj)E (0.5) 0

3.11

  At y  0 and t  0, E  E0 ˆi. The wave travels in the  y( ˆj) direction. As a result, B must be in the kˆ direction. Then,  1 B( x, y, z, t )   E0 kˆ exp[i( ky   t )] C

3.12

  At z  0 and t  0, B  B0 ˆj. The wave travels in the  z (kˆ ) direction. As a result, E must be in the iˆ direction. Then,  E( x, y, z, t )  cB ˆi exp[i(kz   t )] 0

3.13

By Gauss’ law, E   / 0 , where   q /A is the surface charge density. Putting the average value of this electric field into uE   0 E 2 /2 gives uE   2 /8  0 .

3.14 3.15

uB  B 2 /2 0 ; c  1/  0 0 , so c 2  0  1/0 . uB  c 2  0 B 2 /2; E  cB, so uB   0 (cB)2 /2   0 E 2 /2  uE .       t T cos 2 (k  r   t )  (1/T ) cos2 (k  r   t ) dt . Let k  r   t   x; then t    2 cos (k  r   t )  (1/T ) cos2 x dx  (1/2 T ) (1  cos 2 x ) dx   (1/2T )[ x  0.5sin 2 x ]kk rr  t( t T ) .     Similarly use sin 2 (k  r   t )  (1/2) 1  cos 2(k  r   t ) and       sin(k  r   t ) cos (k  r   t )  (1/2) sin 2(k  r   t ) .

1 (1  cos 2 ) we have 2 1 1 1 cos2  t  [1  cos 2 t ]  1  cos 2 t T   [1  (sinc T ) cos 2 t ].

T 2 2 2 T

3.16

Using the identity cos2  

3.17

Using the identity sin 2  

1 (1  cos 2 ) we have 2 1 1 1 sin 2  t  [1  cos 2 t ]  [1  cos 2 t T ]  [1  (sinc T ) cos 2 t ].

T 2 2 2 T

11

Chapter 3 Solutions

3.18

    I  S T  c 2 0 E0  B0 cos2 (k  r  T )      c 2 0 E0  B0 cos2 (k  r   t )  c 2 0 E0 B0 /2; E0  cB0 , c  1/ 0 0 , so

 0 c  1/0 c. I  E02 /2c 0 . If E0  15.0 V/m, I  (15.0 V/m)2 /2(3  108 m/s)(4  10 7 m-kg/C 2 )  .9375/ W/m 2  .298 W/m 2 .

3.19

Start with the following relationship for the irradiance: c I  0 E02 2 P 1  10 3 W c 0 2 3  108 m/s  8.85  10 12 C  V 1 m 1 2 I  E0  E0  A 1  10 4 m 2 2 2 2  10 W/m 2  E02 2.7  10 3 C  V 1s1

Using the definition of a Volt (V  J/C):

E02  7.4  103 V 2 /m 2 E0  86 V/m 2

3.20

3.21

 1  2 A   r2      1.0 cm    40J/s  60 s/min  2.4  103 J/min    1.80  1015 (a) Since E   B, Then use     1.50  108 m/s, to obtain: k 1.20  10 7  B  (1.50  108 m/s)( 100 V/m) ˆi exp[i(kz   t )]

(b)   k E  (c) k E  (d) I  3.22

c



 1.5

 , then   k E  0  (1.5)2  0  2.25 0  1.99  10 11 F/m 0

 2

E02 

(1.99  10 11 F/m)(1.50  108 m/s) (100 V/m)2  14.9 W/m 2 2

Total Power  20 W; Total Area at 1.0 m  4 (10.0 m)2  4 m 2 ; I = Power/Area  (20 W)/(4 m 2 )  5/ W/m 2  1.6 W/m 2 .

3.23

(a)   1/  10 7 s,   c,   c /  C  30 m. (b) E y  0.08 cos[(2   (t  x /c)], Bz  E y /c.) (c) By Eq. (3.44), S T  c 0 E02 /2.

3.24

Will find I, then E0 using Eq. (3.44). Total Power  L  3.9  10 26 W ; Total area at 1.5  1011 m  4  (1.5  1011 m)2  9.0   10 22 m 2  2.8  10 23 m 2 .

I = Power/Area  (3.9  10 26 W) / (2.8  10 23 m 2 )  1.4  103 W/m 2 .

12

Chapter 3 Solutions From Eq. (3.44) I  (c 0 /2)E02 , so E0  2 I /c 0 ; E0 

2(1.4  103 W/m 2 ) (3  108 m/s)(8.85  10 12 s 2 -C 2 / m-kg)

E0  1.0  103 V/m

3.25

  E0  ( E0 / 2)(iˆ  ˆj ), k  (2 / 2 )(i  j ), hence  E  (10/ 2)(i  j )cos[( 2 / )( x  y)   t ] and I  c 0 E02 /2  0.13 W/m 2 .

3.26

(a) l  ct  (3.00  108 m/s)(2.00  10 9 s)  0.600 m. (b) The volume of one pulse is V  l R 2  2.9  10 6 m 3 ; therefore (6.0 J)/V  2.0  10 6 J/m 3 .

3.27

From Eq. (3.44), I  c 0 E02 /2 and so E0  2 I /c 0  7.535  10 22  2.7  1011 V/m.

3.28

u  (power)(t)/(volume)  (10 3 W(t )/[( r 2 )(ct )]  10 3 W/[ (10 3 )2 (3  108 )], u  1.1  10 6 J/m 3 .

3.29

V  Al  At so that N /At  nV /At  n  100 m 3 6 m/60s  10 m 2 s1 .

3.30

I /E  I /h  (19.88  10 2 )/(6.63  10 34 )(100  10 6 )  3  10 24 photons/m s.

2

n  (1/c )( I /E )  1016 photons/m .

3

3.31

N /t  P /h  P /hc  2.8  10 20 s1 .

3.32

Pe  iV  (0.25)(3.0)  0.75 W. This is the electrical power dissipated.

The power available as light is Pl  (0.01) Pe  75  10 4 W. (a) The photon flux is Pl /h  Pl  /hc  2.1  1016 photons/s. (b) There are 2.1  1016 in volume (3  10 8 )(1s)(10 3 ) m 2 . Therefore 2.1  1016 /3  10 5  0.69  1011 is the number of photons per cubic meter.

(c) I  75  10 4 W/10  10 4 m 2  7.5 W/m 2 .

3.33

I  P /4 r 2  100w/4 (1m)2  8W/m 2 , E0  2 I / 0 c  78V/m, and B0  E0 /c  2.6  10 7 T.

3.34

Imagine two concentric cylinders of radii r1 and r2 surrounding the wave. The energy flowing per second through the first cylinder must pass through the second; that is, S1 2 r1  S2 2 r2 , and so S 2 r  constant and S varies inversely with r. Therefore, since S  E02 , E0 varies as

3.35

p  E /c  h /c  2.2  10 23 kg m s1

3.36

dp /dt  dW /dt /c, with area A, P  dp /dt /A  dW /dt /Ac  I /c.

3.37

 V  (a) E( z, t )  i  6.0  cos k ( z  ct ) m    E0 E ˆ , thus B( z, t )   j cos k ( z  ct ) (b) Since B0  c c   (c) P  kˆ 0 E02 cos2 k ( z  ct ) c

1/r .

13

Chapter 3 Solutions

3.38

From Eq. (3.52) the force exerted by the beam of light, AP  p/t, where p(incident) = j /c.

For reflected light at normal incidence, p  twice the incident momentum  2(j /c) j AP  2(j /c)/t , but, I  , so P = 2I/c.

Area  time At an angle q with respect to the normal, only the component of momentum normal to the surface changes, so p (normal)  p cos  , so, P( )  2 I cos /c.

3.39

E  P  t  (300 w)(100s)  3  10 4 J, p  E /c  10 4 kg m/s.

3.40

(a) P  2 S /c  2(1.4  103 W/m 2 )/(3  108 m/s)  9  10 6 N/m 2 . (b) S, and therefore P, drops off with the inverse square of the distance, and hence S  [(0.7  10 9 m)2 /(1.5  1011 m)2 ] (1.4  103 W/m 2 )  6.4  107 W/m 2 , and P  0.21 N/m 2 .

3.41

I (absorded)  I and I (scattered)  (1  ) I; the pressure arises from both contributions, viz.

P   I/c  2 (1 −  ) I/c  (2   ) I/c.

3.42

The reflected component has a momentum change, and thus a pressure, of twice the incident momentum, while the absorbed component has a momentum change of the incident momentum.

P (reflected)  2(70.0%)I /c  2(0.700)(2.00  106 W/m 2 )/(3  108 m/s)  .93  10 2 N/m 2 .

P (absorbed)  (30.0%)I /c  (0.300)(2.00  106 W/m 2 )/(3  108 m/s)  .20  10 2 N/m 2 .

P  P (reflected)  P (absorbed)  1.13  10 2 N/m 2 .

3.43

S  1400 W/m 2 , P  2(1400 W/m 2 /3  108 m/s)  9.3  10 6 N/m 2 .

F  P A  (9.3  10 6 N/m 2 )(2  10 3 m 2 )  1.86  10 2 N

3.44

S  (200  103 W)(500  2  10 6 s)/A(1 s), F  A P  A S /c  6.7  10 7 N.

3.45

F  A P  A S /c  10 W/3  108  3.3  10 8 N, a  3.3  10 8 /100 kg  3.3  10 10 m/s2 ,

3.46

  at  3.3  10 10 t  10 m/s. Therefore t  3  1010 s or t  940 years.      B surrounds  in circles, and E is radial, hence E  B is tangent to the sphere, and no energy radiates outward from it.

3.47

(a)  = 5 × 1014 Hz (b)  = v/ = 0.65c/ = 3.9 × 10–7 m = 390 nm (c) n = c/v = 1.538

3.48

c /  2.42;   1.24  108 m/s.

3.49

0  540 nm; n  0 / ; 0 /n    406 nm.

3.50

n  c /  1/0.90  1.11  1.1.

3.51

n  c /  (3  10 8 m/s)/(1.245  108 m/s)  2.410

3.52

l   t  (c /n)t  (3.00  108 m/s)(1.00 s)/1.333  2.25  108 m.

14 3.53

Chapter 3 Solutions

  0 /n  (500 nm)/1.60  312.5 nm; (1.00  10 2 m)/(312.5  10 9 m)  3.2  10 4 waves.

3.54

t1  (20.0 m)/(c /1.47) and t 2  (20.0 m)/(c /1.63), hence t 2  t1  3.2/c  1.07  10 8 s.

3.55

The number of waves is vacuum is AB /0 . With the glass in place, there are ( AB  L )/0 waves in vacuum and an additional L / waves in glass for a total of ( AB /0 )  L (1/  1/0 ). The difference in number is L (1/  1/0 ), giving a phase shift of f of 2 for each wave; hence, 2 L (1/  1/0 )  2 L (n /0  1/0 )  2 L /20  2000 .

3.56

Thermal agitation of the molecular dipoles causes a marked reduction in K e but has little effect on n. At optical frequencies n is predominantly due to electronic polarization, rotations of the molecular dipoles having ceased to be effective at much lower frequencies.

3.57

From Eq. (3.70), for a single resonant frequency we have   Nqe2  1 n  1   2 2    0 me  0    

1/ 2

;

since for low-density materials n » 1, the second term is  1, and we need only retain the first two terms of the binomial expansion of n. Thus

1  x  1  x /2 and n  1  Nqe2 /[2 0 me (02   2 )].

3.58

(a) The polar molecule, water, in the liquid state, is relatively free to move in response to the incident radiation. In the solid state, the molecules are not free to move. (b) The radar (microwave) interacts strongly with the liquid water in the droplets.

3.59

The normal order of the spectrum for a glass prism is R, O, Y, G, B, V, with red (R) deviated the least and violet (V) deviated the most. For a fuchsin prism, there is an absorption band in the green, and so the indices for yellow and blue on either side ( ny and nB ) of it are extremes, that is, nY is the maximum, nB the minimum, and nY  nO  nR  nV  nB . Thus the spectrum in order of increasing deviation is B, V, black band, R, O, Y.

3.60

Since ( Nqe2 / 0 me )1/ 2 has dimensions of frequency, the right-hand side is dimensionless and the units check.

3.61

With  in the visible, 02   2 is smaller for lead glass and larger for fused silica. Hence n( ) is larger for the former and smaller for the latter.

3.62

Subtract 1 from each side of Eq. (3.70) and then invert both sides: 1/(n 2  1)  ( 0 me /Nqe2 )(02   2 ); since   2 c / the desired result follows.

3.63

C1 is the value that n approaches as  gets larger.

15

Chapter 3 Solutions

3.64

The horizontal values of n () approached in each region between absorption bands increase as  decreases. n1  1.54 1  400nm n2  1.50 2  800nm 1.54  n1  C1  C2 /12 1.50  n2  C1  C2 /22  1 1  n  .04  C2  2  2   1 2 



C2  n12 22 / 22  12 C2  8.53  10 nm 3



2

C1  n1  C2 /12  1.45

3.65

Subtracting the two equations 1.557  n1  C1  C2 /12 and 1.547  n2  C1  C2 /22 gives n  0.01  n1  n2  C2 (1/12  1/22 ) so that C2  n12 22 /(22  12 )  3.78  103 nm 2 . Then C1  n1  C2 /12  1.5345 and n (610 nm)  C1  C2 /32  1.545.

3.66

Binomially expanding n 2  1  A /(1  02 / 2 ) gives n 2  1  A(1  02 /  2 ) or n 2  (1  A)[1  A02 /(1  A) 2 ]. Taking the square root and expanding again gives n  (1  A)1/ 2 [1  A02 /2(1  A) 2 ]. This has the Cauchy form with C1  (1  A)1/ 2 and C2  A02 /2(1  A)1/ 2 .

3.67



E  2.66  1015 Hz h

16

Chapter 4 Solutions

Chapter 4 Solutions 4.1

Eos  V Eoi /r  KV Eoi /r ; thus V K /r must be unitless, and so K has units of (length)2. The only quantity unaccounted for is  and so we conclude that K   2 , and I i /I s  K 2   4 .

4.2

The degree of Rayleigh scattering is proportional to 1/ 4 . But y  1.45 and so 1/y4  (1/1.45 )4 hence violet is scattered (1.45)4  4.42 times more intensely than yellow. The ratio of yellow to violet is 22.6%.

4.3

The sinusoids represent the field, in this case the E-field of the disturbance. The wavefront is a surface of constant phase and it meets each sinusoid at the same point (same phase) in its development. The outward radial lines are rays and they are everywhere perpendicular to the wavefronts.

4.4

(a) On the left-hand side are the inertial, drag force, and elastic force terms; on the right-hand side is the electric driving force. (b) x0 (  2  02  i )  (qe E0 /me ) exp(i ), forming the absolute square of both sides yields x02 [(02   2 )2   2 2 ]  (qe E0 /me )2 and x0 follows by division and taking the square root. (c) As for , divide the imaginary parts of both sides of the first equation above, namely x0  (qe E0 /me )sin  , by the real parts, x0 (02   2 )  (qe E0 /me ) cos  to obtain   tan 1 [ /(02   2 )]. α ranges continuously from 0 to  /2 to .

4.5

(a) The phase angle is retarded by an amount  ny 2 /   y 2 / or (n  1)y /c. Thus E p  E0 exp i[ t  ( n  1)y /c  y /c ] or E p  E0 exp[ i (n  1)y /c ]exp i (t  y /c ). (b) Since e x  1  x for small x, if n  1 or y  1, exp[ i ( n  1)y /c ]  1  i ( n  1)y /c and since exp(i /2)  i, E p  Eu   ( n  1)y( Eu /c ) exp( i /2).

4.6

sin 58  x /(5.0 m), x  4.2 m.

4.7

The statue is 16 m from the point of incidence, and since the ray-triangles are similar, 4 m : 16 m as 3 m : Y and Y  12 m.

4.8

At the first mirror,  r  i . For the second,  i  90   r  90   i and r  i, so  r  90   i .

4.9

 n  60  r 2  75

17

Chapter 4 Solutions

4.10

OQ 

1/ni 1/nt  sin  i sin  t

sin  i  sin  t 

1/ni OQ 1/nt



1 OQ 1

 ni sin i

  nt sin  t OQ OQ nt sin  t  ni sin  i

4.11

ni sin  i  nt sin  t , sin 30  1.52 sin  t ,  t  sin 1 (1/3.04), so  t  1913.

4.12

Ptransverse  mi sin  i

 m f sin  t

s BP . So where “m” is the presumed mass. But i  0 ,  f  t t (s0 )sin  i  ( BP )sin  t sin  i 

The factor

BP sin  t s0

BP corresponds to nti . s0

4.13

nt sin  t  ni sin  i sin  t 

ni 1 sin  i  sin 30  0.333 1.5 nt

 t  19.5  d  i   t  30.0  19.47  10.5 4.14

The slope of the curve is nit  ni /nt . Slope  0.75/1.00, so that nt  1.33.

This suggests that the dense medium is water.

4.15

 t  sin 1 [(sin 45)/2.42]  17, the angular deviation is 45°  17°  28°.

4.16

 t  sin 1 [(nw /ng )sin i ]  sin 1 [(8/9)sin 45]  39. For a ray incident in the glass at this angle,  t  sin 1[( ng /nw )sin 39]  sin 1 [(9/8) sin 39]  45.

18

4.17

Chapter 4 Solutions (a) nti  nt /ni   c /t  /  c /i   i /t  i /t  i /t . Therefore

t  i 3/4  9 cm. (b) sin i  nti sin  t ,  t  sin 1[(3/4)sin 45]  32.

4.18

t  i /nti  600/1.5  400 nm, color depends on frequency, which is the number of peaks that arrive per sec. This doesn’t change when going into a new medium. Thus it is still orange.

4.19

1.00 sin 55°  n sin 40°; n  1.27 or 1.3.

4.20

1.33 sin 35°  1.00 sin t; t  50°.

4.21

For i  0, 10, 20, 30, 40, 50, 60, 70, 80, 90 degrees, t  0, 6.7, 13.3, 19.6, 25.2, 30.7, 35.1, 38.6, 40.6, 41.8 degrees respectively.

4.22

Consider one ray on each side of the beam, with a perpendicular separation D. The width of the beam on the interface is D/cos i. Likewise, the width of the beam at the interface is Dcos t, where D is the perpendicular separation (width) of the rays in the glass, and D/cos i  Dcos t. (4.4) ni sin i  nt sin t so cos  t  (1  sin 2  t )1/ 2  (1  sin 2 i / ng2 )1/ 2

so 1/ 2

 sin 2 i  D 1   nt2  D   cos i

4.23

(4.4) ni sin  i  nt sin  t so sin(60.0°)  nt sin . Diameter of emerging beam (D) is related to the difference in horizontal displacement of red and violet light (h) by D cos(60.0°)  h (See Problem 4.19).

Red: sin  red  sin(60.0)/nred  ( 3 /2)/(1.505),  red  35.1; tan red  hred/10.0 cm so hred  (10.0 cm) tan(35.1°)  7.04 cm.

Violet: sin  violet  sin(60.0)/nviolet  ( 3 /2)/(1.545);  violet  34.1; hviolet  (10.0 cm) tan (34.1)  6.77 cm. D  h / cos(60.0)  (hred  hviolet )/cos(60.0)  (7.04  6.77)/(0.5)  0.54 cm .

4.24

Si nt 1   So ni 1.47 Si 

4.25

10 cm  6.8 cm 1.47

Si nt  So ni Si 1   Si  2 cm 3 cm 1.5 8 cm  2 cm  10 cm

19

Chapter 4 Solutions

4.26

1.00 sin 35°  1.50 sin t1; t1  22.48° and cos 22.48°  (2.00 cm)/L; L  2.16 cm or 2.2 cm.

4.27

sin i  n sin i/2; since sin 2  2 sin  cos  , sin i  2 sin (i /2)cos(i /2) and so setting these two expressions equal we get 1.70 sin(i /2)  2 sin( i /2) cos( i /2); cos i /2  0.85;

31.79°  i /2; i  63.6°.

4.28

The glass will change the depth of the object from dR to dA, where d A /dR  1.00/1.55; but dR  1.00 mm; hence, dA  0.645 mm and the camera must be raised 1.00 mm  0.645 mm  0.355 mm.

4.29

d A1 /dR1  1.50/1.33; dR1  1.00 m; dA1  1.1278 m; dR2  dA1  0.20 m; d A 2 /dR 2  1.00/1.50; dA2  1.3278(1.00/1.50)  0.885 m.

4.30

The number of waves per unit length along AC on the interface equals ( BC /i )/( BC sin i )  ( AD /t )/( AD sin  t ). Snell’s Law follows on multiplying both sides by c/.

4.31

With the origin in the plane of incidence, z  0; with the origin on the   interface y  0 so (ki  r )  kix x   (ki  r   r )  krx x   r   (kt  t   t )  ktx x   t and as  r   t  0, Eq. (4.19) becomes kix  krx  ktx or ki sin  i  kr sin r  kt sin  t . Since k  2 / sin  i

i



sin  t

t

,

which is the condition derived in Problem (4.26) for wave front continuity.

4.32

Let  be the time for the wave to move along a ray from b1 to b2, from a1 to a2, and from al to a3. Thus a1 a2  b1b2  i and a1 a3  t . sin  i  b1b2 / a1b2  i a1b2 , sin  t  a1a3 / a1b2  t a1b2 , sin  r  a1a2 / a1b2  i a1b2 , sin  i / sin  t  i /t  nt /ni  nti and  i  r .

4.33

ni sin i  nt sin  t , ni ( kˆi  uˆn )  nt ( kˆt  uˆn ), where kˆi , kˆt are unit propagation vectors. Thus nt (kˆt  uˆ n )  ni (kˆi  uˆn )  0,  (nt kˆt  ni kˆi )  uˆn  0. Let nt kˆt  ni kˆi    uˆn .  is often referred to as the astigmatic constant;  is the difference between the projections of nt kˆt  and n kˆ on uˆ ; in other words, take the dot product   uˆ : i i

n

n

  nt cos  t  ni cos  i .

4.34

Since i  r , kˆix  kˆrx and kˆiy  kˆry , and since (kˆi  uˆn )uˆn  kˆiy , kˆ  kˆ  2(kˆ  uˆ ) uˆ . i

r

i

n

n

20

Chapter 4 Solutions

4.35

Since SB  SB and BP  BP, the shortest path corresponds to B  coincident with B in the plane of incidence.

4.36

(Refer to Figure 4.35.) Let SP  a, distance along interface  S  B   x, distance of S and P from interface  h. t 

SB

i



BP

i

(h  x 2 )1/ 2 2

i



(h 2  (a  x )2 )1/ 2

i

Minimize t(x) w.r.t. x. dt x ( a  x )   0 2 2 1/ 2 2 dx i (h  x ) i (h  (a  x )2 )1/ 2 sin  i  sin  r or i  r

4.37

The mirrors are set as two sides of the acute triangle. The front of the laser is placed along the third side. The inscribed triangle is found by adjusting the position and the angle of the laser beam until the incoming and reflected beams meet on the third side of the triangle. This follows from format’s principle since the alignment relies on the law of reflection.

4.38

n1 sin  i  n2 sin  t ,  t   i, n2 sin i  n1 sin  t, n1 sin i  n1 sin  t and

i   t. cos  t  d / AB, sin(i   t )  a / AB, sin(i   t )  (a /d ) cos  t , d sin( i   t )/ cos  t  a.

4.39

The left and right beams will be parallel if t (Left)  t (Right) in the final medium (a). Since all interfaces are parallel, the transmitted angle into a medium equals the incident angle at the next medium.

At each interface (4.5) sin i  nti sin t.

Left: sin  i  n1a sin  t1  n1a (na1 sin  ta )  n1a na1 (n2 a sin  t 2 )  n1a na1 n2 a (na 2 sin  ta )  sin  ta .

Right: sin  i  n1a sin  t1  n1a (n21 sin  t 2 )  n1a (n21 sin  ta )  n1a n21 (na 2 sin  ta )  sin  ta .

For each beam,  ta   i .

4.40

Rather than propagating from point S to point P in a straight line, the ray traverses a path that crosses the plate at a sharper angle. Although in so doing the path lengths in air are slightly increased, the decrease in time spent within the plate more than compensates. This being the case, we might expect the displacement a to increase with n21. As n21 gets larger for a given i ,  t decreases,  i   t increases, and from the results of Problem 4.34, a clearly increases.

4.41

(a) In the case of a single thick piece of plexiglass, when looking down, one sees the text from the sheet underneath. Along the vertical surfaces, one sees the text from the page that is reflected off of the plexiglass – air interface. (b) When pressed lightly together, there is a very thin airgap between the two pieces of plexiglass. As a result, looking down through each, one sees the text beneath. However, along the vertical surfaces of each of the two pieces of plexiglass, one sees the text from the page that is reflected off of each of the plexiglass – air interfaces. This image is from total internal reflection due to the shallow angle at which the light from the text is incident on the vertical plexiglass-air interface.

21

Chapter 4 Solutions

(c) In the case of two pieces pressed lightly together with castor oil between the two pieces, one sees an image which is the same as in photgraph a (the case of a single, thick piece of plexiglass). This occurs because the index of refraction of castor oil is very nearly the same as that of plexiglass. As a result, the light from the text incident on the vertical plexiglass-castor oil interface does not undergo total internal reflection, but instead is transmitted through the interface (as if the interface wasn’t even there).

4.42

From Eq. (4.40), r  (1.52 cos 30  cos1913)/( cos1913  1.52 cos30), where from Problem 4.8, t  19°13. Similarly, t  2 cos 30 /(cos1913  1.52 cos 30), r  0.165, t  0.766.

4.43

Starting with Eq. (4.34), divide top and bottom by ni and replace nti with sin i / sin  t to get r 

sin  t cos  i  sin i cos  t , sin  t cos i  sin  i cos  t

which is equivalent to Eq. (4.42). Equation (4.44) follows in exactly the same way. To find r start the same way with Eq. (4.40) and get r 

sin  i cos  i  sin  t cos  t . sin  t cos  t  sin  i cos  i

There are several routes that can be taken now; one is to rewrite r as r 

sin  i cos  t  sin  t cos i cos i cos  t  sin  i sin  t sin  i cos  t  sin  t cos i cos i cos  t  sin  i sin  t

and so r 

sin( i   t ) cos( i   t ) tan( i   t )  . sin( i   t ) cos( i   t ) tan( i   t )

We call find t , which has the same denominator, in a similar way.

4.44

From Snell’s Law t  13.99°; from Eq. (4.43), r  tan 8.01 / tan 35.99  0.194;

using Eq. (4.42),

r   sin 8.01 / sin 35.99  0.237; [ Eor ]  r [ Eoi ]  1.94 V/m; [ Eor ]  r [ Eoi ]  4.74 V/m.

4.45

4.46

For small angles Snell’s Law becomes 1 i  n t : from Eq. (4.42) using the identity sin(   )  sin  cos   cos  sin  and using cos i  cos t  1 (  1  ) (   ) (n  1) . r   i t   i n1 i   (i   t ) ( i  n i ) (n  1) t  [t  ]i  0 

2 ni cosi ni cosi  nt cos t 2ni cos 0 2 ni  ni cos 0  nt cos 0 ni  nt

22

Chapter 4 Solutions

t  [t ]i  0 

2 ni cos  i nt cos i  ni cos  t 2 ni cos 0 2 ni  nt cos 0  ni cos 0 ni  nt

Thus, [t  ]i  0  [t ]i  0 .

4.47

Starting in air (ni  1) and going into glass (nt  1.5) : t 

2 ni 2.0   0.8 ni  nt 2.5

Starting in glass (ni  1.5) and going into air (nt  1) : t 

4.48

r 

2 ni 3.0   1.2 ni  nt 2.5

ni cos  i  nt cos  t ni cos  i  nt cos  t

[r ]i  0 

ni cos 0  nt cos 0 ni  nt  ni cos 0  nt cos 0 ni  nt

[t  ]i  0 

2 ni ni  nt

Air-to-Glass: 0.5  0.2 2.5 2ni 2   0.8 [t  ]i  0  ni  nt 2.5 [r ]i  0  

t   (r )  0.8  0.2  1 Glass-to-Air: 0.5  0.2 2.5 2ni 3   1.2 [t  ]i  0  ni  nt 2.5 [r ]i  0 

t   (r )  1.2  0.2  1 4.49

From (4.47), R  r 2  (n  1/n  1)2  (1.522  1/1.522  1)2  0.043.

T  1 – R  0.957.

4.50

T  1 – R  1 – r2  1  (n  1/n  1)2  1  (1.33  1/1.33  1)2  0.98.

From (4.55), I t  TI i  (0.98)(500 W/m 2 )  490 W/m 2 .

23

Chapter 4 Solutions

4.51

r 

nt cos  i  ni cos  t nti cos  i  cos  t  ni cos  t  nt cos  i cos  t  nti cos  i

Using Snell’s Law: ni sin i  nt sin  t sin 2 i  nti2 sin 2  t cos2  t  1  sin 2  t nti cos  t  nti 1  sin 2  t

Using 1  sin 2  t  1 

1 sin 2 i  0,1 , 2 nti

nti cos  t  nti2  sin 2  i 1 nti r  1 nti cos i  nti

nti cos  i 

r 

nti2  sin 2  i nti2  sin 2  i

nti2 cos  i  nti2  sin 2  i nti2 cos  i  nti2  sin 2  i

Now, looking at r : r 

ni cos i  nt cos  t ni cos i  nt cos  t

r 

cosi  nti cos  t cos i  nti cos  t

r 

4.52

r 

r  r  r  r 

cos i  nti2  sin 2 i cosi  nti2  sin 2 i

nti2 cos i  nti2  sin 2  i nti2 cos  i  nti2  sin 2 i 1.6 2 cos30  1.6 2  sin 2 30 1.6 2 cos30  1.6 2  sin 2 30 2.217  2.31 2.217  2.31

 0.187

cos  i  nti2  sin 2  i cos  i  nti2  sin 2 i cos30  1.6 2  sin 2 30 cos30  1.6  sin 30 2

2



 0.654   0.274 2.386

24

4.53

Chapter 4 Solutions R  r2  0.03497 R  r2  0.0751 T  1  R  0.965 T  1  R  0.925 1 T  T   12  0.965  0.925  0.945 2 R  1  T  0.055

T

4.54

Half of the unpolarized light s its E-field perpendicular to the plane of incidence ( I   500 W/m 2 ), then: I T   0.8  500 W/m 2  400 W/m 2 R  1  T  0.2 I T   0.2  500 W/m 2  100 W/m 2

4.55

Half of the unpolarized light s its E-field perpendicular to the plane of incidence ( I   1000 W/m 2 ), then: I R   300 W/m 2 , then R 

I R  200 W/m 2 , then R 

300  0.3 1000

200  0.2 1000

T  1  R  0.7 T  1  R  0.8 T

4.56

1 (T  T )  0.75 2

I inc  2000 W/m 2  I R   I R   I T   I T   300 W/m2  200 W/m2  (0.7)(1000 W/m2)  (0.8)(1000 W/m2)  300 W/m2  200 W/m2  700 W/m2  800 W/m2  2000 W/m2

4.57

From (4.47), R  r 2  ( nt  ni /nt  ni )2  (1.376  1.33/1.376  1.33)2  0.000289.

T  1 – R  0.999711.

From (4.55), I t  TI i  (0.999711)(400 W/m 2 )  399.884  400 W/m 2 .

4.58

r  nt  ni /nt  ni . Air-water: r 

4/3  1  1/7  0.14. Air-crown glass: 4/3  1

3/2  1  1/5  0.20. More reflectance for glass. From (4.54) and (4.56) 3/2  1 I r /I i  R  r 2 . Air-water: R  (1/7)2  0.02. Air-crown glass:

r

R  1 5   0.04 .

2

25

Chapter 4 Solutions

4.59

sin x  x  x 3 /3!  x 5 /5!   and so sin(   )  (   )[1  (   )2 /6]

using Snell’s Law  t (1   t2 /6  )  ( i /n)(1   t2 /6  ). Use 1 i  n t and the fact that when x is very small (1  x )1  1  x we have  t  ( i /n) (1   i2 /6)(1   i2 /6 n 2 ) dropping terms higher than the third power in  i we get  t  ( i /n) [1  ( n 2  1) i2 /6 n 2 ] and so    1  n2  1 2   i   t   i 1   1  i   .

6n2   n Using Eq. (4.42) and the power series representation of the sine, where terms higher than the third power in i are dropped, r 

n 1 n 1

4.60

i2 6n 2

i2

6n2

[ n 2  1  (n  1)3 ] [ n 2  1  (n  1)3 ]

2  n  1   i   1    n   n 1

cos( i   t )/ cos( i   t )  1  2 i2 /n multiplying by the ratio of the sines

from the previous problem, viz., [( n  1)/( n  1)](1   i2 /n) and dropping higher order terms yields the desired equation.

4.61

From Snell’s Law n sin  t  1sin 90  1 and so with Eq. (4.42) in mind, sin(   )  sin  cos   cos  sin 

and sin(90   t )  sin 90 cos t  cos 90 sin  t ;

then sin(90   t )  1cos  t ,

using sin 2   cos 2   1 and Snell’s Law

cos t  1  sin 2 t  sin(90   t )  1  (1/n)2 and so r  1 at glancing incidence.

4.62

Compute dr /di at i  90; we’ll use d t /d i  0 and then prove it; taking the derivative of Eq. (4.42) we get dr /d i   cos( i   t )/ sin(i   t )  sin(i   t ) cos(i   t )/ sin 2 ( i   t )

and for  i  90 this becomes dr /d i   sin  t / cos  t  sin  t cos  t / cos 2  t  2 tan  t

and using Snell’s Law, i.e., sin  t  1 n when i  90, and the fact that

cos t  1  sin 2  t , dr /di  2 tan  t  2sin  t / cos t  2/n cos t  2/ n 2  1 : this is the rise over the run at the end of the curve where i  90. Thus if

  is the angle made ith the vertical tan    n 2  1/2.

26 4.63

Chapter 4 Solutions [ Eor ]  [ Eoi ]  [ Eot ] ; tangential field in incident medium equals that in transmitting medium, [ Eot /Eoi ]  [ Eor /Eoi ]  1, t   r  1.

Alternatively, from Eqs. (4.42) and (4.44), sin( i   t )  2 sin  t cos  i 1 sin( i   t ) sin  i cos  t  cos  i sin  t  2 sin  t cos  i  1. sin i cos  t  cos  i sin  t

4.64

ni sin  i  nt sin  t so, sin  t  (ni /nt )sin  i  (1.00/1.52)sin(30)  0.33

 t  sin 1 (0.33)  19.2.

(Eq. 4.44) t   2 sin  t cos i / sin(i   t )  2 sin(19.2) cos(30)/ sin(49.2)  0.75.

(Eq. 4.42) r   sin( i   t )/ sin( i   t )   sin(10.8)/ sin(49.2)  0.25. t   (r )  0.75  (0.25)  1.00.

4.65

Let  i   p   /2   t . Reflected beam is polarized if r or r equal zero.

(4.43) r  tan  i   t  / tan  i   t   tan( /2   t   t ) / tan( /2   t   t )  tan( /2  2 t )/ tan( /2).

But tan( /2) is infinite, so r  0.

4.66

i   t  90 when  i   p , ni sin  p  nt sin  t  nt cos  p , tan  p  nt /ni  1.52,  p  5640.

4.67

At  p , r  0. So from (4.38) (nt /t ) cos i  (ni /i ) cos  t  0. Recall (4.4) ni sin  i  nt sin  t . (3.59) n   / 0 0 and cos 2   1  sin 2  . Approach:

solve for tan  p  sin  p / cos  p where  i   p .

4.68

tan  p  nt /ni  n2 /n1 , tan  p  n1 /n2 , tan  p  1/ tan  p .

sin  p / cos  p  cos  p / sin  p . Therefore sin  p sin  p  cos  p cos  p  0, cos( p   p )  0, so  p   p  90.

4.69

From Eq. (4.94), tan  r  r [ Eoi ] /r  Eoi   (r /r ) tan  i and from Eqs. (4.42) and (4.43) tan  r  

4.70

E  (4.56) R   or   Eoi 

cos( i   t ) tan  i . cos( i   r )

2

Eor2  Eor2   Eor2  . Eoi2  Eoi2   Eoi2  .

E  E  (4.34) r   or   (4.38) r   or  .  Eoi    Eoi 

27

Chapter 4 Solutions

Eor2   Eor2 

R 

Eoi2   Eoi2  ( Eor  /Eoi  )2 1  ( Eoi  /Eoi  )2

 

( Eor  /Eoi  )2 ( Eoi  /Eoi  )2  1

r2 r2  1  1/ tan 2 i tan 2  i  1

 R sin 2  i  R cos2  i

(4.57) T 

nt cos t  Eot    ni cosi  Eoi 

2

2

E  as above,  ot   t 2 sin 2  i  t cos 2  i , and using (4.63, 4.64),  Eoi   n cos  t  2 T,   t  t  , .  ni cos i 

T  T sin 2  i  T cos2  i .

4.71 4.72

Note that  e  41.8. Note that R increases steadily, while R has a minimum at i  0. nt cos  t t 2 cos  i from Eq. (4.45) T|| 

 sin i cos t  4 sin 2  t cos 2 i T||    2 2  sin t cos i  sin (i   t ) cos ( i   t ) sin 2i sin 2t T||  sin 2 (i   t ) cos 2 ( i  t ) T  nt t 2 cos  t /ni cos  i . From Eq. (4.44) and Snell’s Law,

 sin i cos  t   4 sin 2  t cos2  i  sin 2 i sin 2 t .

T       2 2  sin  t cos  i   sin  i   t   sin i   t 

4.73

Use (4.62) and (4.43). R  r2  tan 2 (i   t )/ tan 2 i   t   [sin 2 ( i   t )/ cos 2 ( i   t )]  [cos 2 ( i   t )/ sin 2 ( i   t )]. Note that R and T have now the same denominator.

Use (4.61) and (4.42). R  r2  sin 2 ( i   )/ sin 2 ( i   t ). Note that R and T have the same denominator.

4.74

If i is the incident radiant flux or power and T is the transmittance across the first air-glass boundary, the transmitted flux is then T  i . From Eq. (4.68) at normal incidence the transmittance from glass to air is also T. Thus a flux T i T emerges from the first slide, and i T 2 N from the last one. Since T  1  R, Tt  (1  R )2 N from Eq. (4.67).

28

Chapter 4 Solutions R  (0.5/2.5)2  4%, T  96%, Tt  (0.96)6  78.3%.

4.75

T  I ( y )/I 0  e  y , T1  e  , T  (T1 ) y . Tt  (1  R )2 N (T1 )d .

4.76

At i  0, R  R  R  [(nt  ni ) / (nt  ni )]2 . As nti  1, nt  ni and clearly R  0. At i  0, T  T  T 4 nt ni / (nt  ni )2 and since nt  ni ,

lim nti 1 T  4ni2 /(2ni )2  1. From Problem 4.61 and the fact that as nt  ni Snell’s Law says that  t  i , we have lim T  sin 2 2 i / sin 2 2i  1, lim T  1 .

nti 1

nti 1

From Eq. (4.43) and the fact that R  r2 and  t  i , lim nti 1 R  0.

4.77

(4.34) r     r   

4.78

ni cos i  nt cos  t  ni cos  i  nt cos  t cos i  nti cos  t cos i  nti cos  t cos  i  nti 1  sin 2  t cos  i  nti 1  sin 2  t cos  i  nti2  sin 2 i cos  i  nti2  sin 2  i nt cos i  ni cos  t nt cos  i  ni cos  t nti cos  i  1  sin 2  t nti cos i  1  sin 2  t nti2 cos i  nti2  sin 2  t nti2 cos  i  nti2  sin 2  t

For i   c , Eq. (4.70) can be written

  , r  cos  i  sin   n  cos i  i sin 2 i  nti2



2

i

r r* 

i

2 ti

1/2

1/2

cos2 i  sin 2 i  nti2  1. cos2 i  sin 2 i  nti2

Similarly r r*  1.

4.79

 c  sin 1 (1/1.5)  42.

4.80

Light incident from glass to air. t increases as i increases, if ni > nt, then ni/nt > 1, since sin  i  1, then there exists a c such that sin i  nt /ni This max  i   c .

(4.4) ni sin  i  nt sin  t so, sin  t  (ni /nt )sin i .

Maximum i  90° as  i  90, sin i  1 so, sin  t  ni /nt  sin  c .

4.81

1.00/2.417  sin  c ;  c  24 diamond refracts light back out and so looks brilliant.

29

Chapter 4 Solutions

4.82 4.83

sin 48.0  (1.00/n); n  1.35.

i  45   c sin i 

n

nt , where nt  1 ni

1  1.41 sin 45

4.84

Light entering at glancing incidence is transmitted at the critical angle and those rays limit the cone of light reaching the fish; sin  c  1/1.333;  c  49 and the cone-angle is twice this or 98°.

4.85

sin  c  nt /ni ;  c  59.1.

4.86

From Eq. (4.73) we see that the exponential will be in the form k(x  t), provided that we factor out kt sin i /nti , leaving the second term as  nti t /kt sin i , which must be t t. Hence  nt /(2 /t )ni sin i  t , and so t  c /ni sin i  i / sin i .

4.87

From the defining equation,   kt [(sin 2  i /nti2 )  1]1/2  3.702  10 6 m 1 , and since y   1, y  2.7  10 7 cm.

4.88

The penetration depth () is related to the attenuation coefficient ( ) by   1/ .

One first calculates the attenuation coefficient: 2  2 nt  ni    sin 2 i  1   0  nt   

1/2

2  2 1  3  2    sin 60  1 500 nm  2  

1/2

  1.257  10 7 [2.25  0.75  1]1/ 2   1.257  10 7  0.8292  1.04  106   9.6  10 7 m 4.89

4.90

nt ni 1.33 sin( c )  2.42  c  sin 1 (0.55)  33.3 sin( c ) 

sin( c ) 

nt ni

1.46 2.41  c  sin 1 (0.61)  37.3 sin( c ) 

4.91

The beam scatters off the wet paper and is mostly transmitted until the critical angle is attained, at which point the light is reflected back toward the source. tan  c  ( R /2)/d, and so nti  1/ni  sin[tan 1 ( R /2 d )].

30

Chapter 4 Solutions

4.92

1.00029 sin 88.7°  n sin 90°, n  1.00003.

4.93

Can be used as a mixer to get various proportions of the two incident waves in the emitted beams. This could be done by adjusting the gaps. [For some further remarks, see H. A. Daw and J. R. Izatt, J. Opt. Soc. Am. 55, 201 (1965).]

4.94

Light traverses the base of the prism as an evanescent wave, which propagates along the adjustable coupling gap. Energy moves into the dielectric film when the evanescent wave meets certain requirements. The film acts like a waveguide, which will support characteristic vibration configurations or modes. Each mode has associated with it a given speed and polarization. The evanescent wave will couple into the film when it matches a mode configuration.

4.95

From Fig. 4.62 the obvious choice is silver. Note that in the vicinity of 300 nm, nI  nR  0.6, in which case Eq. (4.83) yields R  0.18. Just above 300 nm, nI increases rapidly, while nR decreases quite strongly, with the result that R  1 across the visible and then some.

4.96

CYAN

MAGENTA

YELLOW

WHITE



R



G



B



R



B



R



G



G

R



R

R



G

G





R

R





G

G





B



R

R





G

G





B



R

R





G

G





B



R

R





G

G



G



G

G



G



R

R





B

YELLOW

RED

Y

Y

CYAN

WHITE

YELLOW GREEN MAGENTA

BLACK G GREEN

G

GREEN

CYAN

RED

YELLOW

4.97

GREEN GREEN BLACK

31

Chapter 4 Solutions

4.98

Graph # 1 2 3 4 5 6 7 4.99

Color Magenta Orange Light-Grey Red Plum-Red Yellow-Green Purple

t 

2 sin  2 cos 1 sin(1   2 ) cos(1   2 )

t 

2 sin 1 cos  2 sin(1   2 ) cos( 2  1 )

t t 

sin 21 cos 2 2  T sin 2 (1   2 ) cos 2 (1   2 )

from Eq. (4.100). Similarly t  t   T .

2

 tan(1   2 )    tan( 2  1 )  r2       tan(   )   tan(1   2 )  1 2 

2

2

 tan( 2  1 )  2 r2     r  R .  tan( )    1 2 

4.100 (4.84) Eoi t (1 )t ( 2 )  Eoi r ( p )r ( p )  Eoi .

(4.85) Eoi r (1 )t (1 )  Eoi t (1 )r( 2 )  0 where  2   i   p and r( p )  0. From Problem (4.66),  p  1 . From (4.84),

t ( p )t ( p )  0  1; t ( p )t ( p )  1. From (4.85), r ( p )t ( p )  0  0

Since t ( p )  0, r ( p )  0. From (4.100), T  t t , when T  1 , there is no reflected wave, as T  R  1.

4.101 From Eq. (4.45)    2 sin  p cos  p 2 sin  p cos  p t( p )t ( p )      sin( p   p ) cos( p   p )   sin( p   p ) cos( p   p )  sin 2 p sin 2 p  since  p   p  90 cos2 ( p   p ) 

sin 2 2 p cos2 ( p   p )

since sin 2 p  sin 2 p ,

sin 2 2 p cos2 (2 p  90)

 1.

32

Chapter 5 Solutions

Chapter 5 Solutions

5.1

All OPLs from S to P must be equal, therefore l0 n1  li n2  constant; drop a perpendicular from A to the optical axis, the point where it touches is B. BP  s0  si  x and the rest follows from the Pythagorean Theorem.

5.2

l0  li 3/2  constant, 5  (6)3/2  14. Therefore 2l0  3li  28 when l0  6, li  5.3, l0  7, li  4.66. Note that the arcs centered on S and P have to intercept for physically meaningful values of l0 and li .

5.3

The OPD from F1 to any point D on ∑ must be constant: ( F1 A)n2  ( AD )n1  C and ( F1 A)  ( AD )n12  C /n2  C ; if S corresponds to the directrix of the ellipse, ( F2 A)  e( AD ) where e is the eccentricity; if n12  e we get ( F1 A)  ( F2 A)  C .

5.4

A plane wave impinging on a concave elliptical surface becomes spherical.

If the second spherical surface has that same curvature, the wave will have all rays normal to it and emerge unaltered.

5.5

Recall that the angles of a triangle sum to 180.

 2  (180   )    180; 2     . sin  2  sin(   )  sin  cos(  )  cos  sin(  )  sin   sin   h /R  h /si (180  1 )      180

1     sin 1  sin(   )  sin  cos   cos  sin   sin   sin   h /R  h /s.

(4.4) n1 sin 1  n2 sin  2 ; n1 (h /R  h /so )  n2 (h /R  h /si ), ni /so  n2 /si  (n2  n1 )/R.

5.6

From Snell’s Law n1i  n2 t ; tan  i  yo /so and tan  t   yi /si since yi is negative; thus  i  yo /so and  t   yi /si , therefore MT  yi /yo  (n1 si )/(n2 s0 ).

5.7

From Eq. (5.8), 1/30.0  1.33/si  (1.333  1.000)/5.0; si  40 cm and MT  1.33, thus the image is 3.05 cm tall.

5.8

First surface n1 /so  n2 /si  (n2  n1 )/R, 1/1.2  1.5/si  0.5/0.1, si  0.36 m (real image 0.36 m to the right of first vertex). Second surface so  0.20  0.36  0.16 m (virtual object distance).

1.5/( 0.16)  1/si  0.5/( 0.1), si  0.069 m. The final image is real (si  0), inverted ( MT  0), and 6.9 cm to the right of the second vertex.

33

Chapter 5 Solutions

5.9

At the first surface from Eq. (5.8), 1/30.0  1.33/si  (1.333  1.000)/5.0 and si  40; a real image right of the vertex. For the second surface so  30 cm and the image will be right of the second vertex, so 1.33 / (30)  1/si  (1.000  1.333)/(5.000); and si  9.01 cm to the right of the second surface. The first surface produces a magnification of MT  1.33, thus the intermediate image is 4 cm tall. The second surface produces a magnification of MT  (n1 si )/(n2 so )  (1.333)(9.01)/(1.000)( 30)  0.4

and the total magnification is the product of the two, viz., 0.532. The image is real, inverted and minified.

5.10

(5.16) 1/f  (n  1)(1/R1  1/R2 ) where R2   R1 , so 1/f  ( n  1)(2/R1 ) R1  ( n  1)(2)( f )  (1.5  1)(2)( 10.0 cm)  10.0 cm

(5.17) 1/so  1/si  1/f ; so  1.0 cm; 1/si  1/f  1/so  1/10.0 cm  1/1.0 cm  9.0/10.0; si  1.1 cm.

(5.25) M T   si /so  (1.1 cm) / (1.0 cm)  1.1. Image is virtual, erect, and larger than the object.

5.11

(5.14) In the thin lens limit (d  0) becomes nm /so  nm /si  (nl  nm )(1/R1  1/R2 ) so, 1/so  1/si  1/f  (nl  nm /nm )(1/R1  1/R2 ). For a double concave lens R1  0, R2  0, so that (1/R1  1/R2 )  0. For air lens in water, nl  nm , so that nl  nm  0; 1/f  0, lens is converging.

5.12

(5.15) 1/so  1/si  (nl  1)(1/R1  1/R2 ) so, 1 1 1  1  (0.5)     si  20 10  20 1 1  40 20 si  40 

1/si  (nl  1)(1/R1  1/R2 )  1/so ; 1/si  13.3 cm.

(5.25) MT  si /so  13.3/20.0  0.67. Image is virtual, erect, and smaller than the object.

5.13

1/8  1.5/si  0.5/( 20). At the first surface, si  10 cm. Virtual image 10 cm to the left of first vertex. At second surface, object is real 15 cm from second vertex. 1.5/15  1/si  0.5/10, si  20/3  6.66 cm.

Virtual, to left of second vertex.

34

Chapter 5 Solutions

From (5.15) 1 1 1   1   (1.5  1)    8 si  20 10  1  1  1  (0.5)    si  20  8 1  0.15 si si  6.67 cm

5.14

(a) (5.17) 1/so  1/si  1/f so, 1/si  1/f  1/so  1 / (5.00 cm)  1/(1000 cm); si  5.03 cm  50.3 mm. (b) (5.25) MT  si /so  5.03 cm/1000 cm  .00503.

Image size  MT (object size)  (.00503)(1700 mm)  8.55 mm.

5.15

so  si  so si /f to minimize so  si , (d /dso )(so  si )  0  1  dsi /dso or

d  so si  si so dsi  0.    dso  f  f f dso Thus dsi /dso  1 and dsi /dso   si /so , therefore si  so .

5.16

1/5  1/si  1/10, si  10 cm virtual, MT  si /so  10/5  2 erect.

Image is 4 cm high. Or 5( xi )  100, xi  20, MT   xi /f  20/10  2.

5.17

1/so  1/si  1/f . For so  0, f , , 2 f , 3 f ,  f ,  2 f , f /2, si  0,  , f, 2 f, f 3/2, f/2, f 2/3, −f, respectively.

5.18

Draw a ray at 6.0° to the axis passing through the center of the lens. The image is virtual and on the image plane 50.0 cm in front of the lens. The image height yi is gotten from the fact that tan 6.0°  yi /f and so yi  5.3 cm.

5.19

1 1 1   So Si f 1 1 1   10 30 f f   15 cm

This lens must be negative because Si  So for negative lenses and So  Si for positive lenses

5.20

si  0 because image is virtual. 1/100 + 1/(−50)  1/f, f  −100 cm.

Image is 50 cm to the right as well. MT  si / so  50 / 100  0.5. Ant’s image is half-sized and erect ( MT  0).

35

Chapter 5 Solutions

5.21

1 1 1   So Si f

If So   then Si  f , thus if the object is moved So  18 cm and si  3 f , then: 1 1 1   18 3 f f 1 1 1 2    18 3 f f 3 f

f  12 cm 5.22

(5.16) 1/f  (nl  1)(1/R1  1/R2 )  (1.5  1)(1/20  1/(40))  3/80; f  27 cm. (5.17). 1/so  1/si  1/f so 1/si  1/f  1/so  1/27  1/40, si  83 cm. (5.25) MT  si /so  80/40  2.08. Image is real, inverted,

at 80 cm and twice the size of the object.

5.23

1/f  (nl  1)[(1/R1 )  (1/R2 )]  0.5(1/  1/10)  0.5/10,

f  − 20 cm, D  1/ f  1/0.2  5 D. 5.24

 1 1 1   (nl  1)    f  R1 R2  For this plano-convex lens, nl  1.530, R1  , and f  250 cm.

1 1  1  (1.530  1)    250 cm   R2   R2  (250 cm)(1.530  1) R2  132.5 cm If n was reduced to nl  1.500 :  R2  (250 cm)(1.500  1) R2  125.0 cm 5.25

1 1 1   So Si f

At So  , then Si  f , if the object is then moved So  90 cm and Si  3 f , then: 1 1 1   f 90 3 f 1 1 1 2    90 f 3 f 3 f f  60 cm

5.26

(5.16) 1/f  (nl  1)(1/R1  1/R2 )  (1.5  1)(1/(5.00 cm)  1/)  1/(10.0 cm); f  10.0 cm.

In a medium where nm  0, (5.16) becomes 1/ f  ((nl  nm ) nm )(1/R1  1/R2 ).

So, for water (nm  1.33), 1/ f  ((1.5  1.33)/1.33)(1/(5.00 cm)  1/).

f  39.1 cm (so f increases).

36

Chapter 5 Solutions

5.27

Yes, it is possible to move the lens to a new location and not change the positions S and P. This is done by moving the lens such that the original object distance is now the image distance and the original image distance is now the object distance.

5.28

1 1 1   So Si f

In this case the new positions are Si and So , such that: 1 1 1 1    So Si S0 Si s S 80 cm 40 cm 1  2, and MT   i    Then MT   i   so S0 40 cm 80 cm 2

5.29

MT  

Si S  N i So So

Since Si  So , and So  Si, then, Si S N o So Si N

Si2 S  N  i 2 So So

So N  Si

37

Chapter 5 Solutions

Next use, Si  So  d Si  So  d 1 1 1 1 1     So Si f So Si f 

Si So SS S S   i o  i o Si  So d d

Plug in So N  Si :

f 

so2 N So  So N



So2 N So (1  N )



So N (1  N )

Next, take from above and plug in So N  Si Si  So  d d  So N  So  So (1  N ) So 

d (1  N )

Plug this into the relation for f above, to obtain: f 

N (1  N )2

d

5.30

(5.17) 1/f  1/so  1/si  1/45  1/90  3/90; f  30 cm.

5.31

(a) From the Gaussian lens equation 1/15.0  1/si  1/3.00, si  3.75 m. (b) Computing the magnification, we obtain MT  si /so  3.75/15.0  0.25. Because the image distance is positive, the image is real. Because the magnification is negative, the image is inverted, and because the absolute value of the magnification is less than one, the image is minified. (c) From the definition of magnification, it follows that yi  MT yo  (0.25)(2.25 m)  0.563 m. where the minus sign reflects the fact that the image is inverted. (d) Again from the Gaussian equation 1/17.5  1/si  1/3.00 and si  3.62 m. The entire equine image is only 0.13 m long.

5.32

(5.17) 1/ f  1/so  1/si so, 1/si  1/ f  1/so  1/(30)  1/(10)  4/3. si  7.5 cm. (5.25) MT   si /so  ( 7.5)/30  1/4  0.25.

(Image size)  MT (object size)  (0.25) (6.00 cm)  1.50 cm.

The image is virtual, 7.5 cm in front of the lens, erect, and 1.50 cm tall.

38

5.33

Chapter 5 Solutions R1  R2 , so (5.16) becomes 1/ f  (nl  1)(1/R1  1/( R1 ))  (nl  1)(2/R1 )  1/si  1/so ; so  si  60 cm (Image real). MT  (25 cm)/(5.0 cm)  5  si /so so,

so  60 cm MT 

s s 25 5 i  i 5 so 60

so  300 cm 1 1 1 1 1 6      f so si 60 300 300 f  50 cm 1 1 2   ( nl  1)   f 50 R 2 0.02  (0.5)   R R  50 cm si  5(so );

so  5(so )  60 cm.

so  10 cm;

si  50 cm.

1/ f  1/so  1/si  1/10  1/50  6/50;

f  8.3 cm.

R1  (nl  1)(2)( f )  (1.5  1)(2)(8.3 cm)  8.3 cm.

5.34

1 1 1   So Si f 1 1 1   127 cm Si f MT  

Si So

5.80  

127 cm So

So  21.90 cm 1 1 1   127 cm 21.90 cm f f  18.7 cm

39

Chapter 5 Solutions

5.35

s M  2   i so M  si  2 so

 1 1 1  1  (n  1)     (1.5  1) 100 f R R cm 2   1 f  200 cm 1 1 1   f si so 1 1 1 3    200 2 so so 2 so so  300 cm d  si  so  3so  900 cm

Ray Diagram:

5.36

1 1 1   So Si f

At So  , then Si  f , if the object is then moved So  180 cm and Si  3 f , then: 1 1 1   f 180 3 f 1 1 1 2    180 3 f f 3 f f  120 cm  1 1 1   (nl  1)    f  R1 R2 

40

Chapter 5 Solutions Since this is an equiconvex lense, R1  R2 : 1 1 1 1 2  (nl  1)      (nl  1)   f R R f R 1  2   1  2   (1.5  1)        120 cm  R   2  R  R  120 cm

5.37

1/so  1/si  1/ f and MT  si /so  1/2 hence 1/so  2 / so  1/ f but

so  60.0 cm, hence f  20.0 cm; draw a ray cone from an axial image point, it enters the edges of the lens and focuses at 30.0 cm and then spreads out beyond to create a blur on the screen; from the geometry (0.04 mm)/(10.0 mm)  (R/300 mm), R  1.2 cm so the diameter is 2.4 cm.

5.38

The first thing to find is the focal length in water, using the lensmaker’s formula. Taking the ratio

fw / fa  fw /(10 cm)  (ng  1)/[(ng /nw )  1]  0.56/0.17  3.24; fw  32 cm. The Gaussian lens formula gives the image distance: 1/si  1/100  1/32.4; si  48 cm.

5.39

The image will be inverted if it’s to be real, so the set must be upside down or else something more will be needed to flip the image; M T  3   si /so ; 1/so  1/3so  1/0.60; so  0.80 m, hence

0.80 m  3(0.80 m)  3.2 m.

5.40 5.41

1/ f  (nlm  1)(1/R1  1/R2 ), 1/ fw  (nlm  1)/(nl  1) fa  0.125 / 0.5 fa , fw  4 fa .   1/so  1/si  1/ f hence for A and B, 1/(1.1 f )  1/si  1/ f and so si  11 f , hence MT  si /so  (11 f )/(1.10 f )  10; both vectors are imaged inverted   and 10 times larger than life, viz., 1 f long. A is in the x-direction and B is  in the z-direction. As for C, from (5.28)

M L   MT 2 M L  100

5.42

Image-to- object distance  L  so1  si1  so 2  si 2 . Also, so1  so 2  d  si 2  si1 1/ f  1/si1  1/so1 ; 1/ f  1/si 2  1/so 2  1/(si1  d )  1/(so1  d ).

Approach: With three is independent equations (two for 1/f and L  so1  si1 ) eliminate so1 and si1 , leaving f (L, d).

5.43

Find si1 first, and use this position for so 2 . (5.17) 1/ f  1/so  1/si , so 1/si1  1/ f1  1/so1 ; 1/si1  1/( 30)  1/(50)  (5  3)/150; si1  75 cm, which puts so 2 at (20  75 cm) cm  55 cm.

1/si 2  1/(50)  1/(55 cm)   0.038; si 2  26.2 cm, (real image).

5.44

For the first lens si1  (so1 f1 )/(so1  f1 )  37.5 cm and MT 1  1.50; for the second lens so 2  60.0  37.5  22.5 cm, and si 2  (so 2 f2 )/(so 2  f2 )  9.00 cm and MT 2  9.00/22.5  0.40; the net magnification is MT  MT 1 MT 2  0.60; the image is real, minified, and inverted.

41

Chapter 5 Solutions

5.45

MT 1  si1 /so1   f1 /(so1  f1 ), MT 2  si 2 /so 2  si 2 /(d  si1 ), MT  f1 si 2 /(so1  f1 )(d  sil ). From (5.30), on substituting for si1 , we have MT  f1 si 2 /[(so1  f1 )d  so1 f1 ].

5.46

(a) (5.17) 1/ f  1 / so  1/s2 , so 1/si1  1/ f1  1/so1  1/( 10.0 cm)  1/(15.0 cm)  (3  2)/(30.0); si1  30.0 cm  300 mm. (b) Image is real, inverted, and larger than the object. (c) (5.25) MT 1   si1 /so1  30.0/15.0  2.00. (d) si1 sets so 2 at 5.00 cm (beyond the second lens, virtual object).

1/si 2  1/ f  1/so 2  1/(7.50 cm)  1/(5.00 cm)  (2  3)/(15.00); si 2  15.0 cm  150 mm. (e) MT 2   si 2 /so 2  (15.0)/(5.00)  3.00. MT  ( MT 1 )( MT 2 )  (2.00)(3.00)  6.00. (Image is real, inverted).

5.47

First lens 1/si1  1/30  1/30  0, si1  . Second lens 1/si 2  1/(20)  1/( ), the object for the second lens is to the right at , that is so 2   . si 2  20 cm, virtual, 10 cm to the left of first lens. MT  ( /30)(20/  )  2/3 or from (5.34) MT  30(20)/[10(30  30)  30(30)]  2/3.

5.48

(5.17) 1/ f  1/so  1/si so 1/si1  1/ f1  1/so1  1/(15.0)  1/(25.0)  (5  3)/ 75.0;

si1  37.5 cm, which makes so 2  12.5 cm.

MT 1   si1 /so1  25.0/37.5  0.67.

1/si 2  1/ f2  1/so 2  1/( 15.0)  1/( 12.5); si 2  75.0 cm.

MT 2   si 2 /so 2  ( 75.0)/( 12.5)  6.00.

MT  ( MT 1 )( MT 2 )  ( 0.67)(6.00)  4.00.

Image is real, inverted, 75.0 cm beyond the second lens and 4 times the size of the object.

5.49

For the two positive lenses, note that incoming parallel rays result in outgoing parallel rays.

5.50

(5. 34) MT 

f1si 2 d (so1  f1 )  so1 f1

But si 2  so1  s  .

MT  lim s  MT 

where d  f1  f2 , MT 

f1s f1  . d (s  f1 )  sf1 d  (df1 /s)  f1 f1 , d  f1

f1  f1 / f2 ; ( f1  f2 )  f1

f2  f1 MT  (5.00 cm)(0.80 cm/0.10 cm)  40.0 cm; d  f1  f2  45.0 cm.

42

Chapter 5 Solutions

5.51

Figure 5.110

5.53

In 5.53a, the rays through h2 should bend away from the axis (diverging lens). In 5.53b, ray 4 should be directed at F1; also F2  F2 .

5.54

f 1200 mm  30  M o  40 mm fe f /#

5.55

f 120 cm   40 3 cm D

The angle subtended by L1 at S is tan 1 3/12  14. To find the image of the diaphragm in L1 we use Eq. (5.23): xo xi  f 2 , (6)( xi )  81, xi  13.5 cm, so that the image is 4.5 cm behind L1 . The magnification is  xi / f  13.5/9  1.5, and thus the image (of the edge) of the hole is (0.5)(1.5)  0.75 cm in radius. Hence the angle subtended at S is tan 1 0.75/16.5  2.6. The image of L2 in L1 is obtained from (4)( xi )  81, xi  20.2 cm, in other words, the image is 11.2 cm to the right of L1 . MT  20.2/9  2.2; hence the edge of L2 is imaged 4.4 cm above the axis. Thus its subtended angle at S is tan 1 4.4/(12  11.2) or 9.8°. Accordingly, the diaphragm is the A.S., and the entrance pupil (its image in L1 ) has a diameter of 1.5 cm at 4.5 cm behind L1 . The image of the diaphragm in L2 is the exit pupil. Consequently, 1/2  1/si  1/3 and si  6, that is, 6 cm in front of L2 . M T  6/2  3, so that exit pupil diameter is 3 cm.

5.56

(a) D1 is its own image since there is no lens to the left of it. (b) L is its own image since there is no lens to the left of it. (c)

1 1 1   so si f 1 1 1   4 cm si 12 cm 1 1 1 1    6 cm Si 12 cm 4 cm Si  6 cm MT  

Si 6 cm   1.5 4 cm So

Chapter 5 Solutions

43

(d)

 is the smallest angle, thus D2 is the aperture stop and its image (located 6 cm behind the lens) is the entrance pupil, which is 1.5  8 cm  12 cm in diameter.

5.57

Either the margin of L1 or L2 will be the A.S.; thus, since no lenses are to the left of L1 , either its periphery or P1 corresponds to the entrance pupil.

Beyond (to the left of) point A, L1 subtends the smallest angle and is the entrance pupil; nearer in (to the right of A), P1 marks the edge of the entrance pupil. In the former case P2 is the exit pupil; in the latter (since there are no lenses to the right of L2) the exit pupil is the edge of L2 itself.

5.58

The A.S. is either the edge of L1 or L 2 . Thus the entrance pupil is either marked by P1 or P2 . Beyond Fo1 , P1 subtends the smaller angle; thus 1 locates the A.S. The image of the A.S. in the lens to its right, L 2 , locates P3 as the exit pupil.

5.59

M

Do De

10 

5.0 cm De

De 

5.0 cm  5.0 mm 10

5.60

Draw the chief ray from the tip to L1 such that when extended it passes through the center of the entrance pupil. From there it goes through the center of the A.S., and then it bends at L2 so as to extend through the center of the exit pupil. A marginal ray from S extends to the edge of the entrance pupil, bends at L1 so it just misses the edge of the A.S., and then bends at L2 so as to pass by the edge of the exit pupil.

5.61

Figures P.5.61a and P5.61b.

5.62

No—although she might be looking at you.

44

Chapter 5 Solutions

5.63

The mirror is parallel to the plane of the painting, and so the girl’s image should be directly behind her and not off to the side.

5.64

1/so  1/si  2/R. Let R   : 1/so  1/si  0, so   si , and

MT  1. Image is virtual, same size, and erect.

5.65

For a plane mirror si  so , thus the tree is located next to the woman.

The tree will appear as if it were viewed from 1200 cm away, but will have left-right reversal.

5.66

Q 2 will appear to be located above Q1 . Q2 will appear to be located

at the same point in space as Q3 .

5.67

Q1 A  QA thus Q1 is 16 cm from A

For Q2 A, the light travels from the source to A (8 cm), from A to B (20 cm) so Q2 B  28 cm and Q2 A  28  20  48 cm.

For Q3 A, the light travels from the source to A (8 cm), from A to B (20 cm) and back to B (20 cm), thus Q3 A  48 cm.

5.68

DM 9 cm  DC 12 cm

45

Chapter 5 Solutions

5.69

Triangles GBH ad GFI are similar. As a result DG  FI  2 BH  2(0.20 m)  0.40 m Point is therefore 1.25 m  0.40 m  1.65 m above the floor.

5.70

tan(2   i ) 

20 cm  0.20 100 cm

2   i  11.31 tan( i ) 

5 cm  0.05 100 cm

i  2.86 2  8.45,   4.23 5.71

From Eq. (5.50), 1/100  1/si  2/80, and so si  28.6 cm. Virtual (si  0), erect ( MT  0), and minified. (Check with Table 5.5.)

46 5.72

Chapter 5 Solutions (5.48) 1/so  1/si  2/R, R  0.5 ft.

1/si  2/R  1/so  2/(0.5 ft)  1/( 5 ft), MT  si /so  (0.24)/(5)  0.048.

si  5/21  0.24 ft.

Image is virtual (seen in the mirror), erect, and 0.048 times the object size.

5.73

Ant has 3 images: from lens, from mirror, back out from lens. (i) (5.17) 1/ f  1/so  1/si so, 1/si  1/ f  1 / so  1/50.0  1/250  4/250, si  62.5 cm (between lens and mirror). (ii) (5.48) 1/so  1/si  2/R so, 1/si  2/R  1/so  1/so , ( R  ), si  187.5 cm (virtual image). (iii) 1/si  1/f  1/50  1/50.0  1/(250  187.5), si  56.5 cm. Real image, (left of lens).

5.74

Image on screen must be real, therefore si is positive.

1/25  1/100  2/R, 5/100  2/R, R  40 cm.

5.75

The image is erect and minified. That implies (Table 5.5) a convex spherical mirror.

5.76

1 1 1   So Si f 1 1 1   35 cm si 30 cm 1 1 1   Si 30 cm 35 cm Si  210 cm MT  

210 cm  6 35 cm

The image is magnified, inverted and real. It is 6  1 cm  6 cm.

R 2 R  2 f  60 cm f 

5.77

From Eq. (5. 8), 1/  n /si  (n  1)/R; si  2 R; n /2 R  (n  1)/R; n  2.

5.78

Want MT  si / so  yi /y0  1.0 cm / 100 cm  .01. so  1000 cm. si  so MT  (1000)(.01)  10 cm. Want real image, so si  0, and image

will be inverted. Detector is 10 cm in front of the mirror.

(5.50) 1/so  1/si  1/f ; 1/ f  1/1000  1/10  101/1000; f  9.9 cm.

5.79

1 1 1 2    So Si f R 1 1 2 1    30 cm si 12 cm 6 cm 1 1 1   Si 6 cm 30 cm

Chapter 5 Solutions Si  5 cm MT  

5 cm 1  30 cm 6

yi  0.1 cm

The image is virtual, minified and erect.

5.80

To be magnified and erect the mirror must be concave, and the image virtual; MT  2.0  si /(0.015 m), si  0.03 m, and hence 1/ f  1/0.015 m  1/( 0.03 m); f  0.03 m and f  R/2; R   0.06 m.

5.81

MT  yi /yo   si /so , using Eq. (5.51), si  fso /(so  f ), and since f   R /2, MT   f /(so  f )  ( R /2)/(so  R /2)  R /(2 so  R).

5.82

(5.49) f   R /2 so R  2 f . (5.50) 1/so  1 / si  1/ f . MT  si /so , so si  so MT  (10.0 cm)(0.037)   0.37 cm (image is virtual).

1/ f  1/so  1/si  1/(10.0)  1/(0.37), f  0.38 cm. R  2( 0.38)  0.76 cm.

5.83

(5.50) 1/ f  1/so  1/si ; so / f  1  so /si  (si  so )/si (si /so  1)  ( MT  1)/( MT ). so  f ( MT  1)/MT . (si /so ) (5.50) 1/ f  1/so  1/si ; si /f  si /so  1  ( MT )  1; si   f ( MT  1). 

5.84

MT   si /25 cm  0.064; si  1.6 cm. 1/25  1/1.6  2/R, R  3.0 cm.

5.85

Image size in plane mirror equals object size, so | MT (convex mirror) |  0.5; MT  si / so , so si  so MT  (5.0)(0.5)  2.5 m; si  2.5 m (image is virtual).

(5.50) 1/ f  1/so  1/si  1/(5.0 m)  1/(2.5 m), f  5.0 m.

5.86

(5.49) f  R/2. Primary f p  ((200 cm))/2  100 cm.

(5.50) 1/so  1/si  1/f p , so  , so 1/si  1/f p ; s  100 cm. Object for secondary is at so  25 cm. 1/ fs  1/so  1/si so 1/si  1/ fs  1/so .

Secondary fs   R /2  (60 cm)/2  30 cm.

1/si  1/(30)  1/(25); si  150 cm, or 75 cm behind the primary.

The effective focal length of the “lens” is 75 cm.

5.87

See Table 5.3. For f  so  2 f , a real inverted image is made with   si  2 f . If this image is directed back at the same angles, the final image will occur at the original object. So, it should be placed at the image of the lens (at si).

5.88

MT   si /so , so, si   MT so  1.5so .

(5.50) 1/ f  1 / so  1/si  1/so  1/(1.5so ); 1/10  1/3so ; so  10 / 3  3.3 cm.

Note in Table 5.5 so  f for an erect, magnified image.

5.89

f   R /2  30 cm, 1/20  1/si  1/30, 1/si  1/30  1/20. si  60 cm, MT  si /so  60/20  3. Image is virtual ( si  0), erect ( MT  0), located 60 cm behind mirror, and 9 inches tall.

47

48 5.90

Chapter 5 Solutions

The light passes through the lens twice.

1 1 1 1    f fL fL fM 1 2 2   f fL RM

5.91

Treat the first surface as a mirror with radius of curvature R. (5.49) fm   R /2, which is where the parallel reflected rays converge.

Lens: (5.16) 1/fl  (nl  1)(1/R1  1/R2 ); R1   R, R2   R so 1/ fl  (2  1)(1/( R)  1/R )  2/R; fl   R /2  fm .

5.92

Image is rotated through 180°.

5.93

From Eq. (5.61), NA  (2.624  2.310)1/ 2  0.560,

 max  sin 1 0.560  3405.

Maximum acceptance angle is 2 max  6810. A ray at 45° would quickly leak out of the fiber; in other words, it will strike the interior walls at less than Qc and thus only be partially reflected.

5.94

sin  max  NA  (n 2f  nc2 )1/ 2  (1.4812  1.4612 )1/ 2  0.2426

 max  14 2 max  28 5.95

Considering Eq. (5.62), log 0.5  0.30   L /10, and so L  15 km.

5.96

V-number 

C 

D 2 2 (n  nc )  2.405 C f 1 2

 D(1.4572  1.4512 )

1 2

2.405

C   D(0.0549)  0.604  m 5.97

 (8.0  m)(0.13)   1.358 m 2.405 2.405 c C   2.2  1014 Hz C 

 DNA

C

5.98

From Eq. (5.61), NA  0.232 and N m  9.2  102.

5.99

NA  (n 2f  nc2 )1/ 2  (1.50 2  1.482 )  0.2441

V-number   Nm 

 DNA C (100  10 6 m) (0.2441)  58.989 (1300  10 9 m) V2  1739 2

49

Chapter 5 Solutions 5.100 (5.68)  t  (Ln f /c)(n f /nc  1), so t /L  (n f /c)(n f /nc  1). t /L  1.500/(3  10 4 km/ns)(1.500/1.485  1)  50.51 km/ns.

5.101 M T   f /xo  ( 1/xo ) D. For the human eye D  58.6 diopters.

xo  230,000  1.61  371  103 km, MT  1/3.71  106 (58.6)  4.6  1011 , yi  2160  1.61  103  4.6  1011  0.16 mm.

5.102 Recall that the angles of a triangle sum to 180°. Recall that at both mirrors  r   i . For the triangle made by the three rays, (2 i1 )  (2 i 2 )  (180   )  180 so   2( i1   i 2 ). For the triangle containing “,”   (90  i1 )  (90   i 2 )  180.   ( i1   i 2 ), so,   2 .

5.103 1/20  1/sio  1/4, sio  5 m. 1/0.3  1/soe  1/0.6, soe  0.6 m.

MTo  5/20  0.25, MTe  

0.3  0.5  1.2, MTo MTe  0.125. (0.6)

5.104 From Table 5.3, the types, positions, and sizes of the images are OK, but the rays from one portion of an object do not consistently trace to the same portion of the image.

5.105 The pinhole allows the eye to get much closer to the object and still see it clearly and that creates a larger retinal image. The pinhole works like a magnifier.

5.106 Want same amount of light to reach the film. f / # varies as the square root of the time, so we want f /5.5.

5.107 See figure on page 686 of text, Ray 1 in the figure misses the eye-lens, and there is, therefore, a decrease in the energy arriving at the corresponding image point. This is vignetting.

5.108 Rays that would have missed the eye-lens in the previous problem are made to pass through it by the field-lens. Note how the field-lens bends the chief rays a bit so that they cross the optical axis slightly closer to the eye-lens, thereby moving the exit pupil and shortening the eye relief. (For more on the subject, see Modern Optical Engineering, by Smith.)

5.109 From Table 5.3, image is virtual, erect, and magnified. As thickness of lens approaches 0, si

approaches so , i.e., MT approaches 1. However, the entire bug is imaged, so that this can be used as a field-lens.

5.110 (a) D  1/ f . If so  , 1/ f  1/si . D  1/(0.02 m)  50 m 1 .

If so  0.50 m,1/ f  1/(0.50 m)  1/0.02 m, D  52 m 1 . (b) Accommodation of 2 m 1 . (c) D  1/ f  1/(0.25 m)  1/(0.02 m)  54 m1. (d) Need to add 2 m 1 .

5.111 Unaided eye, D  1/ f  1/so  1/si , ( si  2 cm), D  1/(1.25 m)  1/(0.02 m)  50.8 m 1 .

Want D  1/ f  1/(0.25 m)  1/(0.02 cm)  54 m1. Lens must have a power of (54  50.8)  3.2 m1.

5.112

1 1 1 1 1     f So Si  1

(a)

f  1 m D  1

50

Chapter 5 Solutions

1 1 1   f eye 18 25

(b)

f eye  10.46 mm 1 1 1 1 1     f f eye f c 10.46 1000 f  10.57 mm

new near point  18.3 mm 5.113 Dc  7D DC 7   7.82 D 1  DC d 1  (7)(0.015)

Dl 

5.114 DC 

Dl 9   10.1D 1  Dl d 1  (9)(0.012)

5.115 A far point of 16.67 cm means that Si  16.67 and So  .

1 1 1 1 1 1      So Si f 16.67  f 1 0.1667  6.46 D Dl   1  1  (1.2)    0.1667 



5.116 An object 25 cm away should appear 100 cm away.

1 1 1   So Si f 1 1 1   f 25 cm 100 cm

1 f  33 cm 3 1 1 1   So Si f 1 1 1   1 So  33 3 cm

The new far point is therefore 33 13 cm.

5.117 D1  Dc /(1  Dc d )  3.0 D /[1  (3.0 D)(0.017m)]  2.85D or to two figures 2.9 D. f1  0.345 m, and so

the far point is 0.345  0.017 m  0.33 m behind the eye lens. For the contact lens fc  1/3.0  3.3 m.

Hence the far point at 3.3 m is the same for both, as it indeed must be.

5.118 (a) (5. 77) MP  do  D  1  (0.25 m)(1/.0254 m)  1  10.8.

(b) Size  (MP) (object size)  10.8 (5.0 mm)  54 mm diameter. (c) tan  u  yo /do  (0.0254 m)/(0.25 m),  u  5.80  0.101 rad. (d) tan   yi /L  yi /do  (0.054 m)/(0.25 m),  a  12.19  0.213 rad.

Chapter 5 Solutions

51

5.119 (a) The intermediate image-distance is obtained from the lens formula applied to the objective; 1/27  1/si  1/25 and si  3.38  102 mm. This is the distance from the objective to the intermediate image, to which must be added the focal length of the eyepiece to get the lens separation; 3.38  102  25  3.6  102 mm.

(b) MTo  si /so  3.38  102 /27  12.5, while the eyepiece has a magnification of do D  254/25  10.2  . Thus the total magnification is MP  (12.5)(10.2)  1.3  102 ; the minus just means the image is inverted.

5.120 The x-ray “lens” is a mirrored surface that forms a portion of a non-spherical mirror. The reflected rays converge to the focus ( F1 ) .

5.121 (a) These are a parabola and a hyperbola of two sheets. The parabola and left-hand hyperbola share a common focus, F1 . Rays reflected from the parabola head for that focus. Rays directed at the first focus of a hyperbola reflect toward the second focus. (b) Parallel rays coming off the parabola seem to be leaving its first focus. Because this is also the focus of the ellipse the rays reflect toward its second focus.

5.122 The limit of resolution is 1.22/D; at 0.50mm, 1.22(0.50  106 )/2.4  2.54  107

radians; 1.0  102  R2.54  107 and R  39 km.

52

Chapter 6 Solutions

Chapter 6 Solutions

6.1

(6.7) MT  (si1 /so1 )( si 2 /s02 )   si /so . Let so   so that so  so1 , si1  f1 , so 2  (si1  d ), si  f .

Substituting into (6.7), ( f1 /so )(si 2 /  (si1  d ))   f /so . f  f1 (si 2 /(so 2 ))  f1 (si 2 /(si1  d )).

From 1 / so 2  1/si 2  1 / f2 ; 1/si 2  1/ f2  1/so 2 , si 2  so 2 f2 /(so 2  f2 ). f  ( f1 /so 2 )(so 2 f2 /(so 2  f2 ))   f1 f2 /(so 2  f2 )  f1 f2 /(si1  d  f2 ).

1/ f  (si1  d  f2 )/ f1 f2  1/ f1  (si1  d )/ f1 f2 .

But si1  f1 , so, 1/ f  1/ f1  1/ f2  d / f1 f2 .

6.2

From Eq. (6.8), 1/ f  1/ f   1/ f   d / f  f   2/ f   2/3 f , f   3 f  /4.

From Eq. (6.9), H11 H1  (3 f  /4)(2 f /3)/f   f  /2. From Eq. (6.10),

H 22 H 2  (3 f  /4)(2 f  /3)/f    f  /2.

6.3

From Eq. (6.2), 1/ f  0 when (1/R1  1/R2 )  (nl  1)d /nl R1 R2 . Thus d  nl ( R1  R2 )/(nl  1).

6.4

 1 1 1 (n2  1)d   (n2  1)     f R R n2 R1 R2  2  1

Taking d  1 cm, R1  10 cm and R2   9 cm,  1  1 1 0.5(1 cm)  0.5     f 10 cm 9 cm 1.5(10 cm)(9 cm)  1  0.0037 cm 1 f f  270 cm

6.5

1/ f  0.5[1/6  1/10  0.5(3)/1.5(6)10], f  24;

h1  24(0.5)(3)/10(1.5)  2.4, h2  24(0.5)(3)/6(1.5)  4.

6.6

Since h1  h2 it follows from Eqs. (6.3) and (6.4) that  f (nl  1)dl / R2 nl   f (nl  1)dl / R1 nl and

R2  R1 which means the lens is a sphere.

6.7

f  (1/2)nR /(n  1); h1   R, h2   R.

6.8

This is a thick lens with  R2  R1  R  10 cm and d  2R  20 cm.

(6.2)

1/ f  (nl  1)[1/R1  1/R2  (nl  1)d /nl R1 R2 ]  (1.33  1)[1/10  1/(10)  (1.33  1)(20)/1.33(10)(10)]; f  20.2 cm.

53

Chapter 6 Solutions

6.9

From Problem (6.6) or (6.7), (6.2) becomes 1/ f  (( nl  1)/nl )(2 / R), with R  10 cm.

1/ f  ((1.4  1)/1.4)(2/10)  0.057cm 1 , f  17.5 cm. (6.1) 1/ f  1/so  1/si , where so and si

are measured from the principal planes. 1/si  1/ f  1/so  1/(17.5) cm  1/(400  10) cm; si  18.3 cm.

(6.7) MT  si /so  18.3/390  .047. Image is real, inverted, and 0.047 times the size of the object.

6.10

(6.2) 1/ f  (nl  1)[1/R1  1/R2  (nl  1)d /nl R1 R2 ]; (1.5  1)[1/23  1/20  (1.5  1)(9.0)/1.5(23)(20)]  0,

f  .

Generally, if 1/ f  0, [1/R1  1/R2  (nl  1)d /nl R1 R2 ]  0; nl R2 nR ( n  1) d  l 1  l 0 nl R1 R2 nl R1 R2 nl R1 R2 (nl  1)d  nl ( R1  R2 ); ( R1  R2 )  (nl  1)nl d

for nl  1.5, ( R1  R2 )  ((1.5  1) / 1.5)d  d / 3.

6.11

f  29.6  4.0  33.6 cm; so  49.8  2  51.8 cm;

1 1 1   51.8 si 33.6

si  95.6 cm from H 2 and 91.6 cm from the back face.

6.13

From Eq. (6.2), 1/ f  (1/ 2)[1/ 4.0  1/( 15)  (1/ 2)4.0/(3/ 2)(4.0)( 15)]  0.147

and f  6.8 cm. h1  (6.8)(1/2)(4.0)/( 15)(3/2)  0.60 cm, while h2  2.3. To find the image 1/(100.6)  1/si  1 / (6.8); si  7.3 cm or 5 cm from the back face of the lens.

6.14

For both, –R2 = R = R1, so (6.2) becomes: 1 d  2   n  1    n  1 2  f R nR   2 1 5.0   1.5  1   1.5  1  ; f  50.8 cm f 1.5(50) 2   50

1 1 1 d     26.7 cm f f1 f 2 f1 f 2

6.15

(6.8) 1/ f  1/ f1  1/ f2  d / f1 f2  1/(20)  1/(20)  10/(20)(20); f  40 cm. The principal planes are found from (6.9) and (6.10).

(6.9) H11 H1  fd / f2  (40)(10)/(20)  20 cm.

(6.10) H 22 H 2  fd / f1  (40)(10)/(20)  20 cm.

6.16

1  D1  R1    from (6.16) where 0 1  D1  ( n  1) / R1  (1.5  1) / 2.5 cm  0.2 cm 1 .

0  1 T21    (6.24)  d21 /n 1  0  1 0  1    1.2/1.5 1   0.8 1   1  D2  R2   1  0

54

Chapter 6 Solutions but R2  , so D2  0.

1 0   1 0  1 0.2   1 0.2  (6.29) A  R2 T21 R1       0 1  0.8 1  0 1  0.8 1.16  Check: | A |  1(1.16)  0.2(0.8)  1.

6.17

  nt1  ni1     1.81  1.00   1     1      1 0.0736  R1    11.0      R1     1  0 0   0  1 1     1 1 0   T21   d   3.00  n 1    1.81

0 0  1 1.657 1   1   

  nt 2  ni 2     1.00  1.81   3 1     1       1 6.75  10  R2    R  120     2    1 0  0   0  1 1     1 6.75  10 3   1 0  1 0.0736  A  R2 T21 R1     1  1 0  1.657 1  0  0.989 6.75  10 3   1 0.0736  A  1.657 1  1 1.657 

0.989 0.080  A  1.657 0.876  6.18

When the object and image are both in air, nI  no  1. Since the distance between the vertex and the object is negative, take d0 is taken to be a negative.

0   0  1   y0  a12  I     1 0   0   y   a d  a a12 dI  a22    do 1   y0  21  I   11 I a11  do a12 a12   0  1    y    d (a  a d )  a  d a a12 d1  a22   y0  21 o 22  1   1 11 12 o  0 (a11  do a12 )  a12 y0  I     y    [ d (a  a d )  a  d a ]  y (a d  a )  21 o 22 0 12 I 22   I   0 I 11 12 o

 I   1    yI   d I

0   a11 1   a21 a11

a12   1 a22   do

y1   0 [ dI (a11  a12 do )  a21  do a22 ]  y0 (a12 dI  a22 )

6.19

Equation (6.36) is: dI 2 

a21  a22 d10 a11  a21d10

55

Chapter 6 Solutions

Where the matrix A is given by equation (6.31):

 D2 dl 1 nl a12      a22  dl   nl

a A   11  a21

D1 D2 dl  nl   Dd 1 1 l  nl 

 D1  D2 

In the case of a thin lens, dl  0 a12   1  D1  D2    a22   0 1 

a A   11  a21

Then (6.36) becomes: dI 2 

Using D1 

0  d1o 1  ( D1  D2 )d1o

nl  1 n 1 and D2  l , then R1 R2 ( D1  D2 ) 

1 f

And,

d1o d 1  1o f 1 1 1   So SI f dI 2 

1

So So  SI f So So  1 SI f SI 

This is equivalent to dI 2  6.20

d1o 1  1fo d

So So  So S 1 1 o f f

, when one keeps in mind the fact that when So  0, d1o  0.

Working in centimeters, D1  (2.4  1.9)/R1  0.1 cm 1 , D2  (1.9  2.4)/R2  0.05 cm 1

therefore 0   1  D1   1  D2   1 A  d /n 1    1  l 1  0  0  1 0.05  1 0  1 0.1 1.02 0.052  A  1   0.4 1  0 1   0.4 0.96  0

(1.02) (0.96)  (0.4) ( 0.052)  0.979  0.0208  1.

56 6.21

Chapter 6 Solutions

We have det A  a11 a22  a12 a21  1  ( D1  D2 )d 21 /nt1  D1 D2 d 212 /nt21

( D1  D2 )d21 /ntl  D1 D2 d212 /nt21  1.

6.22

h1  ni1 (1  a11 )/(a12 )  ( D2 d21 /nt1 ) f  (nt1  1)d21 f /R2 nt1 , from

Eq. (5.70) where nt1  nl ; h2  nt 2 (a22  1)/(a12 )  ( D1 d21 /nt1 ) f from Eq. (5.71), h2  (ni1  1)d21 f /R1 nt1 .

6.23

A  R2 F21 R1 , but for the planar surface

1 D2  1 0  R2    and D2  (nt1  1) / ( R2 ) but R2   R2  0 1  0 1     which is the unit matrix, hence A  F21 R1 .

6.24

6.25

0.6 2.6  D1  (1.5  1) / 0.5  1 and D2  (1.5  1)/  ( 0.25)  2, A    0.2 0.8  and | A |  0.48  0.52  1.

From the equation above (6.39), 1  ( nt1  d21 )    1 0.2   a12  (nt1  1)    1  .     R1 R2  R1 nt1

Solving for the reciprocal of the second radius gives  (n  1)  R1 nt1 1 1    a12  t1  4 cm 1 .     R2 R ( n 1) ( n d R n )   t1 1 21 1 t1 t1

Then R2  0.25 cm.

6.26

Starting with the three 2  2 matrices from Eq. (6.33): 1 d  I

0   a11 1   a21

a12   1 a22   do

0  a11 a12   1 0    1   a11 dI  a21 a12 dI  a22   do 1  a11  do a12 a12      dI (a11  a12 do )  a21  do a22 a12 dI  a22 

Then use Eq. (6.35): dI (a11  a12 do )  a21  do a22  0, and Eq. (6.37) M T  a12 d I  a22 : a  d a   11 o 12 0 

a12  MT 

Now take:

a  d a det  11 o 12 0 

a12   (a11  do a12 ) MT  1 MT  MT 

1 (a11  do a12 )

57

Chapter 6 Solutions

6.27

1 D1  R1    from (6.16) where 0 1  D1  (n  1)/R1  (3 / 2  1)/  10.0 cm  0.050 cm 1 .

0  1 0  1 0  1   T21       d21 /n 1  1.00/1.50 1  0.67 1  1  D2  R2   1  0

But R2  , so D2  0.

0  1 0.05  1 0.05 1 0   1  (6.29) A  R2T21 R1         0 1  0.67 1  0 1  0.67 1.03  Check: | A |  1(1.03)  (0.05)(0.67)  1.  nt t   ni i   y   A y   t   i   t  0, ni  1, yt  yi .

0.05  i  0  i   1 y   A  y       i  i   0.67 1.03   y1   0    i  (0.05) yi   y   (0.67)  (1.03) y  i i  i  0   i  (0.05) yi , yi  (0.67) i  (1.03) yi ,

both yield  i  (0.05)(2.0)  0.10 or 0.10 radians above the axis.

6.28

(6.34) 1/ f  a12  ( D1  D2  D1 D2 d /nl );

D1  (nl  1)/R1  (1.5  1)/0.5  1.0; D2  (nl  1)/R2  (1.5  1)/(0.25)  2.0.

1/ f  (1.0  2.0  (1.0)(2.0)(0.3)/1.5),

f  0.71.

V1 H1  (1)(1  a11 )/  a12 , (6.36)

a11  1  D2 d /nl  1  ( 2.0)(0.3)/1.5  1.4.

V1 H 1  (1  1.4) / 1.4  0.29.

(6.37) V2 H 2  (1)( a22  1)/  a12 ; a22  1  ( D1d )/nl  1  (1.0)(0.3)/1.5  0.8; V2 H 2  (0.8  1)/1.4  0.14.

58 6.29

Chapter 6 Solutions

For two reflections

 1 2 / (r )  1 0   1 2 / (r )   1 0  0 1   d 1   0 1   d 1   and this yields the desired matrix. When d  r the matrix for two traversals becomes

 1 0   0 1   and for four it is

 1 0   1 0  1 0   0 1  0 1  0 1       and since this is a unit matrix the light ray is back to where it started.

6.30

See E. Slayter, Optical Methods in Biology.

PC / CA  (n1 /n2 ) R / R  n1 /n2 , while CA / P C  n1 /n2 . Therefore triangles ACP and ACP  are similar; using the sine law sin P AC / PC  sin APC / CA

or n2 sin P AC  n1 sin APC , but  i  P AC , thus

 t  APC  P AC , and the refracted ray appears to come from P .

6.31

From Eq. (5.6), let cos   1   2 /2; then lo  [ R 2  (so  R )2  2 R(so  R )  R(so  R ) 2 ]1/ 2 , lo1  [ s02  R(so  R) 2 ]1/ 2 , li1  [ si2  R(si  R ) 2 ]1/ 2 ,

where the first two terms of the binomial series are used, l01  s01  ( so  R )h 2 / 2 so3 R

where

  h /R, li1  si1  ( si  R )h 2 /2 si3 R. Substituting into Eq. (5.5) leads to Eq. (6.40).

6.33

Because (a) is symmetrical and looks like a somewhat altered Airy pattern; this is spherical aberration. (b) This pattern is asymmetrical as if the Airy system were pulled off to the side, so it corresponds to a little coma. (c) This pattern is asymmetrical along two axes and must be due to astigmatism.

6.34

Fig. P.6.34a is bi-axially asymmetric and therefore corresponds to astigmatism. (b) is elongated along one axis and is due to coma, and because the pattern isn’t very complicated there isn’t much of it.

59

Chapter 7 Solutions

Chapter 7 Solutions 7.1

E02  36  64  2(6)(8) cos  /2  100, E0  10; tan   8/6,   53.1  0.93 rad.

E  10 sin(120 t  0.93).

7.2

E1  E01 cos( t );

E2  E01 cos( t   2 ).

E  E1  E2  E01 cos  t  E01 cos( t   2 )  E01 (2 cos 12 ( t   t   2 )cos 12 ( t   t   2 ))  2 E01 cos( t   2 /2) cos( 2 /2).

Recall cos( )  cos  , so, E  (2 E01 cos( 2 /2))(cos( t   2 /2))  E0 cos( t   ).

To show that this follows from (7.9) and (7.10), recall that cos   sin(   /2) so that

1  1   /2   /2,  2   2   /2.

7.3

In phase: 1   2 cos( 2  1 )  cos(0)  1.

(7.9) E02  E012  E022  2 E01 E02 cos( 2  1 )  E012  E022  2 E01 E02  ( E01  E02 )2 .

Out of phase,  2  1   , cos( 2  1 )  cos   1.

(7.9) E02  E012  E022  2 E01 E02  ( E01  E02 )2 .

7.4

OPL  i ni xi  (c /i ) xi  i cti , where ti is the time spent in medium i. But cti is also the distance the light would travel, in vacuum.

7.5

1 m/500 nm  0.2  107  2, 000, 000 waves. In the glass

0.05/0 /n  0.05(1.5)/500 nm  1.5  105 ; in air 0.95/0  0.19  10 7 ; total 2,050,000 waves.

OPD  [(1.5)(0.05)  (1)(0.95)]  (1)(1), OPD  1.025 1.000  0.025 m,  /0  0.025/500 nm  5  10 4 waves.

7.6

OPLB  nx  (1.00)(100 cm)  100 cm  1.00 m.

OPL A  i ni xi  (1.00)(89 cm)  2(1.52)(0.5 cm)  (1.33)(10 cm)  103.82 cm  1.0382 m.

  OPL A  OPLB  1.0382  1.00  .00382 m.

(7.16)   k0   (2 / 0 )  2 (3.82  10 3 m)/5.00  10 7 m  1.28  10 4  .

An integer multiple of 2 , so waves are in phase.

60 7.7

Chapter 7 Solutions E1  E01 sin[ t  k ( x  x )], so 1   k ( x  x ). E2  E01 sin[ t  kx ], So  2   kx.

(7.9)

E02  E012  E022  2 E01 E02 cos( 2  1 )  E012  E012  2 E012 cos( kx  ( k ( x  x )))  2 E012 (1  cos k x )  2 E012 (cos(0)  cos( k x ))  4 E012 cos2 (k x /2),

(see Problem 7.2), E0  2 E01 cos(k x /2).

(7.10) tan  

E01 sin 1  E02 sin  2 E01 cos 1  E02 cos  2



E01 sin(k ( x  x )  E01 sin( kx )) E01 cos(k ( x  x ))  E01 cos(kx )



2 sin 12 ( k ( x  x )  kx ) cos 12 (k ( x  x )  kx ) 2 cos 12 (k ( x  x )  kx ) cos 12 ( k ( x  x )  kx )

 tan(kx  (k x /2)),   k ( x  (x / 2)).

7.8

E  E1  E2  E01{sin[ t  k ( x  x )]  sin( t  kx )}. Since sin   sin   2 sin(1/2)(    ) cos(1/2)(    ), E  2 E01 cos(k x /2)sin[ t  k ( x  x /2)].

7.9

E  E0 Re ei ( kx  t )  ei ( kx t )   E0 Re eikx 2i sin  t   E0 Re[2i cos kx sin  t  2 sin kx sin  t ]  2 E0 sin kx sin  t.

Standing wave with node at x  0.

7.10

Ei  3cos  t  30, (1  0). E2  4 sin  t , but sin   cos(   /2), so E2  4 cos( t   /2)  4   /2.

E3  E1  E2 .

E  E  E  2 E01 E02 cos( 2  1 ) 2 3o

2 01

2 02

 9  16  2(3)(4) cos( /2), E3o  5.

(7.10) tan   ( E01 sin 1  E02 sin  2 )/( E01 cos 1  E02 cos  2 )  (3(0)  4(1))/(3(1)  4(0))  4/3;   53;

so   53  0.93 rad. Note that 1   , so E1 leads E3 .

7.11

This is a 3 – 4 – 5 right triangle, thus A  5, α  53°.

61

Chapter 7 Solutions

7.12

A  12 2  6 2  13.4

  tan 1 7.13

6  27 12

By Faraday’s law, E /x  B /x. Integrate to get B( x, t )    (E /x ) dt  2 E0 k cos kx  cos  t dt  2 E0 (k / ) cos kx sin  t.

But E0 k /  E0 /c  B0 ; thus B(x, t)  2B0 cos kx sin ω t.

7.14

Fringes are spaces  /2 vertically. sin θ  (fringes/cm) vertical/(fringes/cm) on film; (fringes/cm) on film  (1/( /2))/ sin   (1/5.50 × 107 cm)/ sin(1°)  1.04  108 cml.





7.15

Nodes are spaced at  /2 apart. c   ,

  c /  (3  108 m/s)/(1010 Hz)  0.03 m.

Node spacing is .015 m.

7.16

(7.30) E (standing wave)  2Eol sin kx cos ωt from two wave, El  E0 l sin( kx   t ); E R  E0l sin(kx   t ),

so,

62

Chapter 7 Solutions El  50 sin( 23  x  5 t ); E R  50 sin( 23  x  5 t ).

7.17

EI  E0 sin(kx   t ) E R   E0 sin(kx   t ) E  EI  E R  E0 sin(kx   t )   E0 sin(kx   t )  E0 [sin kx cos  t  cos kx sin  t   sin kx cos  t   cos kx sin  t ]  E0 [(1   )sin kx cos  t  (1   ) cos kx sin  t ]  E0 [(1   )sin kx cos  t  (1   )(sin( kx   t )  sin kx cos  t )]  E0 [(1   )sin kx cos  t  sin kx cos  t   sin kx cos  t  (1   )sin(kx   t )]  2 E0  sin kx cos  t  (1   )E0 sin(kx   t )

7.18

Hear beat frequency   2  1  2 Hz.

7.19

One can see that the relative phase of the two waves varies, and that a maximum occurs (positive or negative), and that a zero occurs when the relative phase is  n (n odd).

Also at the maxima, the relative phase between one wave and the net wave is zero.

At those zeroes where the relative phase between one wave and the net wave is  / 2, the “faster” wave “laps” the slower one, and the relative phase changes abruptly. 

7.20

E1  E01 cos[(kc  k ) x  (c   )t ]; E2  E01 cos[(kc  k ) x  (c   )t ];

1 E  E1  E2  2 E01 cos [(kc  k ) x  (c   )t  (kc  k ) x  (c   )t ]  2 1 cos [(kc  k ) x  (c   )t  (kc  k ) x  (c   )t ] 2  2 E01[cos(kc x  c t )cos(kx   t )]

so that kc  k , c   , k  km ,   m . Wavelength of envelope m  2 /km  2 /k.

Period of envelope Tm  2 /m  2 /. Speed of envelope m /Tm  (2 /k ) / (2 / ).

7.21

E  E0 cos c t  E0 cos m t cos c t  E0 cos c t  ( E0 /2)[cos(c  m )t  cos(c  m )t ].

Audible range  m  20 Hz to 20  103 Hz. Maximum modulation frequency  m (max)  20  103 Hz.  c  m (max)     c   m (max),   2 m (max)  40  103 Hz. 7.22

   /k  ak , g  d /dk  2ak  2 .

7.23

g 

d  d   d  d     dk  d   d    dk 

Use the fact that

d dk 2  2 , and  2 : d d  2  d     2 d         2 d d   

g  (2 ) 

63

Chapter 7 Solutions d d (2 ) d d c 1    dk d (2 / ) d (1/ ) d (1/ )  n   c c d (1/n) g   n  d (1/ )

7.24

g 

7.25

g 

c c d (1/n) c d ( n 1 )    n  d (1/ )  d (1/ ) c dn g    (1)(n 2 )  d (1/ )

g   1 



1 dn  n d (1/ ) 

Insert  g 

d : d (1/ )

 1 d dn   g   1   d (1/ ) d  n 



g   1  

g dn   n d 

 g dn  dn     g 1   n d  n d   g    dn  1  n d      g 

7.26

c 2.99  108 m/s   2.063  108 m/s n 1.449 c 2.99  108 m/s g    2.045  108 m/s ng 1.462



g   7.27

1/g  d ( / )/d and the rest follows.

7.28

From the previous problem 1/ g  (n /c)  ( n 2 /c 2 )[ d (c /n)]/d and the rest follows.

7.29

  g /k 

k     kd /dk , where  g g  k2    d  k k  dk k 2 g /k 





g k  k  2 g/k

k 



 2

64

Chapter 7 Solutions

7.30

We have   2 /k , d  /dk  2 /k 2   /k so that the term kd /dk  k (d  /dk )(d /d  )  k ( /k )d /d    d /d  and the expression for g follows.

7.31

g    kd /dk and d /dk  (d /d )(d /dk )   g d /d. Since   c /n, d /d  (d /dn)(dn /d )  (c /n 2 ) dn /d ,

 g    ( g ck /n 2 ) dn /d   / [1  (ck /n 2 )(dn /d )]  c /[ n   (dn /d )].

7.32

(7.40) ng  c / g . From Problem 7.31  g  c /(n  2(dn /d )), so  dn( )  ng  n   (dn /d )  n( )  2 (dn( )/2 d )  n( )    .  d 

7.33

Starting with: dn( ) d dn d   n  d  d 2  2  4 2 d d      dv  c  c  k 2  ck 2  2  dk   ng  n( )  

using:  

c k 2

2 dn k d dn ng  n   d ng  n 

7.34

Using dimensional analysis:  

g    1 

1 dn  n dk 

g: m/s n: dimensionless k: m1 then we have the following dimensional relationship which cannot be true: 1   m /s  m /s  1  1   m /s (1  m)  m 

7.35

For   a / ,  g     d /d   a /   a / 2  2 a /  2 .

65

Chapter 7 Solutions

7.36

(7.38)

v g  v  k (dv/dk )  c /n  (kc /n 2 )(dn /dk )  c /n  (kc /n 2 )(dn /d  )(d  /dk )  c /n  (kc /n 2 )(dn /d  )(2 /k 2 )  c /n  (2 /k )(c /n 2 )(dn /d  )  c /n  ( c /n 2 )(dn /d  )

7.37

c  c dn  n n2 d 2.99  108 m/s  c  n g   2 n n  n n n  1 2  1.3321 2 1  2  622.8 nm  2

g 

g 

2.99  108 m/s (622.8 nm)(2.99  108 m/s)  1.9  10 3     9 1.3321 (1.3321)2  67  10 m 

g  2.22  108 m/s 

c  2.25  108 m/s n

  g

   /k 

0 sin(kl /2)

7.38

 0 sinc(kl /2); (kl /2) l g  d /dk  0 sin(kl /2).

2

7.39

   /k therefore  2   p2  c 2 ( / )2 and   c /[1  ( p / )2 ]1/ 2 ;  g  d /dk  c 2 k /  c[1  ( p / )2 ]1/ 2 .

7.40

For  2  i2 , n 2  1  ( Nqe2 / 2 0 me ) i fi  1  Nqe2 / 2  0 me . Using the binomial expansion, we have (1  x )1/ 2  1  x /2 for x  1, so that n  1  Nqe2 /2 2  0 me ,

dn /d   Nqe2 / 0 me 3 .

g  c /[ n   (dn /d )]  c /[1  Nqe2 /2 0 me 2 ] and  g  c,   c /n  c /[1  Nqe2 /2 0 me 2 ]. By binomial expansion, (1  x )1  1  x for x  1,   c[1  Nqe2 /2 0 me 2 ];  g  c 2 .

7.41

E1  2 E0 cos  t ;  2 E0 cos  t 

E2 

1 E0 sin 2 t.

2

E  E1  E2

1 E0 sin 2 t  E0 (2  sin  t ) cos  t.

2

Resultant is anharmonic, but periodic with period ω.

7.42

(a) Both sine and cosine terms are required because the function is neither symmetric nor anti-symmetric about the origin.

66

Chapter 7 Solutions

(b) Odd and even multiples are required because the waveform does not have half-wave or quarter-wave symmetry. (c) 1/3 (d) DC Term  A0/2, thus A0  2/3 (e) T  2/ 7.43







0

0

0

 akx sin bkx dx  (1/2k )   cos[(a  b) kx] k dx   cos[(a  b)kx] kdx   0 if a  b. Whereas if a  b, 



 sin akx dx  (1/2k ) (1  cos 2akx)k dx   /2.

2

0

0

The other integrals are similar.

7.44

Even function, therefore Bm  0.

A0  (2/ )

/a

Am  (2/ )

/a

 / a

 / a

dx  (2/ )(2 /a )  4/a, /a

(1) cos mkx dx  (4/mk  )sin mkx 0

 (2/m )sin 2 m /a.  8  x  1  3 x  1  5 x   2 cos   2 cos    cos     2     2  3  2  5  2  

7.45

f ( x)  1 

7.46

A0  0, A1  A, and all other Am  0 moreover Bm  0 so f (x)  A cos(x/L).

7.47

f ( x) 

2 3

 cos(kx ) cos(2 kx ) cos(3kx )   4    2 2 2 1 2 3  

7.48

Am  4/m 2 , m  0; A0  8 2 /3; Bm  4 /m .

7.49

Am  2(1  cos m )/ (m 2  1) where m  1 and A1  0.

7.50

f ( x)  

1

sin kL /2

a

EL cos kx dk  kL /2 0

0

E0 L a sin(kL /2  kx ) E L a sin(kL /2  kx ) dk  0  dk.  2 0 kL /2 2 0 kL /2

67

Chapter 7 Solutions Let kL /2  w,  L /2  dk  dw, kx  wx , f ( x) 

E0

b

 0

E b sin(w  wx  ) sin(w  wx  ) dw  0  dw,  0 w w

where b  aL /2. Let w  wx   t , dw/w  dt/t, 0  w  b and 0  t  ( x   1) b.

Let w  wx   t in the other integral, 0  w  b and 0  t  ( x   1)b.

E ( x  1)b sin t sin t dt  0  dt , 0  0 t t E E f ( x )  0 Si  b( x   1)  0 Si  b( x   1) , f ( x) 

E0

 

( x  1) b





with x   2 x /L. 7.51

2 2 1 1  E (t )  E0   cos  t  cos 2 t  cos 4 t       2 3 15  

7.52

f ( x) 

7.53

(a) The envelopes of frequency spectra become more compressed as the wavelength increases.

This is because the angular spatial frequency, k, is inversely proportional to the wavelength.

1 2  cos(2 kx ) cos(4 kx ) cos(6kx )   sin kx        2   1 3 35 57  1

(b) The Fourier expansion of the square wave consists of only of odd harmonics of the fundamental frequency. Thus if one considers the number of frequency terms present in a range defined by a multiple of the square waves’ fundamental harmonic, then there will be the same number of terms present. (c) There are no even terms present because the expansion of a square wave consists only of odd harmonic terms.

7.54

By analogy with Eq. (7.61), A( )  (t /2)E0 sinc ( p   )t /2. From Table 1 in the appendix, sinc(/2)  63.7%. Not quite 50% actually, sinc(/1.65)  49.8%. ( p   )t /2   /2 or  /t   p     /t ; thus appreciable values of A() lie in a range   2 /t and t  1. Irradiance is proportional to A2(ω), and [sinc ( /2)2 ]  40.6%.

68

7.55

Chapter 7 Solutions xc  ctc , xc  c / . But  /k0   /k0  c; thus  /0   /0 , xc  c0 /0 , xc  02 /0 . Or try using the

uncertainty principle: x  h /p where p  h / and 0  0 .

7.56

0  21  10 9 m

0  446  10 9 m lc 

7.57

c0  2 (446  10 9 m)2 c   0   9.47  10 6 m v 0 v 0 21  10 9 m

xc  ctc  3  108 m/s 10 8 s  3 m. 0  02 /xc  (500  10 9 m)2 /3 m, 0  8.3  10 14 m  8.3  10 5 nm, 0 /0   /  8.3  10 5 /500  1.6  10 7  1 part in 107.

7.58

  54  103 Hz;  1.91  10 9.

 /  (54  103 )(10, 600  10 9 m) / (3  108 m/s) xc  ctc  c / , xc  (3  108 ) / (54  103 )

 5.56  103 m.

7.59

 /  2/1010 ; c   , so

  c /  (3  108 m/s)/(632.8  10 9 m)  4.74  1014 Hz. (7.64) lc  ctc .

Frequency range is ± 2(4.74 × 104 Hz) or 9.48 × 104 Hz, so t  1.05  10 5 s.

7.60

lc  (3  108 m/s)(1.05  10 5 s)  3.15  103 m.

xc  ctc  3  108  10 10  3  10 2 m,   1/tc  1010 Hz,

0  02 /xc (see Problem 7.35), 0  (632.8 nm)2 /(3  10 2 m)  0.013 nm.   1015 Hz, xc  c  10 15  300 nm, 0  02 /xc  1334.78 nm.

7.61

 /   / , (see Table 7.1)  (1  10 10 m)/(600  10 9 m)  1.67  10 4. c   , so   c /  (3  108 m/s)/(600  10 9 m)  5.00  1014 Hz.

  (1.67  10 4 )(5  1014 Hz)

 8.35  1010 Hz, so

t  1.20  10 11 s.

(7.64) lc  ctc  (3  108 m/s)(1.20  10 11 s)  3.60  10 3 m.

7.62

lc  200 . (7.64) lc  ctc , so tc  lc /c  200 /c  20(500  10 9 m)/(3  108 m/s)  3.33  10 15 sec. v  1/tc  3  1014 Hz.

Chapter 7 Solutions

7.63

 /   / , (see Table 7.1)  (1.2  109 m)/(500  109 m)  0.0024. c   , so   c /  (3  108 m/s)/(500  10 9 m)  6.00  1014 Hz. Δ  Frequency Bandwidth  (0.0024)(6.00  1014 Hz)  1.44  1012 Hz.

tc  1/  6.94  10 13 s.

(7.64) lc  ctc  (3  108 m/s)(6.94  10 13 s)  2.08  10 4 m.

69

70

Chapter 8 Solutions

Chapter 8 Solutions 8.1

E x  E0 cos(kz   t ) E y   E0 cos(kz   t ) tan  

Ey Ex



 E0 cos(kz   t )  1 E0 cos(kz   t )

  135°   E  E  ( E02x  E02y ) cos2 (kz   t ) E 2  2 E02 cos2 (kz   t ) E  2 E0 cos(kz   t ) Thus the amplitude is 2 E0 .

8.2

Ex  3cos(ky  t ) Ez  4 cos(ky  t ) tan  

Ex 3cos(ky  t ) 3   Ez 4 cos(ky  t ) 4

q  37° up from the z-axis.

8.3

E x lags E z by  /2. This is right-handed circular polarization. The amplitude is E0  8.

8.4

In each part the x and y components have the same amplitude E 0 .  (a) E  (iˆ  ˆj )E0 cos(kz  t ) is a P state at 135° or 45°.  (b) E  (iˆ  ˆj )E0 sin(kz  t ) is also a P state at 135° or 45°. (c) E x leads E y by  /4. Therefore it is an  state and left-handed. (d) E y leads E x by  /2. Therefore it is an R state.

8.5 8.6

E x leads E y by  /2. Therefore it is a left-handed circularly-polarized standing wave.

 ˆ cos(kz   t )  ˆjE .sin(kz   t ).

ER  iE 0 0  ˆ ˆ EL  iE0 cos(kz   t )  jE0 sin(kz   t ).    E  E  E  iˆ( E  E  ) cos(kz   t )  ˆj ( E  E  )sin(kz   t ).

R

L

0

0

0

0

Let E0  E0  E0x and E0  E0  E0y ;

then  ˆ  cos(kz   t )  ˆjE  sin( kz   t ).

E  iE 0x 0y

From Eqs. (8.11) and (8.12) it is clear that we have an ellipse where    /2 and   0.

71

Chapter 8 Solutions

8.9

E0 y  E0 cos 25; E0 z  E0 cos 25;  E ( x, t )  (0.91 ˆj  0.42kˆ )E0 cos(kx   t   /2).  k  k (iˆ  ˆj )/ 2.  E  E0 [ ˆj sin(kx   t )  kˆ cos(kx   t )].

8.10

  0, I  I1 cos2 0  200 W/m2 .

8.11

Half the energy is removed and half transmitted, hence I  150 W/m2 .

8.12

(8.24) I ( )  I (0) cos 2  so I ( )/I (0)  cos 2 (60)  (1/2)2  0.25.

8.13

Tl  (T0  T90 ) cos 2   T90

8.7

8.8

Tl  T0 cos2   T90  T90 cos2  Tl  T0 cos2   T90 (1  cos2  ) Tl  T0 cos2   T90 sin 2 

8.14

Natural light is unpolarized, thus 50% if the incident irradiance is assumed to vibrate parallel to the transmission axis while the other 50% is absorbed. A perfect filter will thus transmit 50% of the incident irradiance ( I i ). However a real filter will transmit a fraction of that: (1/2) I i T0 .

In the case of an HN-22 polarizer, 1/2 T0  22%, and T0  44%. Thus the transmitted irradiance is:

I t  1/2 Ii T0  (0.5)(1000 W/m2 )(0.44)  220 W/m2 8.15

HN  32, so 32% of incident light is (ideally) transmitted. Since the transmitted light is now polarized in the same direction as the 2nd filter, 64% of light incident on it is transmitted.

I t  (0.32)(0.64) I i  0.2 I i .

8.16

In natural light each filter passes 32% of the incident beam. Half of the incoming flux density is in the form of a P-state parallel to the extinction axis, and effectively none of this emerges. Thus, 64% of the light parallel to the transmission axis is transmitted. In the present problem 32% I i enters the second filter, and with a 30° angle, 1 I  T02 cos 2  I i 2 1  (0.64) 2 cos 2 30 I i 2  0.15 I i

8.17

In the case of an HN38S polarizer, 1/2 T0  38%, and T0  76%.

I t  1/2 (T0 )2 Ii  (0.5)(0.76)2 (1000 W/m2 )  289 W/m2 The resulting transmittance of the pair 

289 W/m 2  29% 1000 W/m 2

8.18

I  I1 cos2   (200 W/m2 )cos2 40  153 W/m2 .

8.19

In the case of an HN32S polarizer, 1/2 T0  32%, and T0  64%. Then for four polarizers;

I t  1/2(T0 )4 Ii  (0.5)(0.64)4 Ii  0.084Ii

72 8.20

Chapter 8 Solutions In the case of an HN-32 polarizer, 1/2 T0  32%, and T0  64% (a) I t  1/2 T0 I i  (0.5)(0.64) I i  0.32 I i 2

(b) I 2 

1 2 1 1  1  T0 cos 2  I i  (0.64)2 cos 2 45 I i  (0.64)2   I i  0.10 I i 2 2 2  2

8.21

1 1 1 I 2  T02T0 I i  T03 I i  (0.64)3 I i  0.13I i 2 2 2

8.22

50% means all the light from the first polarizer, viz. (1/2) I i , is passed by the second.

Hence the angle is 90°.

8.23

I  I1 cos2   (200 W/m2 )cos2 60  (1/4) 200 W/m2  50 W/m2 .

8.24

I  I1 cos2 45  (200 W/m2 )0.50  100 W/m2 .

8.25

The light from the first polarizer is at 10° and has an irradiance of I i cos2 30  0.75Ii ; this makes an angle of 60° with the next filter, hence I 2  (0.75I i )cos2 50  0.31I i .

8.26

I   flux density through middle polarizer  I1 cos2  (8.24).

I   I1 cos2 (45)  I1 /2 I 2  I  cos2 (45)  ( I1 /2)/2  I1 /4 8.27

Without middle polarizer, I t  (Ii /2)cos2 (50)  207 W/m2 . With middle polarizer,

I t  (Ii /2)cos2 (25)cos2 (25)  337 W/m 2 .

8.28

I1  (1/2)Ii ; I 2  I1 cos2 30; I3  I 2 cos2 30; I 4  I3 cos2 30 hence I 4  (1/2)Ii cos2 30 cos2 30 cos2 30  0.21 Ii  0.21(200 W/m2 )  42 W/m2 .

8.29

I 2  I1 cos2   30%Ii and Ii  2I1 ; 30%Ii  (1/2)Ii cos2  ; 0.60  cos2  ,   39.

8.30

With   t, the emergent flux density is 2 2 I  12 E01 sin 2  cos 2   ( E01 /8)(1  cos 2 )(1  cos 2 ) 2 2  ( E01 /8)(1  cos2 2 )  ( E01 /16)(1  cos 4 )  ( I1 /8)(1  cos 4 ).

8.31

No. The crystal performs as if it were two oppositely oriented specimens in series. Two similarly oriented crystals in series would behave like one thick specimen and thus separate the o- and e-rays even more.

8.32

The polarization of the light is lost in the specular reflection from the pencil dot.

8.33

Light scattered from the paper passes through the polaroids and becomes linearly polarized. Light from  the upper left filter has its E -field parallel to the principal section (which is diagonal across the second and fourth quadrants) and is therefore an e-ray. Notice how the letters P and T are shifted downward in an extraordinary fashion. The lower right filter passes an o-ray so that the letter C is undeviated. Note that the ordinary image is closer to the blunt corner.

8.34

(a) and (c) are two aspects of the previous problem. (b) shows double refraction because the polaroid’s axis is at roughly 45° to the principal section of the crystal. Thus both an o- and an e-ray will exist.

73

Chapter 8 Solutions

8.35

 When E is perpendicular to the CO3 plane the polarization will be less than when it is parallel. In the former case, the field of each polarized oxygen atom tends to reduce the polarization of its neighbors.   In other words, the induced field is down while E is up. When E is in the carbonate plane two dipoles reinforce the third and vice versa. A reduced polarizabillity leads to a lower dielectric constant, a lower refractive index, and a higher speed. Thus    .

8.36

For calcite, no  ne . The indices are computed in the usual way, using n  sin[(   m )/2]sin( /2), where  m is the angle of minimum deviation of either beam.

8.37

sin  c  nbalsam /no  1.55/1.658  0.935;  c  69.

8.39

The light initially enters the prism normal to the surface. Thus both the e and o rays travel the same path.

At the 45° interface, the optical axis of the material is rotated by 90°. Thus the e and rays are refracted due to the different indices of refraction.

Refraction of o-ray across interface between two quartz prisms: ni1 sin  i1  nt 2 sin  t 2 (1.5534)sin 45  (1.5443)sin  t 2 1.5534  (0.7071)   45.34  1.5443 

 t 2  sin 1 

Refraction of e-ray across interface between two quartz prisms: ni1 sin  i1  nt 2 sin  t 2 (1.5443)sin 45  (1.5534)sin  t 2  1.5443  (0.7071)   44.66 1.5534 

 t 2  sin 1  Refraction of o-ray exiting Wollaston prism:

ni 2 sin i 2  nt 3 sin  t 3 (1.5534)sin 0.34  (1)sin  t 3

 t 3  sin 1[1.5534(0.0059)]  0.53 Retraction of e-ray exiting Wollaston prism:

ni 2 sin i 2  nt 3 sin  t 3 (1.5443)sin 0.66  (1)sin  t 3

 t 3  sin 1[1.5443(0.0058)]  1.02 The angle separating the two is:   0.53  1.02  1.55 8.40

(c) Undesired energy in the form of one of the P-states can be disposed of without local heating problems. (d) The Rochon transmits an undeviated beam (the o-ray), which is therefore achromatic as well.

8.41

(8.25) tan  p  nt /ni  9.0/1.0,  p  83.7. The dipole is perpendicular to the plane of incidence.

8.42

tan  p  nt /ni  1.33/1.00,  p  53.

8.43

tan  p  nt /ni  1.65/1.33,  p  51.1.

74

8.44

Chapter 8 Solutions

sin  c 

nt ni

Use tan  P 

nt , ni

sin  c  tan  P sin(41)  tan  P  0.656

 P  tan 1 0.656  33.3 8.45

tan  P  nt /ni ; nt  tan 54.30  1.39.

8.46

tan  P  ng /ne  1.65/1.36  1.21 and  p  50.5; ne sin  p  ng sin t ; sin  t  (1.36/1.65)sin 50.50  0.636 and  t  39.5.

8.47

(4.5) sin  i / sin  t  nti ; sin  t  sin  i /nti  sin(40) / 1.5;  t  25.4.

(8.30) R  tan 2 (i   t )/ tan 2 (i  t )  tan 2 (14.6)/ tan 2 (65.4)  0.014.

(8.31) R  sin 2 (i  t )/ sin(i  t )  sin 2 (14.6)/ sin 2 (65.4)  0.077.

(8.32) R  12 ( R  R )  0.0455.

(8.33) V  I p /(I p  I n )  ( R  R )/( R  R  R)  67%.

8.48

If we take I r   Ir  , then I P  I r   I r  . we also have Ir  Ir   I r  . The degree of polarization is:

Vr 

Ir   Ir Ir   Ir 

In the case of natural light, 1 Ii 2 1 I r   R I i   R I i 2 1 I r   R I i   R I i 2 1 Ir   Ir  R I 1RI R  R  Vr  12  i 12  i   Vr  Ir   Ir  R I  2 R I i R  R 2  i I i  I i  

8.49

(4. 5) sin  i / sin  t  nti ; sin  t  sin  t /nti  sin(70)/1.5;  t  38.8.

(8.26) R  tan 2 (i   t )/ tan 2 (i  t )  tan 2 (31.2)/ tan 2 (108.8)  0.0425.

(8.27) R  sin 2 (i  t )/ sin 2 (i  t )  sin 2 (31.2)/ sin 2 (108.8)  0.299.

(8.28) R  12 ( R  R )  0.171.

8.50

The first step is to solve for R .

R 

sin 2 (i  t ) sin 2 (i  t )

75

Chapter 8 Solutions The problem give  i  56, use Snell’s law to obtain  t :

ni sin i  nt sin t sin 56  1.5sin t  t  33.55 R 

sin 2 (56  33.55) sin 2 (22.45) 0.38192    0.1458 sin 2 (56  33.55) sin 2 (89.55) 0.9999 2

R 

tan 2 (i   t ) tan 2 (22.45) 0.413192    1.0531  105 tan 2 ( i   t ) tan 2 (89.55) 127.32 2

Vr 

8.51

R  R R  R



0.14579  0.9999 0.14581

If the reflected light is totally polarized and the total reflectance is 10%:

R  10% 

1 ( R  R ) 2

Set R  0:

R  2(0.10)  0.20 T  1  R  0.80 1 1 (T  T )  (1  0.80)  0.90 2 2 T  90% T

8.52

no  1.6584, ne  1.4864. Using Snell’s law, sin  i  no sin  to  0.766, sin  i  ne sin  te  0.766, sin  to  0.462,  to  2731; sin  te  0.515,  te  311;   330.

8.53

(3.59) n  c /  o /n , so n  o /n.

Ordinary n  o /no  589.3 nm/1.5443  381.6 nm.

Extraordinary n  o /no  589.3 nm/1.5533  379.4 nm.

Same frequency   c /o  (3  108 m/s)/5.893  107 m  5.091  1014 Hz.

8.54

E x leads E y by  /2. They were initially in phase and Ex  Ey . Therefore the wave is left-handed,

elliptical, and horizontal.

8.55

After passing through the first polarizer, the beam is polarized linearly in the vertical direction. The first waveplate rotates the linear polarization from zero to /20 rad. When the beam passes through the second waveplate, its fast axis is at /20 radians with respect to the vertical. As a result, its fast axis is aligned with the polarization of the incident light and thus does nothing. When the beam passes through the next (third) polarizer, its fast axis is oriented at 3/20 rad with respect to the vertical and thus rotates the linear polarization from /20 to /10 rad. The beam’s polarization is now aligned with the orientation of the fourth waveplate (and each even numbered waveplate thereafter) and thus does nothing to the polarization.

Only the odd numbered waveplates rotate the beam, thus at the end of the sequence the polarization of the beam is rotated by a total of /4 radians. The incident intensity is reduced by half as the light passes through the first polarizer (assuming no surface reflections). After passing through the waveplates, the resulting light has been rotated by /4 radians and is thus now horizontally polarized. It is therefore transmitted by the 2nd polarizer, which is horizontal.

8.56

Placing the quarter wave plate first will have no effect on the irradiance. The irradiance will be affected with the quarter wave plate following the polarizer.

8.57

This is left circularly polarized light. Ey leads by    /2  3 /2.

76

Chapter 8 Solutions

8.58

For right circularly polarized light Ey leads by  /2. The quarter wave plate with a vertical fast axis introduces a shift of  /2. As a result, one has linearly polarized light in the 2nd and 4th quadrants, which means that the state shifted 1/4 of the way around the circle in figure 8.42.

8.59

For right circularly polarized light E x leads by 3 /2. The quarter wave plate with a horizontal fast axis introduces a shift of  /2. As a result, one has linearly polarized light in the 1st and 3rd quadrants.

8.60

The a half-wave plate with a fast vertical axis will introduce a shift of , thus linear light in the1st and 3rd quadrants will emerge.

8.61

The half-wave plate with a fast vertical axis will introduce a shift of p, thus left circularly polarized light will emerge.

8.62

The quarter-wave plate with a fast horizontal axis will introduce a shift of  /2, thus left elliptically polarized light will emerge.

8.65

d

(2m  1)0 /2  no  ne

Using m  0

d

0 /2 no  ne



(5.9  10 7 m)/2 2.95  10 7 m   32.4  m 1.5443  1.5534 0.0091

8.66

Emerging wave is elliptically polarized with   ( /2   /4)   /4.

8.67

The polarizers are aligned. The cellophane is a half wave plate, so is seen as “dark” (no beam passing through in this region).

8.68

The R-state incident on the glass screen drives the electrons in circular orbits, and they reradiate  reflected circular light whose E -field rotates in the same direction as that of the incoming beam. But the propagation direction has been reversed on reflection, so that although the incident light is in an R-state, the reflected light is left-handed. It will therefore be completely absorbed by the right-circular polarizer.

8.69

  2 d n /0 but   (1/4)(2 ) because of the fringe shift.

Therefore    /2 and d 

0 4 n



5.89  10 7  2.94  10 5 m.

4(5  10 3 )

8.70

Yes. If the amplitudes of the P-states differ. The transmitted beam, in a pile-of-plates polarizer, especially for a small pile.

8.71

Concentration is 10 g/1000 cm3  .01 g/cm3 , so rotatory power  .01(66.45)/10 cm  0.06645/cm.

Light travels through 1 m = 100 cm, so emerging light is at 6.645° from vertical (clockwise).

8.72

Place the photoelastic material between circular polarizers with both retarders facing it. Under circular illumination no orientation of the stress axes is preferred over any other, and they will thus all be indistinguishable. Only the birefringence will have an effect, and so the isochromatics will be visible.

If the two polarizers are different, that is, one an R, the other an L, regions where n leads to    will appear bright. If they are the same, such regions appear dark.

8.73

From (8.40)   (2 /o )l (| no  ne |) so | no  ne |  o  /2 l.

(8.40) n  o KE 2  o K (V /d )2 so o  /2 l  o K (V /d )2 ;   2 lK (V /d )2 .

8.74

V / 2  0 /2 n03 r63  550  10 9 /2(1.58)3 5.5  10 12  105 /2(3.94)  12.7 kV.

8.75

cos   cos 45  1/ 2  1 1 J      sin 45      sin  2 1     1/ 2 

77

Chapter 8 Solutions

8.76

   * E1  E2*  (1)(e21 )  (2i)(e22 )*  0, E2  (2, i)T .

8.77

(a) E1  (1, 1, 0, 0) has relative irradiance of 1, and is horizontally polarized. E2  (3, 0, 0,3) has relative irradiance of 3, is right circularly polarized. For both, V  1. (b) E  E1  E2  (4, 1, 0, 3), and has both a horizontal P component and an R component. (c) (8.48) V  (S12  S22  S32 )1/ 2 /So  (12  02  32 )1/ 2 /4  0.79. (d) E  (1, 1, 0, 0)  (1, 1, 0, )  (2, 0, 0, 0) and is “natural” light (unpolarized).

8.78

(See Tables 8.5 and 8.6.) 1  1 1 St   2 0  0

1 0 0  1  1     1 0 0  0  1  1   Vertical P 0 0 0  0  2 0       0 0 0  0  0 

Relative irradiance  1/2. (8.48) V  (S12  S22  S32 )1/ 2 /So  1.

8.79

(See Tables 8.5 and 8.6.) 1  1 0 St   2 1  0

0 0 0 0  0 1 0  0 0 0 0 1

1  1  0       1 0   (45P ) 0  2 1      0  0 

Relative irradiance  1/2. (8.48) V  (S12  S22  S32 )1/ 2 /So  1.

0 0

8.80

1 0  0  0

0 1 0 0  0 0 1  0 1 0 

1  1  1  1     0  0      0  0 

8.81

1 0  0  0

0 0 0  0 0 1 0 1 0   1 0 0 

1  1  0  0     1  1      0  0 

Therefore light polarized at 45° is unchanged, as expected. 1 0  0  0

0 0 0  0 0 1 0 1 0   1 0 0 

1  1  1  0     0  0      0  1 

A horizontal P-state is changed to an R state.

78 8.82

Chapter 8 Solutions Fast axis   45, cos 90  0, sin 90  1. Quarter wave plate,    /2; cos( /2)  0, sin( /2)  1.

0 0 0 1  0 C 2  S 2 cos  CS(1  cos  ) S sin     0 CS (1  cos  ) S 2  C 2 cos  C sin     C sin  S sin  cos   0 0 0 0  1 1 0 0  1(0) 0(1)(1  0) 1(1)  0   0 0(1)(1  0) 1  0(0) 0(1)  0    (0)(1) 1(1) 0  0 0

8.83

0 0 0 0 0 1 0 1 0  1 0 0

Quarter wave plate,    /2; cos( /2)  0, sin( /2)  1. Vertical fast axis   0, cos(0)  1, sin(0)  0.

0 0 0 1  0 c 2  s 2 cos  cs (1  cos  )  s sin     0 cs (1  cos  ) s 2  c 2 cos  c sin     c sin  s sin  cos   0 0 0 0  1 1 0 1  0(0) 1(0)(1  0) 0(1)  0   0 1(0)(1  0) 0  1(0) 1(1)  0    1(1) 0(1) 0  0 0

8.84

1 0  0  0

0 0  1 0 1  0 0 1 0  0   1 0 0  0

8.85

(From Problem 8.83).

0 0

0 0 0  1 0 0 1 0  0 1 0  0   1 0 0  0

1 0 St   0  0

0 1 0 0  0 0 1  0 1 0  0

0 0 0 0 0  0 1 0  0 0 1

1

0 0 0  0 0 1 0 1 0   1 0 0 

1  1  1  0        R-state. 0  0      0  1 

E y leads E x by  /2 (see third part from right of Figure 8.7a). 1 0 St   0  0

0 0 0 0 1  0 1 0  1 0 0  0

1  1  1  0        L-state. 0  0      0   1

E x leads E y by  /2 (see third part from left of Figure 8.7a).

0

79

Chapter 8 Solutions

8.86

8.87

1 0  0  0

1 0  0  0

0 0 0  1 1 0 0  0 0 0 1 0  0 1 0  0

0 0 0  1 1 0 0  0  0 0 1 0   0 1 0  0

1 0  0  0

0 0

0  1 0 0  0 1 0   0 0 1

1  1  0  0     0  0      1   1

1 0  0  0

0 0

0  1 0 0  0 1 0   0 0 1

1  1  0  0     0  0       1 1 

1 0  0  0

0 1 0 0  0 0 1  0 1 0 

1 0  0  0

0 0

0 0 0  1  1 0 0  1 0 0 0 1 2 1   0 1 0  0

0  1 0 0  0 1 0   0 0 1 0 0

0  1 1 0 0  0  0 0 1  0   0 1 0  0 0 0

0 1  0 0 0  1 0  0 1 0  2 0   0 0 0 1 0 1

1  1 0 2 0  1

0 1

0 0 0 0  0 0 0  0 1 0

1  1  0       1 0  0  2 0      0  1 

1  1 0 2 0  1

0 1

0 0 0 0  0 0 0  0 1 0

1  1  0       1 0  0  2 0      1  1 

1  1 0 2 0  1

0 0 1 0 0 0  0 0 0  0 0 1

1  1  0       1 0  0  2 0      1  1 

1  1 0 2 0  1

0 1

0 0  0  0 1 0  0  1

1  1  0       1 0  0  2 0       1 1  1  0  0       1 0  0  2 0       1 0 

1  1 0 2 0  1

0 0 0 0 0 1 0 0 0 0 0 0 0 0

fast vertaxis,   0

0  1 0 0  0 1 0   0 0 1 0 0

0 0 0 0  0 0 0  0 1 0 0 1

Half wave plate    , cos( )  1, sin( )  0

80 8.88

Chapter 8 Solutions (8.42)   2 no3r63V /o . At maximum transmission,    . For I t  0, sin 2 ( /2)  0, so ( /2)   . 2  2 no3r63V /o .

V  o /no3 r63  (5.461  107 m)/((1.52)3 (8.5  1012 m/V)  18.29 kV. If polarizers are parallel, I t is a maximum at V  0. Equivalent to I t  I i cos2 ( /2), so

I t /I i  cos2 ( V /2V / 2 ) (from 8.51)  cos2 ( /2)  0.

8.89

 tei  0

0   , where a phase increment of  is introduced into both components as a result of traversing tei 

1 0  0 0  , perfect absorber:  the plate. vacuum:   . 0 1  0 0 

8.90

8.91

t 2  0 0   0

0

0

2

t 0

0 t2

0

0

0  0 0  t 2 

1 0  0  0

0 0 0 0 0 0  0 0 0  0 0 0

V  I p /( I p  I u )  (S12  S22  S32 )1/ 2 /S0 , I p  (S12  S22  S32 )1/ 2 ; I  I p  I u .

So  (S12  S22  S32 )1/ 2  Iu 4 1  5  0  0          0  0  0  0        0  1  1 

5  (0  0  1)1/ 2  Iu .

8.92

8.93

(a)

 cos     sin 

sin   cos   cos cos   sin  sin      cos   sin     cos sin   sin  cos    cos(   )     sin(   ) 

(b)

 cos     sin 

sin   1  1  1  cos   i sin   1  ei    i       cos   2  i  2  cos   i sin   2 e 

(c)

 cos     sin 

sin   1  1  1  cos   i sin   1 i  1 e        cos   2  i  2  cos   i sin   2 i 

 cos2  cos  sin    cos     sin 2    sin    cos  sin   cos cos2   sin  cos  sin     2  cos cos  sin   sin  sin   This is a polarizer with axis given by 

Chapter 8 Solutions 8.94

8.95

(a)

Quarter were plate with fast axis at +45°

(b)

Quarter wave plate with fast axis at –45°

(a)

1 1  3 1  3   1  1 1  3        2 2  1  3 1  3   0  2 2  1  3 

(b)

1 1  3 1  3   0  1  1  3        2 2  1  3 1  3   1  2 3  1  3 

81

82

Chapter 9 Solutions

Chapter 9 Solutions 9.1

      E1  E2  (1/2)( E1e  i t  E1* ei t )  (1/2)( E2 e  i t  E 2* ei t ), where

Re( z)  (1/2)( z  z* ).           E1  E2  (1/4)[ E1  E2 e 2 i t  E1*  E2* e 2it  E1  E2*  E1*  E2 ].   The last two terms are time independent, while E1  E2 e 2 i t  0 and   E1*  E2* e2it  0 because of the 1/T  coefficient. Thus       I12  2 E1  E2  (1/2)( E1  E2*  E1*  E2 ).

9.2

The largest value of r1  r2 is equal to a. Thus if  1   2 ,   k (r1  r2 ) varies from 0 to ka.

If a   , cos  and therefore I12 will have a great many maxima and minima and therefore average to zero over a large region of space. In contrast if a   ,  varies only slightly from 0 to ka  2 . Hence I12 does not average to zero, and from Eq. (9.17), I deviates little from 4 I 0 . The two sources effectively behave as a single source of double the original strength.

9.3

Dropping the common time factor E1  E0 exp(2 iz / ) and E2  E0 exp[(2 i / )( z cos  y sin  )], adding these at the z  0 plane yields E  E0 {1  exp[(2 i / )( y sin  )]}. The absolute square of this is the irradiance viz.

  2  I ( y)  2 E02 1  cos  y sin        and the rest follows from the identity cos 2  2 cos2   1. The cosine squared has zeros at y  m /(2 sin  ) where m is an odd integer. The fringe separation is  / sin  . As q increases, the separation decreases.

9.4

A bulb at S would produce fringes. We can imagine it as made up of a very large number of incoherent point sources. Each of these would generate an independent pattern, all of which would then overlap.

Bulbs at S1 and S2 would be incoherent and could not generate detectable fringes.

9.5

ym  sm /a  14.5  102 m and   0.0145 m:    /  23.7 kHz. This is Young’s Experiment with the sources out-of-phase.

9.6

This is comparable to the “two-slit” configuration, (Figure 9.11), so we can use (9.29) a sin  m  m ( m may not be “small”). Let m = 1, sin   y /(s2  y 2 )1/ 2 , so,

ay   (s 2  y 2 )1/ 2 ; (a 2   2 ) y 2   2 s 2 ; y   s /(a 2  y 2 )1/ 2 . c   , so   c /  (3  108 m/s)/(1.0  106 Hz)  300 m.

y  (300 m)(2000 m)/((600 m)2  (300 m)2 )1/ 2  1.15  103 m

83

Chapter 9 Solutions

9.7

y  s

s  a ay



Using a  1  104 m,   589 nm, y  3.00 mm

s 9.8

yvac 

(1  10 4 m)(3  10 3 m)  0.509m 5.89  10 7 m

s 0 a

Using a  1  103 m, 0  589.3 nm, s  5.000 m yvac  n



5.000 m (5.893  10 7 m)  2.9465 mm (1  10 3 m) c

 0



0 0   

n s s 0 yair    a a n 5.000 m 5.893  10 7 m yair   2.9456 mm 1.00029 (1  10 3 m)

Thus the pattern expands from 2.946 mm to 2.947 mm.

9.9

(a) r1  r2   /2, hence a sin 1   /2 and

1    / 2a   (1/2)(632.8  10 9 m)/(0.200  10 3 m)   1.58  10 3 rad, or since

y1  s1  (1.00 m)( 1.58  103 rad)   1.58 mm. (b) y5  s5 /a  (1.00 m)5(632.8  109 )/(0.200  103 m)  1.582  102 m. (c) Since the fringes vary as cosine-squared and the answer to (a) is half a fringe width, the answer to (b) is 10 times larger.

9.10

y1 

s s  m  m 0 a a n

Using a  1  103 m, 0  589.3 nm, s  3.000 m, n  1.33

y1  9.11

3.000 m 5.893  10 7 m   1.329 mm 1.33 (1  10 3 m)

 m is “small,” so we can use (9.28)  m  m /a,  m is radian, a  m /m  [4(6.943  107 m)] / [1(2 rad/360)]  1.59  104 m.

84 9.12

Chapter 9 Solutions

y  (s /a), so,

s  ay /  [(1.0  104 m)(10  103 m)]/(4.8799  107 m)  2.05 m.

9.13

(9.28)  m  m /a. Want 1, red  2, violet ; (1)red /a  (2)violet /a; violet  390 nm.

9.14

s m a y m m  m  s a s f y1 

ym  f  m 

mf  a

9.15

f  r1  r2 



h 2

 2

 a



2a h f  f  2 2a f h a

9.16

Follow section (9.3.1), except that (9.26) becomes r1  r2  (2m   1)( /2) for destructive interference, where m   1,  2, . . . , so that (2m  1) is an odd integer. This leads to an expression equivalent to (9.28),  m  (2m  1) /2 a.

85

Chapter 9 Solutions

9.17

s m a a   ym ms ym 

Using a  2.7  103 m, s  4.60 m, m  5:

  (5  10 3 m) 9.18

2.7  10 4 m  587 nm 5(4.60 m)

Follow section (9.3.1), except that (9.26) becomes r1  r2    m , where   Optical path differences in beam. Following r1 ,   nd (for  m “small”). (r1  r2 )  m  ; a m  m  nd ;  m  (m  nd )/a.

9.19

As in section (9.3.1), we have constructive interference when OPD  m. There is an added OPD due to the angle, q, of the plane wave equal to a sin q, so (9.26) becomes r1  r2  a sin   m.

(9.24)  m  y /s and (9.25) r1  r2  ay /s are unchanged, for small  m so r1  r2  m  a sin   a( y /s)  a m ;  m  (m /a )  sin  .

9.20

(9.27)

ym  (s /a )m ; y1, red  [(2.0 m)/(2.0  10 4 m)](1)(4  10 7 m)  4.0  10 3 m. y1, violet  [(2.0 m)/(2.0  10 4 m)](2)(6  10 7 m)  12.0  10 3 m.

Distance  8.0  10 3 m.

9.21

r22  a 2  r12  2ar1 cos(90   ). The contribution to cos  /2 from the third term in the Maclaurin expansion will be negligible if (k /2)(a2 cos2  /2r1 )   /2;

therefore

r1  a 2 /.

9.22

E  m 2 /2;   0.42  106 m/s;   h /m  1.73  10 9 m; y  s /a  3.46 mm.

9.23

 /   / ;

   /  1/tc ;

c   , so   c / .

  (c / ) /  c / 2 ; t c   2 /c  ;

lc  ctc  ( 2 / )

 (500 nm)2 /(2.5  10 3 nm)  1  10 8 nm  0.1 m  .

9.24

E  Eo ei t  Eo e t   Eo ei (t  5 / 2) . I  E 2

T

 E  E T , so, as in section 9.1,

I  (3/2)Eo2  2 Eo2 {12 (cos   cos(3 /2)  cos(5 /2))} (three terms of Ei  Ei , 3

cross terms of Ei  E j ). For each beam,

I i  Ei2

T



1 2 Eo , 2

at q  0, so that for all three together I (  0)  32 E 2j . Note that (r2  r1 )  a sin  so that

 2  k (r2  r1 )  k (a sin  ); (r3  r1 )  (5a /2)sin 

86

Chapter 9 Solutions so that  3  k (r3  r1 )  k ( 25 a sin  ) where   ka sin  . So, I ( )  I (0)/3  (2 I (0)/9)(cos   cos(3 /2)  cos(5 /2))

When   0, the second term is zero.

9.25

( R  d ) 2 R ( R  d )  2 R y

y 

Using   6.000  107 m, R  1.000 m, d  3.900 m, y  2  103 m:

 9.26

(1.000 m  3.900 m)(6.000  10 7 m)  0.000735 rad  0.0421° 2(1.000 m)(2  10 3 m)

S  Z  R  d  1 d d 1 ( R  d ) 2 R  5.89  10 7 m   0.00118 rad  y 5  10 4 m y 

9.27

A ray form S hits the biprism at an angle  i (w.r.t normal), is refracted at angle  t , and hits the second face at angle ( t   ).

(4.4) (1) sin  i  (n)sin  t . (n)sin( t   )  (1)sin( /2   ), where angle  is defined in Figure 9.24.

As  i  0,  t  0;  ,  are both “small.” n sin   sin( /2   ), so n  ( /2)   ,   2(n  1) . From the figure tan( /2)  ( a /2)/d , so

 /2  (a /2)/d, 9.28

  a /d .

a /d  2(n  1) ,

a  2d (n  1) .

From Problem 9.19, a  2d (n  1) ; s  2d, so d  1m.

y  (s /a )  s /2d (n  1) ;   s /2d (n  1)y  [(2m)(5.00  10 7 m)]/[2(1 m)(1.5  1)(5  10 4 m)]  0.002 rad.

9.29

y  s0 /2 d (n  n).

9.30

Using   5.893  107 m, s  5.00 m, a  1  102 m: s (5.00 m)(5.893  10 7 m) 2 y  2   2  0.295 mm a 2(1  10 2 m)

9.31

y  (s /a), a  102 cm, a /2  5  103 cm.

9.32

  k (r1  r2 )   Lloyd’s mirror,   k ((a /2)sin   [sin(90  2 )] (a /2)sin  )   ,   ka(1  cos 2 )/2 sin    , maximum occurs for   2 when sin  ( /a)  (1  cos 2 )  2sin 2  .

First maximum   sin 1 ( /2a).

87

Chapter 9 Solutions 9.33

E1r is reflected once. E1r  Eoi r  0 (see 4.47)

 Eoi (n  1) / (n  1)  Eoi (1.52  1)/(1.52  1)  0.206 Eoi .

E2r is transmitted once, reflected once, then transmitted.   air )(t glass   air )  Eoi [2/(1  n)][(1  n)/(1  n)][2 n /( n  1)] E2 r  Eoi (t  0 )(rglass  4 n(1  n)/(n  1)3  Eoi [4(1.52)(1  1.52)]/(1  1.52)3  0.198Eoi ,

(see 4.48) ( indicates  phase changed).

E3r is transmitted, reflected 3 times (internally), and then transmitted.

E3r  Eoi t (r )3 t   Eoi [2/(1  n)][(1  n) / (1  n)]3 [(2n)/(n  1)]  [4 n(1  n)3 ] / (n  1)5  Eoi [4(1.52)(1  1.52)3 ] / (1.52  1)5  0.008Eoi

for water in air.

E1r  Eoi (1.333  1) / (1.333  1)  0.143Eoi .

E2 r  Eoi [4 1.333  (1  1.333)]/(1  1.333)3  0.140 Eoi .

E3r  Eoi [4 1.333  (1  1.333)3 ]/(1.333  1)5  0.003Eoi .

9.34

Here 1.00  1.34  2.00, hence from Eq. (9.36) with m  0, d  (0  1/2)(633 nm)/2(1.34)  118 nm.

9.35

(9.36) d cos  t  (2 m  1)( f )/4 for a maximum at (near) normal incidence, and taking m = D (lowest value)

d   f /4  o /4n  (5.00  107 m)/4(1.36)  9.2  106 m.

9.36

9.37

  d cost  2 m  m  for minimum reflection  2 m (0 /4 n)  4  2 nd [2(1.34)(550.0 nm)] (1474 nm) at   0, 0    m m m for m  1,2,3,; 0  1474 nm, 737 nm, 368.5 nm, In this case, one drops the relative phase shift of  from (9.34):



4 n f

d cos  i 0 4 n f 2  d cos  i 0 cos  t 

0 2n f d



(4.60  10 7 m) 25 nm  2.5  10 8 m 2(1.333)(2.50  10 8 m)

 t  46.356 sin  i  (1.333) sin (46.356)  0.9646 i  74.7 9.38

Eq. (9.37) m  2n f d /0  10,000. A minimum, therefore central dark region.

9.39

The fringes are generally a series of fine jagged bands, which are fixed with respect to the glass.

9.40

x   f /2 ,   0 /2 n f x,   5  10 5 rad  10.2 seconds.

88 9.41

Chapter 9 Solutions (9.40) x   f /2 for fringe separation where   d /x.

x   f /2(d /x)  x f /2d. Number of fringes  (length)/(separation)  x /x so, x /x  2 d / f  [2(7.618  10 5 m)]/(5.00  10 7 m)  304.72  304 fringes.

9.42

9.43

1  f  dm   m   2 2  1  (5.893  10 7 m)   50.8  m d172   172   2 2 

x 2  d1[( R1  d1 )  R1 ]  2 R1d1  d12 . Similarly x 2  2 R2 d2  d22 . d  d1  d2  ( x 2 /2)(1/R1  1/R2 ), d  m f /2. As R2  , xm

approaches Eq. (9.43).

9.44

(9.42) xm  [(m  1/2) f R ]1/ 2 , air film, n f  1, so  f  o .

R  xm2 /(m  1/2)o  (0.01 m)2 /(20.5)(5  107 m)  9.76 m. xm2  xm2 1   f R( mm  mm 1 )

9.45

R

xm2  xm2 1  f R ( m m  m m 1 )

Use

1  2d m   m   0 2  1 2d m  2 m

0

Since the offset is a constant d : R

R

xm2  xm2 1 f  1  1  R  2d m  d    2d m 1  d    2  2  0  n f ( xm2  xm2 1 ) 2 R (d m  d m 1 )

Thus the radius of curvature can be measured independent of d .

9.46

xm  ( m f R )1/ 2

xm 1  xm  ( f R)1/ 2 ( m  1  m ) xm  2  xm 1  ( f R)1/ 2 ( m  2  m  1) ( f R)1/ 2 ( m  1  m ) xm 1  xm   xm  2  xm 1 ( f R)1/ 2 ( m  2  m  1)

m 1  m m  2  m 1

89

Chapter 9 Solutions Expand the square roots for large m (keeping only the first few terms): m  m1/ 2 1 1 m  1  m1/ 2  m 1/ 2  m 3/ 2 2 8 1 3/ 2 1/ 2 1/ 2 m2  m m  m 2 1 1 m1/ 2  m 1/ 2  m 3/ 2  m1/ 2 x m 1  x m 2 8  1 3/ 2 1 1 x m  2  x m 1 1/ 2 1/ 2  m  m1/ 2  m 1/ 2  m 3/ 2 m m 2 2 8 1 1/ 2 1 3/ 2 1 1  m m m 1/ 2  m 3/ 2 1  m 1 4m  1 1 8 4 4  2     1  1 1/ 2 3 3/ 2 3 3 4m  3 2m  m m m 1/ 2  m 3/ 2 1  m 1 2 8 4 4

For m  50

50  1  50 50  2  50  1 1 1  1.01 2(50) 9.47

 1.0099

A motion of  /2 causes a single fringe pair to shift past, hence

92( /2)  2.53  105 m and   550 nm.

9.48

d  N (o /2)  (1000)(5.00  107 m)/2  2.50  104 m.

9.49

d  N N

9.50

0

2 2 d

0



2(1  10 4 m)  400 5  10 7 m

  d  N (o /2);   (nair x  nvacuum x );

N  2 /o  [2(1.00029  1.00000)(0.10 m)]/(6.00  107 m)  97.

9.51

Differentiating  

 c

:

2 c   0  2 c v 1   t lc  ct

0 

0 

02 1 c t



02 lc

2 D  lc 0  D

02 2D

02 (6.43847  10 7 m)2   0.1594 m 20 2(0.0013 nm)

90

Chapter 9 Solutions

9.52

Fringe pattern comes from the interference of two beams, one that passes through the lower medium (n1), and is reflected off its mirror, one that passes through the top medium (n2) and is reflected off its mirror. The two beams reflect off the front surface of the other medium.

It might be used to compare n1 and n2 (especially if one changes, such as due to pressure or temperature), or compare the flatness of one surface, to a known optically flat surface.

9.53

Et2  Et Et*  E02 (tt )2 /(1  r 2 ei )(1  r 2 ei ), I t  I i (tt )2 /(1  r 2 ei  r 2 ei  r 4 ).

9.54

(a) R  0.8944, therefore F  4 R /(1  R)2  321.





(b)   4 sin 1 1/ F  0.223. (c) F  2 /0.223. (d) C  1  F. 2/[1  F ( /4)2 ]  0.81[1  1/(1  F ( /2)2 )],

9.55

F 2 ( )4  15.5F ( )2  30  0.

9.56

I  I max cos2  /2, I  I max /2 when    /2, therefore    . Separation between maxima is 2 . F  2 /  2.

9.57

(4.47) r i  0  (nt  ni )/(nt  ni ). Bare substrate: r  (ns  1)/(ns  1). Substrate with film: r   to  f rf  s t f o .

(4.48) t i  0  2 ni /(ni  nt ), so, r   [2/(1  n f )][(ns  n f )/(ns  n f )][2n f /(n f  1)], where nf = n. Note that for ns  nf  1, both r and r  are positive. But, with thickness  f /4, a  phase shift occurs due to the OPD in the r  beam, so rnet  r  r .

Thus, the r  beam (partially) cancels the r beam.

9.58

At near normal incidence ( i  0) the relative phase shift between an internally and externally reflected beam is  rad. That means a total relative phase difference of (2 / f )[2( f /4)]   or 2 . The waves are in phase and interfere constructively.

9.59

n0  1,

ns  ng ,

n1  ng .

1.54  1.24, d   f /4  0 /4n1  540/4(1.24) nm  167 nm.

No relative phase shift between two waves.

9.60

9.61

The refracted wave will traverse the film twice, and there will be no relative phase shift on reflection.

Hence d = 0/4nf  (550 nm)/4(1.38)  99.6 nm.

  d cost  (2m  1)  m  . Let t  0, m  0, (minimum thickness).  4  d

9.62

0 4n f



5.00  10 7 m  96 nm 4(1.30)

Note that in the triangle including θ and r1, the length of the side from P1 to a plane, parallel to the surface, and containing point z(x) is r1 cos θ. So, from zero elevation, h  r1 cos   z(x) or z(x)  h  r1 cos .

(9.108) can be demonstrated on the triangle (a, r1, r2), where a is the length of the boom: r22  r12  a 2  2r1 a cos   90     sin      cos  90   

and   k (r2  r1 )  (2 / )(r2  r1 ).

91

Chapter 10 Solutions

Chapter 10 Solutions 10.1

( R  l )2  R 2  a 2 ; therefore R  (a 2  l 2 )/2l  a 2 /2l, lR  a 2 /2, so for   l,  R  a 2 /2. Therefore R  (1  103 )2 10/2  10 m.

10.2

A “constant” phase shift is added due to the angle of the incident wave reaching the ends of the slit at different phase, so that (10.11) becomes r  R  y(sin   sin  i )  . . . . This constant carries through the integration, so that the definition of  in 10.18 (or 10.14) becomes   ( kb /2)(sin   sin  i ).

10.3

d sin  m  m ,   N  /2   , 7 sin   (1)(0.21),   2 /N  kd sin  , sin   0.03 so   1.7.

For sin   0.0009,   3 min .

10.4

Converging spherical wave in image space is diffracted by the exit pupil.

10.5

 sin   I ( )  I (0)     

2

From figure 10.13, the first secondary maximum is at   1.43 :

 sin(1.43 )   0.9579   I (0)  I    1.43   I (0)     0.0455I (0)  1.43   4.492  2

2

Then, I (0) I (0)   21.98 I (   1.43 ) 0.0455I (0)

10.6

   , sin     /b   , L   L  /b, L   f 2  /b.

10.7

Is R  b2/?, b  slit width. b2/  (1  104 m)2/(461.9  109 m)  0.02 m  1 m

Fraunhoffer: (Half) angular width from the central maximum



    (kb/2) sin 1, sin 1  2/kb  /b  (4.619  107 m)/(1  104 m);   0.26°.

To what order Young’s fringe does 1 correspond?

  m  (ka/2) sin 1. m  (ka/2) sin 1  (a/)sin 1  (2  104 m)/(4.619  107 m)sin (0.26°)  2  a/b So there are 4 “Young’s Fringes” in the central maximum.

10.8

b sin  m  m  , so,

b  m / sin  m  10(1.1522  10 6 m)/ sin(6.2)  1.07  10 4 m.

In water, b sin  m  m  , where   nair o /nwater . sin  m  mnair o /(nwater b);

 m  sin 1 [10(1.00029)(1.1522  10 6 m)/(1.33)(1.07  10 4 m)]  4.65.

92 10.9

Chapter 10 Solutions

  (20 cm)sin 36.87  12 cm.

10.10 Starting with,

ym 

s m b

For plane waves, s  f,

f m b y b (1.2  10 3 m)(2.5  10 4 m)  0.145 m f  m  m 4(5.1836  10 7 m)

ym 

10.11 The first maximum is located at   1.4303  kb   sin   2 

  1.4303   Use  

2

:

 For small angles: b b 1.4303    sin      y  tan    f

Then,

1.4303

f b

y

b sin  m  m 

10.12 For small angles,

sin  

y f

ym 

m f b

The width of the central irradiance is the distance between the 1 minumum  2y1  2y 

f b



2(5.461  10 7 m)(0.62 m)  0.00451 m 1.5  10 4 m

10.13 ym  f sin  m 

y1 

m f b

f

b 2 f y2  b The distance between the 1st and 2nd zeroes is:

93

Chapter 10 Solutions

y2  y1 

2 f  f  f (4.861  10 7 m)(0.60 m)     0.00146 m b b b 2.0  10 4 m

10.14   ( ka /2) sin  ,   ( kb /2) sin  . a  mb,   m  ,   m 2 , N  number of fringes   /  m2 /  2m.

10.15 Is R  b2 / ?, b  slit width.

b2   (1  104 m)2 /(5  10 7 m)  .02 m  2.5 m.

Fraunhofer.

(Half) angular width of central maximum from

    (kb /2)sin 1 . sin 1  2 /kb   /b  (5  10 7 m)/(1  10 4 m); 1  0.29.

To what order Young’s fringe does 1 correspond?

  m  (ka /2)sin 1 . m  (ka /2 )sin 1  (a / )sin 1  (2  10 4 m)/(5  10 7 m)sin(0.29)  2.

So there are 4 “Young’s Fringes” in the central maximum.

10.16 For single slit diffraction: b sin  mD  m D 

For two slit interference: a sin  mI  m I 

Since there are 11 maxima, there must be 5 peaks on either side of the central peak (mI  5). The outer edge of the central diffraction (1D) then corresponds to the location of the edge of the 5th interference fringe (5I).

sin  5 I  sin 1D

 5 I  1D 5  a



b a 0.100 mm b   0.020 mm 5 5

10.17   3 /2 N   /2, I ( )  I (0)[(sin  )/ ]2 /N 2 and I /I (0)  1/9.

10.18 For the principle maximum, the phasors line on a straight line (  ); the amplitude is 3E0. The phasors at the subsidiary maximum (  ) have an amplitude of E0. The irradiance ratio is: 2

 E0  1    3 E 9  0

10.19 For 16 slits, the amplitude is 16 E0, thus the irradiance ratio is: 2

 E0  1    256  16 E0 

94

Chapter 10 Solutions For 8 slits, the amplitude is 8 E0 and the irradiance ratio is: 2

 E0  1    8 E 64  0

The ratio between the two is 64/256  0.25. The 8-slit screen will have wider m  0 peaks.  sin   10.20 I ( )  I (0)     



2

  a  b sin   sin   4  4

For the 2nd order maximum, m  2, thus    /4   /2.

 sin  /2   1  I ( )  I (0)      0.405   /2   1.5707  2

Where we set I(0)  1. Thus: I ( )  0.405 I (0)

10.21 (a)

(b)

The phasors form an octagon.

2

95

Chapter 10 Solutions (c)

(d)

(e) The angles for b) are 45°, c) are 180°, d) are 90°.

10.22 (10.17) I ( )  I (0)(sin  / )2 , where    kD 2  sin  . “Miniscule Area” corresponds to the limit D  0. As D  0,   0, so lim I ( )  lim( I (0)(sin  / )2 );  0

D 0

As   0, sinc(  )  1, so lim I ( )  I (0), i.e., same in all directions.

D 0

10.23 The minima occur at m  1.

Y Z

R b

R a

Using R  f :

Y Z

f b

f b



(5.43  10 7 m)(1 m)  5.43 mm 1  10 4 m



(5.43  10 7 m)(1 m)  2.73 mm 1.99  10 4 m

 sin     sin    I ( )  I (0)           2

10.24

kaZ , 2R kbY   , 2R

 

a  0.1 mm b  0.2 mm

2

96

Chapter 10 Solutions When Y  0 and Z  1 mm:

   0,   

 (1  10 4 m)(1  10 3 m) (5.43  10 7 m)(10 m)

 0.05786

 sin     sin(0.05786)  I ( )  I (0)   I (0)     0.999 I (0)     0.05786  2

2

When Z  0 and Y  1 mm:

   0,   

 (2  10 4 m)(1  10 3 m) (5.43  10 7 m)(10 m) 2

 0.1157

 sin     sin(0.1157)  I ( )  I (0)    I (0)    0.996 I (0)    0.1157    2

10.25 (from 10.41) E    eik (Yy  Zz )/ R dS. and I (Y , Z )   E 2 . If E is an even function

of (Y , Z ), E (Y ,  Z )  E (Y , Z ). If E is an odd function of (Y , Z ), E (Y ,  Z )   E (Y , Z ), but I (Y ,  Z )  I (Y , Z ).

10.26 If the aperture is symmetrical about a line, the pattern will be symmetrical about a line parallel to it.

Moreover, the pattern will be symmetrical about yet another line perpendicular to the aperture’s symmetry axis. This follows from the fact that Fraunhofer patterns have a center of symmetry.

10.27 For the solution to this problem, please refer to the solutions at the end of the book.

10.28 Three parallel short slits, note the N – 2 = 1 subsidiary maximum between each peak.

10.29 Two parallel short slits.

10.30 An equilateral triangular hole.

10.31 A cross-shaped hole.

10.32 The E-field of a rectangular hole.

10.33 q1  1.22

R 2a

Use R  f  1 m:

2q1  1.22 10.34

q1  1.22

(1 m)(6.56  107 m)  6.67  105 m (1.2  102 m)

f (1.4 m)(5.40  10 7 m)  1.22  6.15  10 6 m 2 D (1.5  10 m)

2q1  1.23  10 5 m

If we doubled the lens diameter (D), then the disk would shrink to: 2q1  0.62  105 m 10.35 Using:

0

5.50  10 7 m  4.11  10 7 m 1.337 n (2.1  10 2 m)(4.11  10 7 m) R  1.22  1.755  10 6 m q1  1.22 2a (6  10 3 m) 2q1  3.51  m





97

Chapter 10 Solutions 10.36 From section 10.2.5, first “ring” (maximum) occurs for u  kaq /R  5.14.

Interpolating from Table 10.1, J1 (5.14)   0.33954 2

2

 2 J (u)   2(0.33954)  From (10.55) I /I (0)   1      0.0175 5.14    u  2

I (0)  2(sin u  cos u)   2 J (u)  10.37 I ( )  I (0)  1   2   u  u  u  

2

This is minimum when sin u  cos u,

u

 5 9

, , , 4 4 4



 (4m  3) , 4

m  1, 2, 3, 

u  ka sin   (4 m  3)



4 (4m  3)  (4 m  3) sin     2 a 4 8a d d 4 (sin  )  (cos )  dm dm 8a 4  (cos  )  m 8a If Δm  1,

 

 2 a cos 

10.38 From Eq. (10.58), q1  1.22( f /D )   .

10.39 For the solution to this problem, please refer to the solutions in the back of the textbook.

10.40 (10.57) q1  1.22( R /2 a )

 1.22[(3.76  108 m)(6.328  10 7 m)]/2[1  10 3 m]  1.45  105 m.

10.41 (10.59) ( )min  1.22 /D  [1.22(5.50  10 7 m)]/7.5  10 4 m

 8.9  10 4 rad, This is about 5 times worse resolution.

1.22 D 1.22 D ( )min

10.42 ( )min 

D

1.22(5.5  10 7 m)  0.134 m 5  10 6 rad

98

Chapter 10 Solutions

10.43  

0 n



( )min 

5.50  10 7 m  4.11  10 7 m 1.337

1.22 1.22(4.11  10 7 m)   2.51  10 4 rad  0.0144 D 2  10 3 m

10.44 ( )min  2.51  10 4 rad 

L L  D 2.5  10 1 m

L  6.28  10 5 m 10.45 1 part in 1000. 3 yd  100 inches.   2.51 104 rad 

D

L D

L  0.1 inch

0.1  398 inches 2.51 104

10.46 (10.59) min  1.22 /D  [1.22(5.50  10 7 m)] /5.08 m

 1.32  10 7 rad; or   1.32  10 7 rad(360/2 rad)  7.55  106 ; or

  (7.55  106 )(3600 sec/degree)  2.72  10 2 arc sec.

To be resolved, s  r  ( in radians).

s  (3.844  108 m)(1.32  10 7 )  50.7 m.

To be resolved by eyes, s  r   r (1.22 /D)  (3.844  108 m)[1.22(5.50  10 7 m) / (4  10 3 m)]  6.44  10 4 m.

10.47 (a) ( )min  (b)

1.22 1.22(5.50  10 7 m)   6.71  10 6 rad D 1  10 1 m

L D L  D( )min  (1  106 m)(6.71  10 6 rad)  6.71 m

( )min 

10.48 Resolution is proportional to wavelength, as a result, blue lasers are able to read smaller locations on the storage media. Thus blue lasers allow one to store more information on each disk (e.g., higher resolution “HD” movies may be recorded on an individual disk).

1.22 D 1.22 D ( )min

10.49 ( )min 

Use (  )min  L /r : D

1.22 r 1.22(5.50  10 7 m)(1.61  10 5 m)   2.16 m L 5  10 2 m

99

Chapter 10 Solutions

10.50 ( )min 

1.22 1.22(5.50  10 7 m)   2.976  10 7 rad D 2.4 m

L  D( )min  (6  10 5 m)(2.976  10 7 rad)  0.168 m 10.51 (10.32) a sin  m  m ; sin 1  y /R, so

a(Y /R)   ; Y  R /a  (2.0 m)(6.943  107 m)/(3.0  106 m)  0.46 m.

10.52 (10.32) a sin  m  m ; sin 3  3 /a  3(5.00  107 m)/(6.0  106 m);  3  16.

10.53 (10.32) a sin  m  m  .

a  2 / sin  2  2(5.50  107 m)/ sin(25)  2.6  106 m.

10.54 a sin   m a

m 2(6.562816  10 7 m)   1.9616  10 6 m sin  sin(42.00)

1  509, 789 lines/m  5, 098 lines/cm a m 2(4.861327  10 7 m) sin    0.4956 a 1.9616  10 6 m   29.71

10.55 From Eq. (10.32), where a  1/(1000 lines per cm)  0.001 cm per line (center to center), sin  m  1(650  109 m)/(0.001  102 m)  6.5  102 and 1  3.73. a sin m  m. sin m ≃ Ym/R; Ym  (m/a)R.

10.56

10,000 lines/cm  106 lines/m so a  106 m.

Y1(589.5923 nm)  [1(5.895923  107 m)/106 m](1.00 m)  0.5895923 m

Y1 (588.9953 nm)  0.5889953 m Y1  Y1  5.97  10 4 m

10.57 N 

 589.2938 nm   987 m( )min (1)(0.597 nm)

This is twice the number of grooves for m  2.

10.58 a sin   m

a

1  1.695  10 6 m/groove 5.9  10 groove/m 5

For   400 nm: sin  400 

m 1(4.00  10 7 m)  a 1.695  10 6 m/groove

 400  13.65

100

Chapter 10 Solutions For   720 nm: sin  720 

m 1(7.20  10 7 m)  a 1.695  10 6 m/groove

 720  25.14    720   400  11.49 10.59 (10.32) a sin  m  m , so sin  m  m /a; is  2 (red)   3 (violet)?

5000 lines/cm  5  105 lines/m; a  2  10 6 cm. sin  2 (red)  2(7.8  10 7 m) / (2  10 6 m);  2 (red)  51.3. sin 3 (violet)  3(3.90  10 7 m)/(2  10 6 m)  35.8.

Spectra do not overlap. Note: Can see “by inspection” by comparing factor of 2 in wavelength to factor of 3/2 in m’s.

10.60 sin  

m 2(7.00  10 7 m)   0.826 a 1.6949  10 6 m

  55.69 

kb sin  2

 b sin    7.00  10 7 m b   8.47  10 7 m sin  0.826



10.61 The largest value of m in Eq. (10.32) occurs when the sine function is equal to one, making the left side of the equation as large as possible, then m  a /  (1/1  106 )/(3.0  108 m/s  4.0  1014 Hz)  1.3, and only the first-order spectrum is visible.

10.62 (10.32) a sin  m  m , where   o /n.

sin  m  m /a; sin 1 (vacuum)/ sin 1 (Mongo)  [(1)o /a]/[(1)o /na]; n  sin(20.0)/ sin(18.0)  1.11.

10.63 sin  i  n sin  n Optical path length difference is m , a sin  m  na sin  n  m . a (sin  m  sin  n )  m .

10.64 R 

  mN  2(60, 000)  120, 000 ( )min

( )min 

 R



5.40  10 7 m  0.0045 nm 120, 000

10.65 R  mN  106 , N  78  103. Therefore m  106 /78  103 ,  fsr   /m  500 nm/(10 6 /78  103 )  39 nm.

R  Fm  F 2b f d /  10 6  fsr   2 /wn f d  0.0125 nm.

10.66 R   /  5892.9/5.9  999, N  R /m  333.

10.67 The diffraction pattern will be the same as for a single hole.

101

Chapter 10 Solutions 10.68 y  L /d, d  12  106 /12  10 2  104 m.

10.69 (From 10.75) El  [  Kl  A  / (   r )][sin(t  k   kr )]rr rrll1 ; sin( t  k   krl )  sin( t  k   krl 1 )  sin( t  k   k (ro  l  /2))  sin( t  k   k (ro  (l  1)( /2)))  sin( t  k (   ro )  2 l  /2  )  sin( t  k (   ro )  (2 (l  1)  /2  )).

Recall sin(   )  sin  cos   cos  sin  and sin  l  sin(l  1)  0.

cos(l  1)   cos l ; cos l  (1)l  , so El  (1)l 1 (2 Kl  A  )/(   ro )sin(t  k (   ro )). 

10.70 A  2 2  sin  d  2 2 (1  cos  ), 0

cos   [   (   r0 )2  rl2 ] /2  (   r0 ), rl  r0  l /2.

2

Area of first l zones A  2 2   (2  2  2  r0  lr0  l 2  2 /r ) / (   r0 ), Al  A  Al 1   (r0  (32l  1) /4)(   r0 ).

10.71 (10.78) becomes

E  E1 2   E1 2  E2  E3 2      Em 1 2  Em  , so that (10.80) becomes E  E2 2  Em 2 and (10.82) becomes E  E1 2  Em 2 so that (10.84) E  E  E1 2  Em 2 .

10.72 N F 

(   r0 ) 2  R 2 r0 R 2 R 2 R 2 R      r0   r0   r0  r0  

For a collimated light source,   , 1/  0:

NF 

R2 (3  10 3 m)2  3 r0  (6.00 m)(5.00  10 7 m)

This would appear as a bright spot. When NF  1, one obtains a Fresnel diffraction pattern.

10.73 (10.91) Rm2  mro  so Rm  (mro  )1/ 2 ;

R1  (1(1.00 m)5.6819  107 m)1/ 2  7.54  104 m.

10.74 The full first zone has a radius q1  1.22 R /2 a. Since area   q 2 , half the first zone corresponds to q  q1 / 2  1.22 R /2 2 a; I o   A2 A 2 , for a plane wave, so (10.55) becomes I  J (kaq /R)  I  J (ka(1.22 R /2 2 a)/R )   o2  1 I  o2  1   2 R  kaq /R  2 R  ka(1.22 R /2 2 a)/R  2

2



2

I o  J1 (1.22 / 2)  Io I (0.026)  o2 (0.013) (using Table 10.2)    R 2 R 2  1.22 / 2  2R2

102

Chapter 10 Solutions

10.75 The primary wave at point P has the form given by equation (10.86):

E

0

(   r0 )

cos[ t  k (   r0 )]

This follows from the fact that the first zone (l  1) is positive (l  1  2), thus El  (1)l 1  1, and from El  2 E. 10.76 For 3 zones: E  E1  E2  E3  E1 This is twice the amplitude of the unobstructed wave: Eu  12 E . Thus,

1 I 4 I  4Iu Iu 

(   r0 ) 2  R 2 r0 R 2 R 2 R 2 R      r0   r0   r0  r0   For a collimated light source,   , 1/  0 :

10.77 N F 

NF 

R2 r0 

r0 

R2 NF

For the first bright spot, NF  1:

r0 

(2.5  10 3 m)2  11.4 m 1(5.50  10 7 m)

The first dark spot, NF  1: r0  11.4 m/2  5.7 m

10.78 N F 

(   r0 ) 2 (1 m  1 m) R  (1  10 3 m)2  4  r0  (5.00  10 7 m)

The point P would be dark.

10.79 The disk obscures zones:

NF 

(   r0 ) 2 (1 m  1 m) R  (.5  10 3 m)2  1  r0  (5.00  10 7 m)

The disk blocks the first zone, the annulus passes zones 2, 3 and 4.

E  2 Eu I 4 Iu

103

Chapter 10 Solutions 10.80 Rm2  mr0   mf 

f  10.81

Rm2 (3.00  10 3 m)2   1.5 m m 10(6.00  10 7 m)

Rm2  mr0   mf  R1  1(2.00 m) (6.47  10 7 m)  1.14 mm R30  (30)(2.00 m)(6.47  10 7 m)  6.23 mm

 2( 0  r0 )  10.82 u  y    0 r0  

1/ 2

1/ 2

1/ 2

  2 r  2   2 0   y u  y 2    r0     r0   r0    For a collimated light source,   , 1/  0:

 2  u  y   r0    2    r0  

1/ 2

1/ 2

  z

 2  u1  y1    r0  

1/ 2

 2  u2  y2    r0    2    r0  

1/ 2

1/ 2

1  z1 

 2    r0  

 2  z2 

1/ 2

  2  (1  10 3 m)   7  (5m)(5.00  10 m)    2  (1  10 m)   7  (5m)(5.00  10 m) 

1/ 2

 0.8944

1/ 2

3

 0.8944

  2  (5  10 4 m)   7 (5m)(5.00 10 m)      2  (5  10 4 m)   7  (5m)(5.00  10 m) 

1/ 2

 0.4472

1/ 2

 0.4472

Iu {[2C (0.8944)]2  [2 L (0.8944)]2 {[2C (0.4472)]2  [2 L (0.4472)]2} 4 I I  u {[2(0.761)]2  [2(0.331)]2}{[2(0.445)]2  [2(0.049)]2} 4 Iu I  {2.316  0.438}{0.792  0.0096} 4 I  0.55 I u  16.6 W I

10.83 (From 10.42 and 10.43), I (0)  12 ( A A /R )2 , recall (3.46) I   o c E 2 so, T I (0)  12  o c( A A /R )2 ; I (incident)  12  o c( A A )2  (flux) (area)  10 W/m2 (5.0  103 m)2  2.5  104 W; I(0)  I(incident) A/R2  (2.5  104 W)(5.0  103 m)2  6.25  109 W.

104

Chapter 10 Solutions

10.84 For the solution to this problem, please refer to the textbook.

10.85 I  ( I 0 /2)([1/2  C (1 )]2  [1/2  S(1 )]2 ),

I  ( I 0 /2)(1 / 1 )2 [sin 2 (12 /2)  cos2 (12 /2)]  I 0 /2(1 )2 .

10.86 Fringes in both the clear and shadow region [see M. P. Givens and W. L. Goffe, Am. J. Phys., 34, 248 (1966)].

10.87 u  y[2/r0 ]1/ 2 ; u  y  103  2.5.

10.88 For the solution to this problem, please refer to the textbook.

10.89 We should see symmetry through the x-y plane in both patterns. The keyhole should bear some resemblance to the combined patterns of a circle and a rectanglar aperture. The image of the triangle should have nearly 3-fold symmetry.

10.90 As the slit widens, the pattern becomes more like that of a rectangular aperture (see Figure 10.49).

1/ 2

 2  10.91   z     r0 

1/ 2



 2  7  (5 10 m)(2 m)  

  z

 z (1414) 1 z   (1 104  ) 2

V1  0.1414

V2  0.1414

B12  0.28 IP 1 2  B12 Iu 2  0.0392 1/ 2

 2  10.92   z    r0   Using z1  0 and z2  0.70 mm:



 2  7 (1 m) (6.00 10 m)   

1  0, 2  (7.0  10 4 m)  B12  0.93 I P 1  (0.93)2  0.432 Iu 2 I P   43 W

1/ 2

 1.278

105

Chapter 10 Solutions

 2  10.93   z    r0  

1/ 2

Using z1  0 and z2  0.70 mm:



 2  7 (1m) (6.00 10 m)   

1  0, 2  (7.0  10 4 m)  B2   0.230 B1  0.707 These are anti-parallel, 0.707  0.230  0.477.

IP 1  (0.477)2  0.114 Iu 2

1/ 2

 1.278

106

Chapter 11 Solutions

Chapter 11 Solutions 11.1

E0 sin k p x  E0 (exp(ik p x )  exp(ik p x ))/2i ; L E0  L exp(i(k  k p ) x ) dx   exp (i(k  k p ) x ) dx     L 2i   L iE0 sin(k  k p )L iE0 sin(k  k p )L   (k  k p ) (k  k p )

F (k ) 

F (k )  iE0 L[sinc(k  k p )L  sinc(k  k p )L ].

11.2

(11.5) F (k )  





L

f ( x ) exp(ikx )dx   sin 2 k p exp(ik p x ) dx L

L

L

  sin k p x cos k p xdx  i  sin 3 k p x dx 2

L



11.3

L

L 1 sin 3 k p x  0  (2/3k p )(sin 3 k p L ).  L 3k p

cos 2  p t  1/2  (1/2) cos 2 p t  1/2  (1/4)(exp(2i p t )  exp(2i p t )).

1 T 1 T 1 T exp(i t ) dt   exp(i(  2 p t )) dt   exp(i(  2 p )t ) dt  T T   2 4 4 T 1 1 1 sin (  2 p )T  sin (  2 p )T  sin T  2 (  2 p ) 2(  2 p ) 

F ( ) 

F ( )  T sincT  (T /2)sinc(  2 p )T  (T /2)sinc (  2 p ) T .

11.4

Show that F 1 F ( K )  f ( x ), where

f ( x )  1, F ( K )  2 ( K ). 

f ( x )  (1/2 )  F ( K ) exp( iKx ) dK

(11.4)

 

 1/2   2 ( K ) exp( iKx ) dK  exp(0)  1. 

11.5

f ( x )  A cos K 0 x 

A (exp(iK 0 x )  exp(iK 0 x )) 2

A  A (exp(iK 0 x )  exp(iK 0 x )) exp(iKx ) dx  [ ( K  K 0 )   ( K  K 0 )] 2  2 K  K  A      f x  0     f x  0   2  2  2   

F K  

11.6

E ( )   ( E0 exp(t 2 /2 2 ) exp(i0 t ))exp(i t ) dt E ( )  E0  exp(t 2 /2 2  i(  o ) t ) dt E ( )  E0 (2 2 ) 2 exp( 14 [(i(  0 ))2 2 2 ]) 1

E ( )  E0 (2 2 ) 2 exp( 14 (  0 )2 2 2 ) 1

E ( )  2 E0 exp( 2 (  0 )2 /2)

107

Chapter 11 Solutions

11.7

2  ( E0 exp( 2 (  0 )2 /2)) exp(i t ) d 2  E  2 2 E (t )  0  exp( 2  2  ( 20  it )  2 02 ) d 2  E (t ) 

Using the integral identity from problem 11.6:

  2 2  exp(ax  bx  c) dx   a  exp( 14 (b /a)  c) 1/2



E (t ) 

E0  2  2 2 2 2  2 1  2  exp( 4 ( 0  it )/( /2)  2 0 ) 2   

E (t ) 

E0  2  i 2 0 t 1 402 2 t2  2  2  exp( 2 2  2 2   2  2 0 ) 2   

E (t ) 

E0  2  t2  2  exp( 2 2  i0 t) 2   

1/2

1/2

1/2

11.8

F (k )   f ( x )e ikx dx   f ( x ) cos kx dx   f ( x )i sin kx dx   f R ( x ) cos kx dx   fI ( x )sin kx dx   fI ( x ) cos kx dx   f R ( x )sin kx dx If f (x) is real, then fI (x)0:   f R ( x ) cos kx dx   f R ( x )sin kx dx If f (x) is even:

F (k )   f R ( x )cos kx dx 11.9

F [ af ( x )  bh( x )]  aF (k )  bH (k ).

11.11 F ( K )  L sinc 2 kL /2. F (0)  L, F (  2 /L )  0. or, F ( f x )  L sinc 2 ( f x L ),where k  2 f x .

11.12 F ( K )  





f ( x ) exp(ikx ) dx, (11.5). Let x  x/a; F ( K )  





f ( x /a ) exp(ik ( x /a)) d ( x /a ).

So, K   Ka, and F{ f ( x /a)}  F ( Ka). If, a  1, F{ f ( x )}  F ( K ).

11.13 F ( K )  F{ f ( x )}  





f ( x ) exp(ikx ) dx, (11.5) (a function of K). 

F{F{ f ( x )}}  F{F ( K )}   F ( K ) exp(ikx ) dK . 



(11.4) f ( x )  1/2   F ( K ) exp(iKx ) dK , 



so, 2 f ( x )   F ( K ) exp(iKx ) dK  F{F{ f ( x )}}  f ( x ). 

108

Chapter 11 Solutions





11.14 F rect  x  x0  /a  





rect  x  x  /a  exp(iKx) dx  0

1/ 2

1/ 2

exp(iKx ) dx

1/ 2



1 1 2 exp(iKx ) (exp(iK /2)  exp(iK / 2))  sin( K /2)  sinc( K /2)  iK iK K 1/ 2

 x F  rect   a sinc( f x a) a   Now apply the shift theorem by x0: F rect ( x  x0 )/a   a sinc ( f x a) exp(i 2 f x x0 )  a sinc (Ka /2) exp(iKx0 ) where, K  2 fx.

11.15 F{rect | x |}  sinc (K /2), from Problem 11.10, F{F{ f ( x )}}  2 f ( x ).

So, F {(1/2 ) F{ f ( x )}  f ( x ), let

f ( x )  rect | x |, F{rect | x |}  sinc (K /2); F{(1/2 ) F{rect | x |}}  F{(1/2 )sinc ( K /2)}  f ( x)  rect | x |, since sinc (x)  sinc (x). 







11.16 F 1{F{ f ( x )}}  (1/2 )  exp( iKx ) dK  







f ( x ) exp(iKx  ) dx 

  dx (  exp(iK ( x   x )) dK ) f ( x ) 

   ( x  x ) f ( x ) dx   f ( x ), 

since the integral is zero except at x  x .

11.17 F  f ( x  x0 )  





f ( x  xo ) exp(iKx ) dx. Change variables, x   x  xo ,

dx   dx. F  f ( x )  



f ( x )exp(iK ( x   xo )) dx   exp(iKxo ) 







f ( x ) exp(iKx ) dx . So

that F  f ( x  xo ) differs from F  f ( x ) by only the phase factor exp(iKxo ).

11.18









 f ( x) h( X  x) dx  



f ( X  x ) h( x  ) dx    h( x ) f ( X  x  ) dx  

where x   X  x, dx   dx . f * h  h * f or

F ( f * h)  F ( f )  F (h)  F (h)  F ( f )  F (h * f ).

11.19

 ( f  h)dt     (u) h(u  t ) du  dt   f (u)   h(u  t ) dt  du    f (u) du    h(t ) dt      

11.20 g(X) is the area under the product function f (x)h(X  x). The shape of g(X) reflects the overlap between f and h: It is non-zero in regions where the two functions overlap; it is maximum at the coordinate X where the two functions have their largest overlap. Since f and h are symmetrical about X  0, their product (and thus g) will be symmetrical and have its maximum at X  0. The width of g(X) is the sum of the widths of the functions f (x) and h(x): Thus the width is 3. The peak value is 0.75.

109

Chapter 11 Solutions 11.22 A point on the edge of f (x, y), for example, at (x  d, y  0), is spread out into a square 2l on a side centered on X = d. Thus it extends no farther than X  d  l, and so the convolution must be zero at X  d  l and beyond.

11.23:

11.24

f ( x  x0 ) * h ( x )  





becomes 





f ( x  x0 ) h( X  x ) dx, and setting x  X0  , this

f ( ) h( X    x0 ) d  g ( X  x0 ).

11.25 The convolution is asymmetric because the sawtooth function itself is asymmetric (i.e., the shape of the convolution reflects the overlap between the two sawtooth functions). The convolution begins at zero because there is no overlap at positions x < 0. The width of the convolution is the sum of the widths of the functions; it is twice the width of the sawtooth function and is thus 2.

110

Chapter 11 Solutions

11.26 The width of g(X) is the sum of the widths of the functions f(x) whose width is 2, and h(x) whose width is 4: Thus the width is 6.

11.27 g( X )  





f ( x ) h  X  x  dx, (11.52)









   ( x )h( X  x ) dx  h( X  0)  ( x ) dx, (see Section 11.2.3),  h( X ),



since   ( x ) dx  1. 

11.28 For the solution to this problem, please refer to the textbook.

111

Chapter 11 Solutions 11.29 F{ f ( x )cos K o x}  F{ f ( x )(1/2)(exp(iK o x )  exp(iK o x ))}    1/2    f ( x ) exp(i( K  K o ) x ) dx   f ( x ) exp(i( K  K o ) x ) dx     

 1/2  [ F ( K  K o )  F ( K  K o )].

F{ f ( x )sin K o x}  F{ f ( x ) 1/2i  (exp(iK o x )  exp(iK o x ))}    1/2i    f ( x ) exp(i( K  K o ) x ) dx   f ( x ) exp(i( K  K o ) x ) dx     

 1/2i  [ F ( K  K o )  F ( K  K o )].

11.31 The solution is symmetrical because f (x)  h(x). The peak occurs when there is maximum overlap of f (x) and h(x) (i.e., at 0). The width of g (X) is the sum of the widths of the functions f (x) and h (x) whose widths are 1: Thus the width is 2. The graphical convolution is seen below.

112

Chapter 11 Solutions

11.32 We see that f (x) is the convolution of a rect-function with two -functions, and from the convolution theorem,

F (k )  F{(rect( x ) *[ ( x  a )   ( x  a )])}  F[rect( x )]  F{[ ( x  a )   ( x  a )]}  a sinc (ka /2)  (exp(ikx )  exp(ika ))  a sinc (ka /2)  2 cos ka.

11.33

f ( x ) * h( x)  [ ( x  3)   ( x  2)   ( x  5)]* h( x)  h( x  3)  h( x  2)  h( x  5).

11.34

11.35 F rect x /(d /2)  



d  kd  sinc    F (k ) 2  4 



F  n   ( x  nd )   n  exp(iKnd )  H ( K ); G  K   H ( K ) F ( K ); 



F ( K )  F  f ( x ) is zero at knd  n or kd   .

11.36 For the solution to this problem, please refer to the textbook.

11.37 A( y, z)  A( y,  z).

E Y , Z , t    A( y, z ) exp(i(kY y  kZ z )) dy dz.

Change Y to Y, Z to Z, y to y, z to z, then kY goes to kY and kZ to kZ.

E Y , Z , t    A( y,  z ) exp(i(kY y  kZ z )) dy dz.

Therefore E(Y, Z)  E(Y, Z).

11.38 From Eq. (11.63), E Y , Z    A( y, z ) exp(ik (Yy  Zz )/R ) dydz.

E  Y , Z    A( y,  z ) exp(ik (Yy  Zz )/R ) dydz;

now let y   y, z    z :

E (Y , Z ) 

1

 

A  y, z   e(ik[(Y / ) y  ( Z / )z ]/R) dy dz 

or E (Y , Z )  (1/ ) E (Y / , Z / ).

1 T A sin ( t   ) A sin ( t     ) dt T  2T  T A2 T  1 1   lim cos( )  cos(2 t    2 )  dt , T  2T  T  2 2  

11.39 C ff  lim

since cos   cos   2 sin (1/2)(   ) sin (1/2)(   ).

Thus C ff  ( A 2 /2) cos ( ).

113

Chapter 11 Solutions

11.40

E (kz )  

b/ 2

b / 2

A0 cos( z /b) exp(ik Z z ) dz

 A0  cos( z /b) cos(k Z z ) dz  iA0  cos( z /b)sin(k Z z ) dz E (k Z )  A0 cos

 bk Z  1 1   .

2   /b  k Z  /b  k Z 

11.41 The width of g(X) is the sum of the widths of the functions f (x) and h (x) whose widths are 1: Thus the width is 2. The peak will be at x  1.

114

Chapter 11 Solutions



11.42 c ff ( x )   cos(ku   ) cos(ku    kx ) du 

Here x was replaced with u  x. Using the identity:

1  cos( A  B)  cos( A  B)  2 1  1 c ff ( x )   [cos(2 ku  2  kx )  cos(kx )] du  cos kx  2 2

cos A cos B 

11.43 The width of the autocorrelation function is twice the width of the function, thus the width of cff (x) is 4x0. The function is an even function which starts at 2x0 and ends at 2x0.

11.44 Symmetry is a fundamental property of autocorrelation. The maximum overlap occurs at the when X  x  0, thus the correlation function cff (x) will be symmetric about the origin. The width of the correlation function is twice that of the original function. Since it is symmetrical about the origin, it will stretch from 1 to 1.

11.45 (From 11.52). h( X )  f ( x ) * g( x )  





f ( x ) * g( x )  h ( x )  





f ( x ) g ( X  x ) dx, so,

f ( x )g  X  x  dx, which is the form of (11.86),

so f ( x ) * g( x )  f ( x )  g( x ).

11.46 The width of the correlation function is twice that of the original function. The width of each individual peak is twice the width of each individual square pulse.

11.47 The autocorrelation is periodic.

11.48 There will be three spots in the autocorrelation function. The first results from the overlap of only one spot from each pair (the original pair and its duplicate). Because f(x, y) is symmetrical, mirroring it has no effect. Thus one just sweeps one circle over the other and records the product at each displacement.

The resulting irradiance can be seen in Fig. 11.37b. The second spot will be separated from the first by the same distance as that between the two original spots. This second spot will be brighter than the first because it results from the overlap of both spots from each pair. The resulting irradiance pattern is the same as that for the first except it will be brighter (i.e., it can be seen in Fig. 11.37b). The third spot will be identical to the first spot, and will be located the same distance from the bright spot as the first spot is from the bright spot. The width of the autocorrelation function is twice the width of the original pattern.

11.50

f (t )  g (t )  h (t )  A cos(o t )  e( io t ).

F ( )  G( ) H ( );

G( )  ( A /2)(2 /(  o )  2 /(  o ));

H ( )  2 /  o , so, F ( )  2 2 A/(  o )(2 / 2  o2 ).

115

Chapter 12 Solutions

Chapter 12 Solutions 12.1

I  I1  I 2  2 I1 I 2 cos  Imax occurs at   0: I max  I1  I 2  2 I1 I 2  ( I1  I 2 )2

Since I1 = 100I2: I max  (10 I 2  I 2 )2  121 I 2

Imin occurs at    : I min  I1  I 2  2 I1 I 2  ( I1  I 2 )2  (10 I 2  I 2 )2  81 I 2



12.2

I max  I min 40 I 2   0.198 I max  I min 202 I 2

I  I 0 cos2   I 0 cos2 (   /2) Use cos(   /2)  cos  cos  /2  sin  sin  /2   sin 

I  I 0 cos2   I 0 sin 2   I 0 2

12.3

2

      d  Ac    c   dc2   0  ~  0  4 4  s   2  s  As ~ ds2  l 2 s2

s 

ds l 2

 l  (l  )2 Ac ~  0   0 As  ds 

2

116

Chapter 12 Solutions

2

12.4

 l  (2.0 m)2 (5.00  10 7 m)2 Ac ~  0   (1.00  10 6 m 2 )  As 

Ac ~ 1.00  106 m 2 12.5

Starting with the result from problem 12.3, 2

 l  (l  )2 Ac ~  0   0 As  ds 

Use, As  l 2  s:

(l0 )2 l 2 02  2 As l s

Ac ~

02

Ac ~ 12.6

Ac ~  s

s

02 s As  r 2  2 l2 l

Using r  lθ, s 

Ac ~

 s

l

2



 l 2 2 l2

  2

02 02 (5.50  10 7 m)2 A  ~   4.45  10 9 m 2  4.45  10 3 mm 2 c 2 3 s  2  9.3  10 rad    

12.7

r2

2



As l2

Using Eq. (12.3):

Ac ~  s

(l0 )2 As

As (l0 )2 02  2  Ac l2 l Ac

12.8

At low pressures, the intensity emitted from the lamp is low, the bandwidth is narrow, and the coherence length is large. The fringes will initially display a high contrast, although they’ll be fairly faint. As the pressure builds, the coherence length will decrease, the contrast will drop off, and the fringes might even vanish entirely.

12.9

Over a long time interval, E1  E2 averages to zero. So, (E1  E2 )2  T   E12  T   E22  T .

12.10 The net irradiance becomes more uniform as more waves are added. There will be a less distinct pattern, which corresponds to a smaller coherence length. The irradiance will become constant as the bandwidth goes to infinity.

117

Chapter 12 Solutions

12.11 Each sine function in the signal produces a cosinusoidal autocorrelation function with its own wavelength and amplitude. All of these are in phase at the zero delay point corresponding to   0.

Beyond that origin the cosines soon fall out of phase, producing a jumble where destructive interference is more likely. (The same sort of thing happens when, say, a square pulse is synthesized out of sinusoids—everywhere beyond the pulse all the contributions cancel.) As the number of components increases and the signal becomes more complex—resembling random noise—the autocorrelation narrows, ultimately becoming a -spike at   0.

12.12 (12.1)   ( I max  I min )/( I max  I min )  2 sinc(a w /s ) /2 (from 12.8, 12.9),  sinc(5  104  /1  103 )  sinc( /2)  0.64.

12.13 The irradiance at  0 arising from a point source is 4 I 0 cos2 ( /2)  2 I 0 (1  cos  ). For a differential source element of width dy at point S , y from the axis, the OPD to P at Y via the two slits is   (S S1  S1 P )  (S S2  S2 P  (S S1  S S2 )  (S1 P  S2 P  ay /l  aY /s from Section 9.3. The

contribution to the irradiance from dy is then dI  (1  cos k  ) dy, I  

b/2

b / 2

I b

(1  cos k  )dy,

d   aY ab   aY ab   sin    sin    I  b 2l  2l   ka   s  s

 (d /ka) sin(kaY /s) cos(kab /2l )  cos(kaY /s)sin( kab /2l )  sin(kaY /s) cos( kab /2l )  cos(kaY /s)sin(kab /2l ) ,

I  b  (2l /ka)sin(kab /2l ) cos(kaY /s).

12.14   ( I max  I min )/( I max  I min ), I max  I1  I 2  2 I1 I 2 12 ,

I min  I1  I 2  2 I1 I 2 12 ,   4 I1 I 2 12 /2( I1  I 2 ).

12.15 When S S1O   S S1O   /2, 3 /2, 5 /2,  , the irradiance due to S  is given by I   4 I 0 cos2 ( /2)  2 I 0 (1  cos  ), while the irradiance due to

S  is I   4 I 0 cos2 ( /2)  4 I 0 cos2 (    )/2  2I 0 (1  cos  ). Hence I   I   4 I 0 .

12.16 Fringes disappear when w  s /a so, a   (s /w), from Figure 12.3, l /b  s /w, a   (l /b)  (5.893  10 7 m)(1 m)/(1  10 4 m)

 5.893  10 3 m.

12.17 dc 

1.220

s

For small angles,  s  0.1 mm /1 m  104 rad

dc 

1.22(5.00  10 7 m)  6.1  10 3 m 4 10 rad

118

Chapter 12 Solutions

12.18 I1 (t )  I1 (t )   I1 ; hence

I1  t    I 2 (t )   I1  I1 (t   )  I 2  I 2 (t ) since I1 is independent of time.

I1  t    I 2 (t )  I1 I 2  I1 (t   )I 2 (t ) if we recall that I1 (t )  0. Eq. (12.34) follows by comparison with Eq. (12.32).

12.20 From Eq. (12.22),   2 (10 I )I /(10 I  I )  2 10 /11  0.57.

12.21 Fringes disappear when w  s /a, so, a   (s /w) , from Figure 12.3, l /b  s /w where l  (mean) distance to sun; b  diameter of sun.

a   (l /b)  [(5.50  107 m)(1.50  1011 m)] /2(6.96  108 m)  5.93  10 5 m.  a b  12.22   sinc  0  l  a b  l ab 1 l l (1 m)(5.00  10 7 m)   5 mm a b 0.1  10 3 m

 a b  12.23   sinc    0.9  l 

Since sinc( /4)  0.9, a b   l 4 l (1 m)(5.50  10 7 m)   6.875  10 4 m  0.69 mm b 4a 4(2.0  10 4 m)

12.24 Using the van Cittert-Zernike theorem, we can find 12 (0) from the diffraction pattern over the apertures, and that will yield the visibility on the observation plane:   12 (0)  sinc .

From Table 1, sin u /u  0.85 when u = 0.97, hence  by /l  0.97, and if y  P1 P2  0.50 mm, then b  0.97(l / y)  0.97(1.5 m)(500  10 9 m)/ (0.50  10 3 m)  0.46 mm.

12.25 (12.23)   |  12 ( ) | .

(12.1)   ( I max  I min )/( I max  I min )  2 sinc (a w /s ) /2 (from 12.8, 12.9),

  0.90  sinc (a w /s )  sinc (a (1.0  103 m)/(10.0 m)(5.00  107 m))  sinc(200 a) ; sin x  x  x 2 /3!, so sinc( x )  1  x 2 /3!; 0.90  1  [(200 a )2 /6]; a  1.23  10 3 m.

119

Chapter 12 Solutions 12.26   sinc ( a b /l ) ; as shown in Figure 12.6,  is a minimum when (a b /l )  m , (m  0). b /l  sin ( 2  1 )  ( 2  1 ) for small angle, so minimum  when [ a( 2  1 ) / ]  m ; a( 2  1 )  m.

12.27 From the van Cittert-Zernike theorem, the degree of coherence can be obtained from the Fourier transform of the source function, which itself is a series of -functions corresponding to a diffraction grating with spacing a, where a sin θm  m. The coherence function is therefore also a series of -functions. Hence the P1 P2 , the slit separation d, must correspond to the location of the first-order diffraction fringe of the source if  is to be maximum. a1   , and so

d  l1  l /a  (500  109 m)(2.0 m)/(500  10 6 m)  2.0 mm.

12.28 h  0.32

R0  0.32 / s D

Use   9.3  103 rad,

  550 nm

h  0.32(5.5 107 m)/(9.3 103 )

 1.89  105 m 12.29 From example 12.1,  s  0.0093 rad, 0  5.00  107 m h  0.32

0 5.00  10 7 m  0.32  1.7  10 5 m  0.017 mm 0.0093 rad s

Compare this to the result in the example ~0.05 mm.

12.30 Two stars: 1  s   m   0 2 h 

From Eq. (12.29), one star:

  s  1.22 0 h

120

Chapter 13 Solutions

Chapter 13 Solutions 13.1

T  673 K, area of each face is A  102 m2,   5.67  108 W m2 K4, then 0.97AIe  0.97 AT4  110 W.

13.2

0.97Ie  0.97  (T 4  TE4 )  76.9 W/m 2 with T  306 K and Te  293 K is the temperature of the environment. Then 0.97AIe  108 W for the radiated power.

13.3

I e  22.8  10 4 W/m 2 , T  ( I e / )1/ 4  1420 K.

13.4

E  T 4 , so the energy radiated increases by a factor of 104.

13.5

T  306 K, max  2.8978  10 3 mK /T  9.45  10 6 m  9.5 m (in the infrared).

13.6

If the blackbody is at T  293 K, then max  2.8978  10 3 m K/T  9.9 m (in the IR).

13.7

T  4.0  104 K,  max  c /max  cT /2.8978  10 3 m K  4.1  1015 Hz (in the UV).

13.8

T  2.8978  103 m K/max  2.8978  10 3 mK/4.65  10 7 m  6230 K.

13.9

T  2.8978  103 m K/max  4300 K.

13.10 We have for the total radiated power per unit area of the blackbody 



0

0

P(T )   I  d   2 hc 2 

d  5 (ehc /  k T  1) B

4

 k T   x 3 dx  2 hc 2  B   , x  hc  0 (e  1) by putting x  hc / k BT , d   hcdx /kBTx 2 . The value of the integral over x is (4) (4)  3!  4 /90   4 /15. Therefore the Stefan-Boltzmann law follows,

P(T ) 

2 5 (k BT )4 .

15 h 3 c 2

13.11 From Eq (13.4):   2 hc 2  1    5   khcBT  e  1  2 hc 2  3.7418  10 16 hc  0.00144 kB I 

I 

   25  3.7418  10 16  1 1  W/m 3  3.7418  10   W/m 2  nm 5 0.00144  0.00144    5   e T  1   e T  1 

13.12 E  hc /  1.99  10 25 J m/. Since 1 eV  1.602  1019 J and 1 nm  10–9 m, this gives E  1240 eV nm/. Therefore the energy of a 600 nm photon is 2.1 eV. 13.13 (min)  300 nm, E  h  hc /  1240 eV nm /300 nm  4.14 eV  6.63  10 19 J.

121

Chapter 13 Solutions

13.14 If the P  100 W light bulb has an efficiency of   2.5%, then the radiated power is  P  Nh /t , where N is the number of photons, v is their frequency, and t  1 ms. In terms of the wavelength   550 nm, this gives N   Pt /h   Pt  /hc. The solid angle subtended by the d  3 cm diameter aperture at distance r = 100 s is: aperature area/r2   d2/ 4r2. Making the assumption that the light bulb emits isotropically, this is the fraction of photon that passes through the aperature, N  d 2 /4r 2   Pt  d 2 /4r 2 hc  4.9  108 photons.

13.15 Nhv  Nhc /  (1.4  103 W /m 2 )(1 m 2 )(1 s) gives N  49  1020.

13.16 The number of Ar atoms present in the chamber is N  pV /kBT  2.69  1017. Taking 1% of this number and using the given excited-state lifetime yields 0.01 N /1.4  10 8 s  1.9  1023 transitions per second.

13.17 With energy density

 ( ) 

A21 /B21 ( B12 /B21 ) eh / kBT  1

and B12  B21, the ratio of interest is B21  ( )/A21  1/(eh / kBT  1).

13.18 kBT  4.14  1021 J, E  3.2  10 19 J, E /kBT  77.3. Since exp( E /k BT )  1, the ratio

1/(exp( E /kBT )  1)  exp( E /kBT )  2.7  10 34 , extremely low.

13.19 kBT  4.14  10 19 J, E  3.2  10 19 J, E /kBT  0.773. Then 1/(exp( E /k BT )  1)  0.86, much higher than in the previous problem. Stimulated emission is quite likely at this high temperature (as at the surface of a hotter star).

13.20 This statement gives the time rate of change of the population in from level 2 as it drops down to level 1.

In thermal equilibrium, the rate of excitation to level 2 must equal the rate of downward transitions to level 1: dN 2 dN1  dt dt

Using: dN 2  B12 uv N1  B21uv N 2  A21 N 2 dt dN1   B12 uv N1  B21uv N 2  A21 N 2 dt 2 B21uv N 2  2 A21 N 2  2 B12 uv N1 A21 N 2  B21uv N 2  B12 uv N1

13.21 Each photon may be said to be the result of a stimulated emission event.

E  hc 

hc



 4.499  10 19 J

# of events 

100  10 3 J/s  2.22  1017 photons/sec.

4.499  10 19 J

122

Chapter 13 Solutions

13.22 N j /N i  e

 ( E j  Ei )/ k B T

 1  ( E j  Ei )/k BT for E j  Ei  k BT .

Therefore as T  , N j /N i , tends to 1.

T  3.0 K, kBT  4.14  1023 J,

13.23

E  hc /  1.99  10 25 J m/  9.51  10 25 J, E /k BT  2.15  10 2 and the ratio 1/(exp( E /k BT )  1)  46, so stimulated emission is very likely, if not dominant.

(The number of significant figures is important in such a case.)

13.24 From the example, ET  2.59  107 J/m3 .

P

E 2.59  107 J/m 3   1.13  1011 W/m 3 t 230  10 6 s

13.25  

1 1   1.06  s 5  Aji 7.8  10  1.6  105

13.26  

1 10 5   1.63  10 8 s  A ji 259  2.83  2.00  2.26  6.09  6.39  33.9  3.45  13.9  2.55

13.27   2.44 /D  5.15  10 4 rad, s  r   5.15  10 2 m or the diameter of the spot on the wall is 5.1 cm.

13.28 The volume of the crystal is V  ( /4) D 2 L  9.8  10 7 m3. Therefore the mass of Cr2O3 present is 0.05  102(3.7  103)9.8  107  1.81  106 Kg. The mass of one Cr2O3 molecule is 152 amu or 2.52  1025 Kg. Therefore approximately 7.17  1018 Cr2O3 molecules. are present. Assuming that each contributes two Cr3+ ions to lasing, Nions  1.4  1019 ions participate in the lasing action, at E  2.87  1019 J. Then Etot  Nions E  4.0 J; the corresponding power is Etot/t  4.0 J/5.0  106s  8  106 W = 800 kW. 13.29 N /t  P /E  1.0  10 3 J s 1 /(1.96)(1.602  10 19 J  3.2  1015 transitions per second.

13.30 V   r 2 h   (5  10 3 )2 (0.2) m 3  1.5708  10 5 m 3 # of photons  (# of ions) (V) (efficiency)  (4.0  1019 ions/cm3)(15.708 cm3)(0.02)

 1.2566  1019 photons E  hc 

hc



 2.834  1019 J

(1.2566  1019 photons)(2.834  1019 J/photon)  3.56 J 13.31 0  02  /c  8.0  10 5 nm.

13.32   c /2 L  6  108 Hz, using   c when n  1.

Chapter 13 Solutions

13.33  

123

c 3  108 m/s   1.5  108 Hz 2L (2)(1 m)

2.9  109 Hz  18 intervals 1.5  108 Hz 19 modes 13.34  m  m

m mc c   2 L 2nL  2 nL





2(0.04 m)(1)  1.33  10 6 6.00  10 7 m

13.35 The condition   1.4  10 9  c /2 L for n  1 gives L  c /2   11 cm.

  1  1   1  1 1 ln  13.36 gth       10 cm    ln(6.25)  40.5 cm  2(0.03 cm)  2 L   R1 R2    13.37 I  ( /2)  E02  (n /2)( 0 /0 )1/ 2 E02 , where   0 , E02  2( 0 / 0 )1/ 2 I /n, ( 0 / 0 )1/ 2  376.730, so

E0  27.4( I /n)1/ 2 .

13.38   2.44 /D  2.6  10 3 rad.

13.39 The three crossed gratings form a type of triangular lattice. The diffraction spots will appear along the directions of the dual lattice, which are directions connecting the centroids of the original lattice. As usual, there will be a central spot of highest irradiance (intensity) and the irradiance decreases with distance from this central spot. Reciprocals of multiples of lattice constants of the original lattice are proportional to the spatial frequencies present in the diffraction pattern. (Strictly speaking, the lattice should have infinite extent in order to consider it a mathematically periodic structure.)

13.40 In this case, the four crossed gratings form a sort of rectangular lattice, whose dual lattice is again rectangular. Therefore the diffraction spots will be located along horizontal and vertical lines. The central spot has the highest irradiance and the irradiance of the others decreases with distance from the center. A horizontal slit filter will keep the vertical lined grating the same in the altered image, but blur the horizontal lines.

13.41 The horizontal grating gives a row of diffraction spots, with the central spot of highest irradiance. The details in the picture image are contained in many high spatial frequency components. The picture can be enhanced by using a filter which blocks out the diffraction pattern of the horizontal grating.

13.42 The circular grating present will generate a central spot of highest irradiance, together with successive rings. In order to enhance the picture, a spatial filter which blocks out these contributions should be used.

13.43 The filter is a long slit, perpendicular to the observed image.

13.44 From the geometry, ft  fi  : kO  k sin  and k1  k sin , hence sin     kO  /2 and sin     kI  /2 , therefore  /  kO /kI and kI  kO ( / )  kO ( ft /fi ). When fi  ft the image will be larger than the object, the spatial periods in the image will also be larger, and the spatial frequencies in the image will be smaller than in the object.

124

Chapter 13 Solutions

13.45 a  (1/50) cm: a sin   m, sin    , hence   5000 m and the distance between orders on the transform plane is f  5000 f  2.7 mm.

13.46 (a) As in Figure 11.10, the transform of the cosine function will be a pair of -functions, at

fx  1/d, where d is the spatial period of the cosine plus a zero-order term at x  0. To pass only the first order terms, we need a filter with holes at these positions, for the specific wavelength, as given by x/f ≃ sin 1  (1/d); x  f/d  [(2.0 m)(5  107 m)/(1  105 m)]  0.1 m, above and below center. (b) Any “DC” components, and all higher order components, are removed. A smoothly varying cosine function should be seen in the image. (c) A filter with a hole in the center would pass only the “DC” term, resulting in a lower intensity, uniform image.

13.47 Each point on the diffraction pattern corresponds to a single spatial frequency, and if we consider the diffracted wave to be made up of plane waves, it also corresponds to a single-plane wave direction.

Such waves, by themselves, carry no information about the periodicity of the object and produce a more or less uniform image. The periodicity of the source arises in the image when the component plane waves interfere.

13.49 The relative field amplitudes are 1.00, 0.60, and 0.60; hence E  1  0.60 cos( ky)  0.60 cos( ky)  1  1.2 cos ky. This is a cosine oscillating about a line equal to 1.0. It varies from +2.2 to –0.2. The square of this will correspond to the irradiance, arid it will be a series of tall peaks with a relative height of (2.2)2, between each pair of which there will be a short peak proportional to (0.2)2; notice the similarity with Fig. 11.32.

13.50 a Sin   , here f   50 f  0.20 cm; hence   0.20/50(100)  400 nm. The magnification is 1.0 when tile focal lengths are equal, hence the spacing is again 50 wires/cm.

13.51 The random (dots will add considerable “noise” to the pattern. The spatial frequency is 1/(0.1 mm)  10 mm1. A filter that is the transform of the regular pattern will remove the random dots.

13.52 The array of top hats corresponds to the pixels, so that each “selects” the amplitude (density) of the picture within its radius. The transform will look like a regular array of dots of varying amplitude.

As in Figure 13.39, filtering out the higher frequency components will yield a continuous image.

13.53 The pinhole blocks the high-frequency components, which correspond to the rapid spatial variations in the beam.

13.54 The randomly, but more or less uniformly, distributed particles in the milk will tend to block the “regular” part of the beam, and thus enhance the relative intensity of the speckle.