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22.1. Visualize: Please refer to Figure Ex22.1. Solve:

(a)

(b) The initial light pattern is a double-slit interference pattern. It is centered behind the midpoint of the slits. The slight decrease in intensity going outward from the middle indicates that the light from each of the individual slits is not uniform but slowly decreases toward the edges of the screen. If the right slit is covered, light comes through only the left slit. Without a second slit, there is no interference. Instead, we get simply the spread-out pattern of light diffracting through a single slit, such as in the center of the photograph of Figure 22.2. The intensity is a maximum directly behind the left slit, and—as we discerned from the intensities in the double-slit pattern—the single-slit intensity fades gradually toward the edges of the screen.


22.2. Model: Two closely spaced slits produce a double-slit interference pattern.

Visualize: The interference pattern looks like the photograph .of Figure 22.3(b). It is symmetrical, with the m = 2 fringes on both sides of and equally distant from the central maximum. Solve: The two paths from the two slits to the m = 2 bright fringe differ by ∆r = r2 − r1 , where

∆r = mλ = 2λ = 2(500 nm ) = 1000 nm Thus, the position of the m = 2 bright fringe is 1000 nm farther away from the more distant slit than from the nearer slit.


22.3. Model: Two closely spaced slits produce a double-slit interference pattern.

Visualize: The interference pattern looks like the photograph of Figure 22.3(b). It is symmetrical with the m = 2 fringes on both sides of and equally distant from the central maximum. Solve: The bright fringes occur at angles θm such that

d sin θ m = mλ ⇒ sin θ 2 =

2(500 × 10

(50 × 10

−9

−6

m)

m)

m = 0, 1, 2, 3, …

= 0.02 ⇒ θ 2 = 0.020 rad = 0.020 rad ×

180° = 1.15° π rad


22.4. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The fringe spacing is

∆y =

−9 −2 λL λL (589 × 10 m )(150 × 10 m ) = 0.221 mm ⇒d= = d 4.0 × 10 −3 m ∆y


22.5. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The dark fringes are located at positions given by Equation 22.9: λL m = 0, 1, 2, 3, … ym′ = ( m + 12 ) d

⇒ y5′ − y1′ = (5 +

1 2

)

λL − (1 + d

1 2

)

4λ (60 × 10 −2 m ) λL ⇒ 6.0 × 10 −3 m = ⇒ λ = 500 nm 0.20 × 10 −3 m d


22.6. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The formula for fringe spacing is L L λL ⇒ = 3000 ∆y = ⇒ 1.8 × 10 −3 m = 600 × 10 −9 m d d d The wavelength is now changed to 400 nm, and L d , being a part of the experimental setup, stays the same. Applying the above equation once again, λL ∆y = = 400 × 10 −9 m (3000) = 1.2 mm d

(

(

)

)


22.7. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: In a span of 12 fringes, there are 11 gaps between them. The formula for the fringe spacing is

∆y =

−9  52 × 10 −3 m  (633 × 10 m )(3.0 m ) λL ⇒ ⇒ d = 0.40 mm = 11 d   d

Assess: This is a reasonable distance between the slits, ensuring d L = 1.34 × 10 −4 << 1.


22.8. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15: d sin θ m = mλ m = 0, 1, 2, 3, …

⇒ sin θ1 = Likewise, sin θ 2 = 0.110 and θ 2 = 6.32°.

(1)(550 × 10 −9 m)

(1.0 × 10

−2

m ) 1000

= 0.055 ⇒ θ1 = 3.15°


22.9. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: The bright constructive-interference fringes are given by Equation 22.15:

d sin θ m = mλ ⇒ d =

(2)(600 × 10 −9 m) mλ = = 1.89 × 10 −6 m sin θ m sin(39.5°)

The number of lines in per millimeter is (1 × 10 −3 m) (1.89 × 10 −6 m ) = 530 .


22.10. Model: A diffraction grating produces an interference pattern. Visualize:

Solve:

The interference pattern looks like the diagram in Figure 22.8.

The bright fringes are given by Equation 22.15:

d sin θ m = mλ

m = 0, 1, 2, 3, … ⇒ d sin θ1 = (1)λ ⇒ d = λ sin θ1

The angle θ1 can be obtained from geometry as follows: (0.32 m) 2 tan θ1 = = 0.080 ⇒ θ1 = tan–1 (0.080) 4.57° 2.0 m Using sin θ1 = sin 4.57° = 0.07968 ,

d=

633 × 10 −9 m = 7.94 µm 0.07968


22.11. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: (a) A grating diffracts light at angles sin θ m = mλ d . The distance between adjacent slits is d = 1 mm 600 = 1.667 × 10 −6 m = 1667 nm. The angle of the m = 1 fringe is

500 nm  θ1 = sin −1  = 17.46°  1667 nm  The distance from the central maximum to the m = 1 bright fringe on a screen at distance L is

y1 = L tan θ1 = (2 m ) tan 17.46° = 0.629 m (Note that the small angle approximation is not valid for the maxima of diffraction gratings, which almost always have angles >10°.) There are two m = 1 bright fringes, one on either side of the central maximum. The distance between them is ∆y = 2 y1 = 1.258 m . (b) The maximum number of fringes is determined by the maximum value of m for which sinθm does not exceed 1 because there are no physical angles for which sinθ > 1. In this case, mλ m(500 nm ) sin θ m = = 1667 nm d We can see by inspection that m = 1, m = 2, and m = 3 are acceptable, but m = 4 would require a physically impossible sin θ 4 > 1. Thus, there are three bright fringes on either side of the central maximum plus the central maximum itself for a total of 7 bright fringes.


22.12. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: The bright interference fringes are given by

d sin θ m = mλ

m = 0, 1, 2, 3, …

The slit spacing is d = 1 mm 500 = 2.00 × 10 −6 m . The angles of diffraction are −9 m  (1)λ  −1  510 × 10 θ1 = sin −1   = sin   = 14.77° −6  d   2.00 × 10 m 

 (2)λ  θ 2 = sin −1   = 30.66°  d 

Likewise, θ 3 = 49.9° . The next order (m = 4) is not seen because

sin θ 4 = Thus, three diffraction orders are visible.

4 × 510 × 10 −9 m > 1.0 2 × 10 −6 m


22.13. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram of Figure 22.8. Solve: The bright interference fringes are given by d sin θ m = mλ m = 0, 1, 2, 3, … The slit spacing is d = 1 mm 500 = 2.00 × 10 −6 m and m = 1. For the red and blue light,

 656 × 10 −9 m  θ1 red = sin −1   = 19.15°  2.00 × 10 −6 m 

 486 × 10 −9 m  θ1 blue = sin −1   = 14.06°  2.00 × 10 −6 m 

The distance between the fringes, then, is ∆y = y1 red − y1 blue where

y1 red = (1.5 m ) tan 19.15° = 0.521 m y1 blue = (1.5 m ) tan 14.06° = 0.376 m So, ∆y = 0.145 m = 14.5 cm.


22.14. Model: A narrow single slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The minima occur at positions

yp = p So ∆y = y2 − y1 =

Lλ . a

−9 λL (633 × 10 m )(1.5 m ) 2λL 1λL λL ⇒a= = = 2.0 × 10 −4 m = 0.20 mm − = 0.00475 m ∆y a a a


22.15. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a = 200 λ is

w=

−9 2λL 2(500 × 10 m )(2.0 m ) = = 0.0040 m = 4.0 mm 0.0005 m a


22.16. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: The width of the central maximum for a slit of width a = 200 位 is

w=

2位L 2位 (2.0 m ) = = 2.0 cm a 200位


22.17. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14. Solve: Angle θ = 0.70° = 0.0122 rad is a small angle (<< 1 rad). Thus we use Equation 22.20 to find the wavelength of light. The angles of the minima of intensity are

θp = p ⇒λ=

aθ p p

=

λ a

(0.10 × 10

p = 1, 2, 3, K −3

m )(0.0122 rad ) 2

= 611 nm


22.18. Model: The spacing between the two building is like a single slit and will cause the radio waves to be diffracted. Solve: Radio waves are electromagnetic waves that travel with the speed of light. The wavelength of the 800 MHz waves is

λ=

3 × 10 8 m / s = 0.375 m 800 × 10 6 Hz

To investigate the diffraction of these waves through the spacing between the two buildings, we can use the general condition for complete destructive interference: a sin θ p = pλ (p = 1, 2, 3, …) where a is the spacing between the buildings. Because the width of the central maximum is defined as the distance between the two p = 1 minima on either side of the central maximum, we will use p = 1 and obtain the angular width ∆θ = 2θ1 from

 λ  0.375 m  a sin θ1 = λ ⇒ θ1 = sin −1   = sin −1 = 1.43°  15 m   a Thus, the angular width of the wave after it emerges from between the buildings is ∆θ = 2(1.43°) = 2.86° .


22.19. Model: The crack in the cave is like a single slit that causes the ultrasonic sound beam to diffract. Visualize:

Solve:

The wavelength of the ultrasound wave is

λ=

340 m / s = 0.0113 m 30 kHz

Using the condition for complete destructive interference with p = 1,

0.0113 m  asin θ1 = λ ⇒ θ1 = sin −1  = 2.165°  0.30 m  From the geometry of the diagram, the width of the sound beam is

w = 2 y1 = 2(100 m × tan θ1 ) = 200 m × tan 2.165° = 7.56 m Assess: The small-angle approximation is almost always valid for the diffraction of light, but may not be valid for the diffraction of sound waves which have a much larger wavelength.


22.20. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: The width of the central maximum for a circular aperture of diameter D is

w=

−9 2.44λL (2.44)(500 × 10 m )(2.0 m ) = 4.88 mm. = 0.50 × 10 −3 m D


22.21. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24, the diameter of the circular aperture is

D=

−9 2.44λL 2.44(633 × 10 m )( 4.0 m ) = = 0.25 mm w 2.5 × 10 −2 m


22.22. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: From Equation 22.24,

L=

(0.12 × 10 −3 m)(1.0 × 10 −2 m) = 77.7 cm Dw = 2.44λ 2.44(633 × 10 −9 m )


22.23. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Visualize: The intensity pattern will look like Figure 22.15. Solve: According to Equation 22.23, the angle that locates the first minimum in intensity is

θ1 =

−6 1.22λ 1.22(2.5 × 10 m ) = 0.01525 rad = 0.874° = D 0.20 × 10 −3 m


22.24. Model: An interferometer produces a new maximum each time L2 increases by

length difference ∆r to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the wavelength is

λ=

−6 2 ∆L2 2(100 × 10 m ) = = 4.0 × 10 −7 m = 400 nm ∆m 500

1 2

λ causing the path-


22.25. Model: An interferometer produces a new maximum each time L2 increases by length difference ∆r to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the number of fringe shifts is ∆m =

−2 2 ∆L2 2(1.00 × 10 m ) = = 30, 467 656.45 × 10 −9 m λ

1 2

λ causing the path-


22.26. Model: An interferometer produces a new maximum each time L2 increases by length difference ∆r to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: From Equation 22.33, the distance the mirror moves is

1 2

λ causing the path-

−9 ∆mλ (33,198)(602.446 × 10 m ) = 0.0100000 m = 1.00000 cm = 2 2 Assess: Because the wavelength is known to 6 significant figures and the fringes are counted exactly, we can determine ∆L to 6 significant figures.

∆L2 =


22.27. Model: The gas increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: From Equation 22.36, the number of fringe shifts is

∆m = m2 − m1 = (n − 1)

(2)(2.00 × 10 −2 m) 2d = (1.00028 − 1) = 19 λ vac 600 × 10 −9 m


22.28. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Fig. 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.28 corresponds to a double-slit aperture. This is because the fringes are equally spaced and the decrease in intensity with increasing fringe order occurs slowly. (b) From Figure P22.28, the fringe spacing is ∆y = 1.0 cm = 1.0 × 10 −2 m . Therefore, λL d λL (6.00 × 10 −9 m )(2.5 m ) ⇒d= = = 0.15 mm 0.010 m ∆y ∆y =


22.29. Model: Two closely spaced slits produce a double-slit interference pattern with the intensity graph looking like Fig. 22.3(b). The intensity pattern due to a single slit diffraction looks like Figure 22.14. Both the spectra consist of a central maximum flanked by a series of secondary maxima and dark fringes. Solve: (a) The light intensity shown in Figure P22.29 corresponds to a single slit aperture. This is because the central maximum is twice the width and much brighter than the secondary maximum. (b) From Figure P22.29, the separation between the central maximum and the first minimum is y1 = 1.0 cm = 1.0 × 10 −2 m . Therefore, using the small-angle approximation, Equation 22.21 gives the condition for the dark minimum: yp =

−9 Lλ (2.5 m )(600 × 10 m ) pLλ = 0.15 mm ⇒a= = y1 d 1.0 × 10 −2 m


22.30. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference pattern looks like the photograph of Figure 22.3(b). Solve: The bright fringes are located at positions given by Equation 22.4, d sin θ m = mλ . For the m = 3 bright

(

)

orange fringe, the interference condition is d sin θ 3 = 3 600 × 10 −9 m . For the m = 4 bright fringe the condition is d sin θ 4 = 4λ . Because the position of the fringes is the same,

d sin θ 3 = d sin θ 4 = 4λ = 3(600 × 10 −9 m ) ⇒ λ =

3 4

(600 × 10

−9

m ) = 450 nm


22.31. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize: The interference fringes are equally spaced on both sides of the central maximum. The interference pattern looks like Figure 22.3(b). Solve: In the small-angle approximation

∆θ = θ m +1 − θ m = ( m + 1) Since d = 200 λ , we have

∆θ =

λ λ λ −m = d d d

λ 1 = rad = 0.286° d 200


22.32. Model: Two closely spaced slits produce a double-slit interference pattern. Solve:

The light intensity of a double-slit interference pattern at position y is πd  Idouble = 4 I1 cos 2  y  λL 

where I1 is the intensity due to each wave. The intensity of a bright fringe is 4I1. For the central maximum, m = 0. So, ycentral = 0 m . The first minimum occurs at y1′ = λL / 2 d . Thus, the halfway position between the maximum and the minimum is y = λL 4 d . Substituting into the intensity equation, the intensity at the halfway position is

 πd λL  2 π  Idouble = 4 I1 cos 2  = 2I1  = 4 I1 cos  4  λL 4 d 


22.33. Model: Two closely spaced slits produce a double-slit interference pattern. Solve:

The light intensity of a double-slit interference pattern at a position y on the screen is

 y  πd  y = 4 I1 cos 2  π  Idouble = 4 I1 cos 2   λL   ∆y  where ∆y = λL / d = 4.0 mm is the fringe spacing. Using this value for λL d , we can find the position on the interference pattern where Idouble = I1 as follows: −3 π π  y = cos −1  1  = π rad ⇒ y = 4.0 × 10 m = 1.33 mm 4 I1 cos 2  y = I1 ⇒  − − 3 3  4.0 × 10 m   4.0 × 10 m   2 3 3


22.34. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is ∆y = λL d . ∆y can be obtained from Figure P22.34. The separation between the m = 2 fringes is 2.0 cm implying that the separation between the two consecutive fringes is 14 (2.0 cm) = 0.50 cm. Thus,

−3 −2 d∆y (0.20 × 10 m )(0.50 × 10 m ) λL = = 167 cm ⇒L= −9 600 × 10 m λ d Assess: A distance of 167 cm from the slits to the screen is reasonable.

∆y = 0.50 × 10 −2 m =


22.35. Solve: According to Equation 22.7, the fringe spacing between the m fringe and the m + 1 fringe is

∆y = λL d . ∆y can be obtained from Figure P22.34. Because the separation between the m = 2 fringes is 2.0 cm, two consecutive fringes are ∆y = 14 (2.0 cm ) = 0.50 cm apart. Thus, ∆y = 0.50 × 10

−2

−3 −2 d∆y (0.20 × 10 m )(0.50 × 10 m ) λL ⇒ λ = = = 500 nm m= 2.0 m L d


22.36. Solve: The intensity of light of a double-slit interference pattern at a position y on the screen is πd  Idouble = 4 I1 cos 2  y  λL  where I1 is the intensity of the light from each slit alone. At the center of the screen, that is, at y = 0 m, I1 = 14 Idouble . From Figure P22.34, Idouble at the central maximum is 12 mW/m2 . So, the intensity due to a single slit is I1 = 3 mW / m 2 .


22.37. Model: A diffraction grating produces an interference pattern. Visualize: The interference pattern looks like the diagram in Figure 22.8. Solve: 500 lines per mm on the diffraction grating gives a spacing between the two lines of d = 1 mm 500 = (1 × 10 −3 m) 500 = 2.0 × 10 −6 m . The wavelength diffracted at angle θ m = 30° in order m is −6 d sin θ m (2.0 × 10 m ) sin 30° 1000 nm = = m m m We’re told it is visible light that is diffracted at 30°, and the wavelength range for visible light is 400–700 nm. Only m = 2 gives a visible light wavelength, so λ = 500 nm.

λ=


22.38. Model: A diffraction grating produces an interference pattern like the diagram of Figure 22.8. Visualize:

Solve: (a) A key statement is that the lines are seen on the screen. This means that the light is visible light, in the range 400 nm–700 nm. We can determine where the entire visible spectrum falls on the screen for different values of m. We do this by finding the angles θm at which 400 nm light and 700 nm light are diffracted. We then use ym = L tan θ m to find their positions on the screen which is at a distance L = 75 cm. The slit spacing is d = 1 mm 1200 = 8.333 × 10 −7 m . For m = 1,

λ = 400 nm: θ1 = sin −1 ( 400 nm / d ) = 28.7° ⇒ y1 = 75 cm tan 28.7° = 41 cm λ = 700 nm: θ1 = sin −1 (700 nm / d ) = 57.1° ⇒ y1 = 75 cm tan 57.1° = 116 cm For m = 2,

λ = 400 nm: θ1 = sin −1 (2 ⋅ 400 nm / d ) = 73.8° ⇒ y2 = 75 cm tan 73.8° = 257 cm For the 700 nm wavelength at m = 2, θ 2 = sin −1 (2 ⋅ 700 nm / d ) = sin −1 (1.68) is not defined, so y2 → ∞ . We see that visible light diffracted at m = 1 will fall in the range 41 cm ≤ y ≤ 116 and that visible light diffracted at m = 2 will fall in the range y ≥ 257 cm. These ranges do not overlap, so we can conclude with certainty that the observed diffraction lines are all m = 1. (b) To determine the wavelengths, we first find the diffraction angle from the observed position by using

y y  θ = tan −1   = tan −1   L  75 cm  This angle is then used in the diffraction grating equation for the wavelength with m = 1, λ = d sin θ1 . θ λ Line 36.85° 56.2 cm 500 nm 41.30° 65.9 cm 550 nm 51.27° 93.5 cm 650 nm


22.39. Model: Each wavelength of light is diffracted at a different angle by a diffraction grating. Solve: angle

Light with a wavelength of 501.5 nm creates a first-order fringe at y = 21.90 cm. This light is diffracted at

21.90 cm  θ1 = tan −1  = 23.65°  50.00 cm  We can then use the diffraction equation dsinθm = mλ, with m = 1, to find the slit spacing:

d=

λ 501.5 nm = = 1250 nm sin θ1 sin(23.65°)

The unknown wavelength creates a first order fringe at y = 31.60 cm, or at angle

31.60 cm  θ1 = tan −1  = 32.29°  50.00 cm  With the split spacing now known, we find that the wavelength is

λ = d sin θ1 = (1250 nm) sin(32.29°) = 667.8 nm Assess: The distances to the fringes and the first wavelength were given to 4 significant figures. Consequently, we can determine the unknown wavelength to 4 significant figures.


22.40. Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m determined by Equation 22.15. Solve: We have

d sin θ m = mλ

m = 0, 1, 2, 3, K ⇒ d sin 20.0° = 1λ and d sin θ 2 = 2λ

Dividing these two equations,

sin θ 2 = 2 sin 20.0° = 0.6840 ⇒ θ 2 = 43.2°


22.41. Model: A diffraction grating produces a series of constructive-interference fringes at values of θ m that are determined by Equation 22.15. Solve: For the m = 3 maximum of the red light and the m = 5 maximum of the unknown wavelength, Equation 22.15 gives

d sin θ 3 = 3(660 × 10 −9 m )

d sin θ 5 = 5λ unknown

The m = 5 fringe and the m = 3 fringe have the same angular positions. This means θ5 = θ3 . Dividing the two equations,

5λ unknown = 3(660 × 10 −9 m ) ⇒ λ unknown = 396 nm


22.42. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. The visible spectrum spans the wavelengths 400 nm to 700 nm. Solve: According to Equation 22.16, the distance ym from the center to the mth maximum is ym = L tan θ m . The angle of diffraction is determined by the constructive-interference condition d sin θ m = mλ , where m = 0, 1, 2, 3, … The width of the rainbow for a given fringe order is thus w = yred − yviolet. The slit spacing is

d=

1 mm 1.0 × 10 −3 m = = 1.6667 × 10 −6 m 600 600

For the red wavelength and for the m = 1 order,

d sin θ1 = (1)λ ⇒ θ1 = sin −1

λ 700 × 10 −9 m = sin −1 = 24.83° d 1.6667 × 10 −6 m

From the equation for the distance of the fringe, yred = L tan θ1 = (2.0 m ) tan(24.83°) = 92.56 cm Likewise for the violet wavelength,  400 × 10 −9 m  θ1 = sin −1   = 13.88° ⇒ yviolet = (2.0 m ) tan(13.88°) = 49.42 cm  1.6667 × 10 −6 m  The width of the rainbow is thus 92.56 cm − 49.42 cm = 43.1 cm.


22.43. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. The visible spectrum spans the wavelengths 400 nm to 700 nm. Solve: The slit spacing is

d=

1.0 × 10 −3 m = 3.333 × 10 −6 m 300

The constructive-interference condition is d sin θ m = mλ , where m = 0, 1, 2, 3, … For the red and violet light in the third-order rainbow,

 3 × 700 × 10 −9 m  θ 3 red = sin −1   = 39.05°  3.333 × 10 −6 m 

 3 × 400 × 10 −9 m  θ 3 violet = sin −1   = 21.10°  3.333 × 10 −6 m 

That is, the angular spread of the third-order rainbow is from 21.10° to 39.05°. For the fourth-order rainbow,

 4 × 700 × 10 −9 m  θ 4 red = sin −1   = 57.14°  3.333 × 10 −6 m 

 4 × 400 × 10 −9 m  θ 4 violet = sin −1   = 28.68°  3.333 × 10 −6 m 

We clearly see that the overlap of wavelengths is from an angle of 28.68° to an angle of 39.05°. Since we are asked to find wavelengths in the third-order rainbow that overlap the wavelengths in the fourth-order rainbow, we will calculate the wavelengths in the third-order that correspond to an angle of 28.68°. Using the constructiveinterference condition,

d sin θ 3 = 3λ ⇒ λ =

1 3

(3.333 × 10

−6

m ) sin(28.68) = 533 nm

Thus, the wavelengths in the range 533–700 nm overlap with some of the wavelengths of the fourth order.


22.44. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: (a) If blue light (the shortest wavelengths) is diffracted at angle θ, then red light (the longest wavelengths) is diffracted at angle θ + 30°. In the first order, the equations for the blue and red wavelengths are

sin θ =

λb d

d sin(θ + 30°) = λ r

Combining the two equations we get for the red wavelength,

 λ2  λ  λ r = d (sin θ cos 30° + cosθ sin 30°) = d (0.8660 sin θ + 0.50 cosθ ) = d  b  0.8660 + d (0.50) 1 − b2   d d   ⇒ (0.50)d 1 −

λ2b 2 2 = λ r − 0.8660λ b ⇒ (0.50) ( d 2 − λ2b ) = (λ r − 0.8660λ b ) d2 2

 λ − 0.8660λ b  2 ⇒d=  r  + λb   0.50 Using λ b = 400 × 10 −9 m and λ r = 700 × 10 −9 m , we get d = 8.125 × 10 −7 m . This value of d corresponds to

1 mm 1.0 × 10 −3 m = = 1231 lines / mm d 8.125 × 10 −7 m (b) Using the value of d from part (a) and λ = 589 × 10 −9 m , we can calculate the angle of diffraction as follows:

(

)

d sin θ1 = (1)λ ⇒ 8.125 × 10 −7 m sin θ1 = 589 × 10 −9 m ⇒ θ1 = 46.5°


22.45. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: An 800 line/mm diffraction grating has a slit spacing d = (1.0 × 10 −3 m) 800 = 1.25 × 10 −6 m . Referring to Figure P22.45, the angle of diffraction is given by

tan θ1 =

y1 0.436 m = = 0.436 ⇒ θ1 = 23.557° ⇒ sin θ1 = 0.400 L 1.0 m

Using the constructive-interference condition d sin θ m = mλ ,

λ=

d sin θ1 = (1.25 × 10 −6 m )(0.400) = 500 nm 1

We can obtain the same value of λ by using the second-order interference fringe. We first obtain θ2:

tan θ 2 =

y2 0.436 m + 0.897 m = = 1.333 ⇒ θ 2 = 53.12° ⇒ sin θ 2 = 0.800 L 1.0 m

Using the constructive-interference condition,

λ=

−6 d sin θ 2 (1.25 × 10 m )(0.800) = = 500 nm 2 2

Assess: Calculations with the first-order and second-order fringes of the interference pattern give the same value for the wavelength.


22.46. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the wavelength used. Solve: From Figure P22.45,

tan θ1 =

0.436 m = 0.436 ⇒ θ1 = 23.557° ⇒ sin θ1 = 0.400 1.0 m

Using the constructive-interference condition d sin θ m = mλ ,

d sin 23.557° = (1)(600 × 10 −9 m ) ⇒ d =

600 × 10 −9 m = 1.50 × 10 −6 m sin(23.557°)

Thus, the number of lines per millimeter is

1.0 × 10 −3 m = 667 lines / mm 1.50 × 10 −6 m Assess: The same answer is obtained if we perform calculations using information about the second-order bright constructive-interference fringe.


22.47. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: The dark fringes are located at

yp =

pλL a

p = 0, 1, 2, 3, …

The locations of the first and third dark fringes are

y1 =

λL a

y3 =

3λL a

Subtracting the two equations,

( y3 − y1 ) =

2(589 × 10 −9 m )(0.75 m ) 2λL 2λL ⇒a= = 0.118 mm = y3 − y1 7.5 × 10 −3 m a


22.48. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: 45° is not a small angle, so we can’t use equations based on the small-angle approximation. As given by Equation 22.19, the dark fringes in the pattern are located at a sin θ p = pλ , where p = 1, 2, 3, … For the first minimum to be at 45°, p = 1 and the width of the slit is

a=

pλ 633 × 10 −9 m = = 895 nm sin θ p sin 45°


22.49. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: These are not small angles, so we can’t use equations based on the small-angle approximation. As given by Equation 22.19, the dark fringes in the pattern are located at a sin θ p = pλ , where p = 1, 2, 3, … For the first minimum of the pattern, p = 1. Thus,

1 a p = = λ sin θ p sin θ1 For the three given angles the slit width to wavelength ratios are

a 1 (a):   = = 2,  λ  30° sin 30°

a 1 (b):   = = 1.15,  λ  60° sin 60°

a 1 (c):   = =1  λ  90° sin 90°

Assess: It is clear that the smaller the a/λ ratio, the wider the diffraction pattern. This is a conclusion that is contrary to what one might expect.


22.50. Model: A narrow slit produces a single-slit diffraction pattern. Visualize: The diffraction-intensity pattern from a single slit will look like Figure 22.14. Solve: As given by Equation 22.19, the dark fringes in the pattern are located at a sin θ p = pλ , where p = 1, 2, 3, … For the diffraction pattern to have no minima, the first minimum must be located at least at θ1 = 90°. From the constructive-interference condition a sin θ p = pλ , we have

a=

pλ λ ⇒a= = λ = 633 nm sin θ p sin 90°


22.51. Model: A narrow slit produces a single-slit diffraction pattern. Solve:

The dark fringes in this diffraction pattern are given by Equation 22.21: pλL p = 1, 2, 3, … yp = a We note from Figure P22.51 that the first minimum is 0.50 cm away from the central maximum. Using the above equation, the slit width is

a=

−9 pλL (1)(500 × 10 m )(1.0 m ) = = 0.10 mm yp 0.50 × 10 −2 m

Assess: This is a typical slit width for diffraction.


22.52. Model: A narrow slit produces a single-slit diffraction pattern. Solve:

The dark fringes in this diffraction pattern are given by Equation 22.21: pλL p = 1, 2, 3, … yp = a We note from Figure P22.51 that the first minimum is 0.50 cm away from the central maximum. Thus,

L=

ay p pλ

(0.15 × 10 m)(0.50 × 10 (1)(600 × 10 m) −3

=

Assess: This is a typical slit to screen separation.

−9

−2

m)

= 1.25 m


22.53. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small angle approximation, which is almost always valid for the diffraction of light, the width of the central maximum is λL w = 2 y1 = 2.44 D From Figure P22.51, w = 1.0 cm, so

D=

−9 2.44λL 2.44(500 × 10 m )(1.0 m ) = = 0.122 mm w (1.0 × 10 −2 m)

Assess: This is a typical size for an aperture to show diffraction.


22.54. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: Within the small-angle approximation, the width of the central maximum is

w = 2.44

λL D

Because w = D, we have

D = 2.44

λL ⇒ D = 2.44λL ⇒ D = (2.44)(633 × 10 −9 m )(0.50 m ) = 0.879 mm D


22.55. Model: Light passing through a circular aperture leads to a diffraction pattern that has a circular central maximum surrounded by a series of secondary bright fringes. Solve: (a) Because the visible spectrum spans wavelengths from 400 nm to 700 nm, we take the average wavelength of sunlight to be 550 nm. (b) Within the small-angle approximation, the width of the central maximum is

w = 2.44

(550 × 10 −9 m)(3 m) ⇒ D = 4.03 × 10 −4 m = 0.40 mm λL ⇒ (1 × 10 −2 m ) = (2.44) D D


22.56. Model: The antenna is a circular aperture through which the microwaves diffract. Solve: (a) Within the small-angle approximation, the width of the central maximum of the diffraction pattern is w = 2.44λL/D. The wavelength of the radiation is

λ=

2.44(0.025 m )(30 × 10 3 m ) c 3 × 10 8 m / s = = ⇒ w = = 915 m 0.025 m f 12 × 10 9 Hz 2.0 m

That is, the diameter of the beam has increased from 2.0 m to 915 m, a factor of 458. (b) The average microwave intensity is

I=

P 100 × 10 3 W 2 = 2 = 0.152 W / m area π [ 12 (915 m )]


22.57. Model: The laser beam is diffracted through a circular aperture. Visualize:

Solve: (a) No. The laser light emerges through a circular aperture at the end of the laser. This aperture causes diffraction, hence the laser beam must gradually spread out. The diffraction angle is small enough that the laser beam appears to be parallel over short distances. But if you observe the laser beam at a large distance it is easy to see that the diameter of the beam is slowly increasing. (b) The position of the first minimum in the diffraction pattern is more or less the “edge” of the laser beam. For diffraction through a circular aperture, the first minimum is at an angle −9 1.22λ 1.22(633 × 10 m ) = 5.15 × 10 −4 rad = 0.0295° = D 0.0015 m (c) The diameter of the laser beam is the width of the diffraction pattern:

θ1 =

−9 2.44λL 2.44(633 × 10 m )(3 m ) = = 0.00309 m = 0.31 cm D 0.0015 m (d) At L = 1 km = 1000 m, the diameter is −9 2.44λL 2.44(633 × 10 m )(1000 m ) w= = = 1.03 m = 103 cm D 0.0015 m

w=


22.58. Model: The laser light is diffracted by the circular opening of the laser from which the beam emerges. Solve:

The diameter of the laser beam is the width of the central maximum. We have

−9 8 2.44λL 2.44(532 × 10 m )(3.84 × 10 m ) 2.44λL ⇒ D= = = 0.498 m = 49.8 cm w 1000 m D In other words, the laser beam must emerge from a laser of diameter 49.8 cm.

w=


22.59. Model: Two closely spaced slits produce a double-slit interference pattern. Visualize:

Solve:

(a) The m = 1 bright fringes are separated from the m = 0 central maximum by

−9 λL (600 × 10 m )(1.0 m ) = = 0.0030 m = 3.0 mm 0.0002 m d (b) The light’s frequency is f = c/λ = 5.00 × 1014 Hz. Thus, the period is T = 1/f = 2.00 × 10−15 s. A delay of 5.0 × 10−16 s = 0.50 × 10−15 s is 14 T . (c) The wave passing through the glass is delayed by 14 of a cycle. Consequently, the two waves are not in phase as they emerge from the slits. The slits are the sources of the waves, so there is now a phase difference ∆φ 0 between the two sources. A delay of a full cycle (∆t = T) would have no effect at all on the interference because it corresponds to a phase difference ∆φ 0 = 2π rad. Thus a delay of 14 of a cycle introduces a phase difference ∆φ 0 = 14 (2π ) = 12 π rad. (d) The text’s analysis of the double-slit interference experiment assumed that the waves emerging from the two slits were in phase, with ∆φ 0 = 0 rad. Thus, there is a central maximum at a point on the screen exactly halfway between the two slits, where ∆φ = 0 rad and ∆r = 0 m. Now that there is a phase difference between the sources, the central maximum—still defined as the point of constructive interference where ∆φ = 0 rad—will shift to one side. The wave leaving the slit with the glass was delayed by 14 of a period. If it travels a shorter distance to the screen, taking 14 of a period less than the wave coming from the other slit, it will make up for the previous delay and the two waves will arrive in phase for constructive interference. Thus, the central maximum will shift toward the slit with the glass. How far? A phase difference ∆φ 0 = 2π would shift the fringe pattern by ∆y = 3.0 mm, making the central maximum fall exactly where the m = 1 bright fringe had been previously. This is the point where ∆r = (1)λ, exactly compensating for a phase shift of 2π at the slits. Thus, a phase shift of ∆φ 0 = 12 π = 14 (2π ) will shift the fringe pattern by 14 (3 mm ) = 0.75 mm . The net effect of placing the glass in the slit is that the central maximum (and the entire fringe pattern) will shift 0.75 mm toward the slit with the glass.

∆y =


22.60. Model: A diffraction grating produces an interference pattern, which looks like the diagram of Figure 22.8. Solve: (a) Nothing has changed while the aquarium is empty. The order of a bright (constructive interference) fringe is related to the diffraction angle θm by d sin θ m = mλ , where m = 0, 1, 2, 3, … The space between the slits is d=

1.0 mm = 1.6667 × 10 −6 m 600

For m = 1,

sin θ1 =

 633 × 10 −9 m  λ ⇒ θ1 = sin −1   = 22.3°  1.6667 × 10 −6 m  d

(b) The path-difference between the waves that leads to constructive interference is an integral multiple of the wavelength in the medium in which the waves are traveling, that is, water. Thus, 633 nm 633 nm λ 4.759 × 10 −7 m λ= = = 4.759 × 10 −7 m ⇒ sin θ1 = = = 0.2855 ⇒ θ1 = 16.6° 1.33 nwater d 1.6667 × 10 −6 m


λ causing the pathlength difference ∆r to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: The path-length difference between the two waves is ∆r = 2L2 − 2L1 . The condition for constructive interference is ∆r = mλ, hence constructive interference occurs when

22.61. Model: An interferometer produces a new maximum each time L2 increases by

1 2

2( L2 − L1 ) = mλ ⇒ L2 − L1 = 12 mλ 2 = 1200( 12 λ ) = 600λ where λ = 632.8 nm is the wavelength of the helium-neon laser. When the mirror M2 is moved back and a hydrogen discharge lamp is used, 1200 fringes shift again. Thus,

L2′ − L1 = 1200( 12 λ ′) = 600λ ′ where λ ′ = 656.5 nm . Subtracting the two equations,

( L2 − L1 ) − ( L2′ − L1 ) = 600(λ − λ ′) = 600(632.8 × 10 −9 m − 656.5 × 10 −9 m) ⇒ L2′ = L2 + 14.2 × 10 −6 m

That is, M2 is now 14.2 µm closer to the beam splitter.


22.62. Model: An interferometer produces a new maximum each time L2 increases by length difference ∆r to increase by λ. Visualize: Please refer to the interferometer in Figure 22.20. Solve: For sodium light of the longer wavelength (λ1) and of the shorter wavelength (λ2),

1 2

λ causing the path-

λ1 λ ∆L = ( m + 1) 2 2 2 We want the same path difference 2(L2 − L1) to correspond to one extra wavelength for the sodium light of shorter wavelength (λ2) . Thus, we combine the two equations to obtain: ∆L = m

m

λ2 λ1 λ 589.0 nm = = 981.67 ≅ 982 = ( m + 1) 2 ⇒ m(λ1 − λ 2 ) = λ 2 ⇒ m = λ1 − λ 2 589.6 nm − 589.0 nm 2 2

Thus, the distance by which M2 is to be moved is

∆L = m

λ1  589.6 nm  = 982 = 0.289 mm   2 2


22.63. Model: The water increases the number of wavelengths in one arm of the interferometer.

Solve: (a) The incident light has λvac = 500 nm and f = c / λ vac = 6.00 × 1014 Hz . Water doesn’t affect the frequency, which is still fw = f = 6.00 × 1014 Hz . The wavelength changes to λ w = λ vac / nw = 376 nm . (b) Light travels in a layer of thickness L twice— once to the right and then, after reflecting, once to the left—for a total distance 2L. The number of wavelengths in distance 2L is N = 2L/λ. If the 1 mm layer is a vacuum, the number of wavelengths is

N vac =

2(0.001 m ) 2L = = 4000 λ vac 500 × 10 −9 m

If the vacuum is replaced by 1 mm of water, the number of wavelengths is

Nw =

2(0.001 m ) 2L = 5319 = λ w 376 × 10 −9 m

So with the water added, the light travels 1319 extra wavelengths. (c) Each additional wavelength of travel shifts the pattern by 1 fringe. So the addition of the water shifts the interference pattern by 1319 fringes.


22.64. Model: The piece of glass increases the number of wavelengths in one arm of the interferometer. Each additional wavelength causes one bright-dark-bright fringe shift. Solve: We can rearrange Equation 22.36 to find that the index of refraction of glass is

n = 1+

(500 × 10 −9 nm)(200) = 1.50 λ vac ∆m = 1+ 2d 2(0.10 × 10 −3 m )


22.65. Model: The number of wavelengths in one arm of the interferometer increases as the refractive index of the electro-optic crystal increases. Each bright-dark-bright fringe shift corresponds to an additional path-difference increase of one wavelength. A path-difference of one-half wavelength thus will cause the bright fringe to shift to a dark fringe. Visualize: Please refer to the interferometer in Figure 22.20. Solve: We will follow the treatment of Section 22.6 in the textbook. For an initial index of refraction n of the electro-optic crystal, the number of wavelengths inside the crystal is (including motion in both directions)

m1 =

2d λ vac n

Similarly, the number of wavelengths inside the crystal when its refractive index increases to n′ is

m2 =

2d λ vac n ′

The increase in the number of wavelengths due to a change in refractive index from n to n′ is

∆m = m2 − m1 =

2d (n ′ − n) λ vac

Because the external voltage causes a bright fringe to shift to a dark one, ∆m = 12 . Thus,

λ 0.981 × 10 −6 m 1 2d ⇒ ∆n = n ′ − n = 0.0025 = (n ′ − n) ⇒ n ′ = n + vac = n + 4d 2 λ vac 4(0.100 × 10 −3 m ) Thus, n ′ = 1.5500 + 0.0025 = 1.5525.


22.66. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8. We also assume that the small-angle approximation is valid for this grating. Solve: (a) The general condition for constructive-interference fringes is

d sin θ m = mλ

m = 0, 1, 2, 3, …

When this happens, we say that the light is diffracted at an angle θm. Since it is usually easier to measure distances rather than angles, we will consider the distance ym from the center to the mth maximum. This distance is ym = L tan θ m . In the small-angle approximation, sin θ m ≈ tan θ m , so we can write the condition for constructive interference as

d

ym mλL = mλ ⇒ ym = L d

The fringe separation is

ym +1 − ym = ∆y =

λL d

(b) Now the laser light falls on a film that has a series of “slits” (i.e., bright and dark stripes), with spacing

d′ =

λL d

Applying once again the condition for constructive interference: y′ mλL mλL = = md d ′ sin θ m = mλ ⇒ d ′ m = mλ ⇒ ym′ = d′ λL d L The fringe separation is ym′ +1 − ym′ = ∆y ′ = d . That is, using the film as a diffraction grating produces a diffraction pattern whose fringe spacing is d, the spacing of the original slits.


22.67. Model: Two closely spaced slits produce a double-slit interference pattern. The interference pattern is symmetrical on both sides of the central maximum. Visualize:

In figure (a), the interference of laser light from the two slits forms a constructive-interference central (m = 0) fringe at P. The paths 1P and 2P are equal. When a glass piece of thickness t is inserted in the path 1P, the interference between the two waves yields the 10th dark fringe at P. Note that the glass piece is not present in figure (a). Solve: The number of wavelengths in the air-segment of thickness t is t m1 = λ The number of wavelengths in the glass piece of thickness t is t t nt m2 = = = λ glass λ n λ The path length has thus increased by ∆m wavelengths, where

t λ From the 10th dark fringe to the 1st dark fringe is 9 fringes and from the first dark fringe to the zeroth bright fringe is one-half of a fringe. Hence, ∆m = m2 − m1 = (n − 1)

19 λ 19 (633 × 10 m ) 1 19 19 t = = 12.0 µm = ⇒ = (n − 1) ⇒ t = 2 (n − 1) 2 1.5 − 1.0 2 2 2 λ −9

∆m = 9 +


22.68. Model: Two closely spaced slits produce a double-slit interference pattern. The fringe intensity decreases because the intensity of the two individual slits is not uniform. Visualize:

Solve: To a good approximation, each slit illuminates an area on the screen that is the width of a single-slit diffraction pattern. (A single slit diffraction pattern does have secondary maxima, but they are very weak in comparison with the central maximum.) The width of the single-slit pattern is

w=

2λL 2(600 × 10 −9 m)(2.0 m) = = 6.0 cm a 4.0 × 10 −5 m

Thus the fringes of the double-slit pattern can be seen over a range of 6.0 cm, from y = –3.0 cm to y = +3.0 cm. How many fringes “fit” within this distance? The fringe spacing of the double slit pattern is

∆yfringes =

λL (600 × 10 −9 m)(2.0 m) = = 0.60 mm d 0.00020 m

The central maximum is at y = 0, then on both sides there are fringes at y = 0.6 cm, 1.2 cm, 1.8 cm, and 2.4 cm. The next fringe, which would be at 3.0 cm, won’t be seen because that is where the intensity has dropped to zero. Thus a total of 9 fringes are seen. Assess: In practice, with a sufficiently strong light source such as a laser, a few fringes may be seen farther away from the center where the secondary maxima of the single-slit pattern restore some intensity, but they will be very dim in comparison with these 9 fringes.


22.69. Model: A diffraction grating produces a series of constructive-interference fringes at values of θm that are determined by Equation 22.15. Solve: (a) The condition for bright fringes is d sin θ = mλ . If λ changes by a very small amount ∆λ, such that θ changes by ∆θ, then we can approximate ∆λ/∆θ as the derivative dλ/dθ: 2

λ=

d d d ∆λ dλ d  mλ   d  − λ2 1−  sin θ ⇒ ≈ = cosθ = 1 − sin 2 θ =  =  m  d  m m m ∆θ dθ m ⇒ ∆θ =

2

∆λ 2

 d  − λ2  m

(b) We can now obtain the first-order and second-order angular separations for the wavelengths λ = 589.0 nm and λ + ∆λ = 589.6 nm. The slit spacing is 1.0 × 10 −3 m d= = 1.6667 × 10 −6 m 600 The first-order (m = 1) angular separation is

∆θ =

0.6 × 10 −9 m 2.7778 × 10 −12 m 2 − 0.3476 × 10 −12 m 2

=

0.6 × 10 −9 m 1.5589 × 10 −6 m

= 3.85 × 10 −4 rad = 0.022° The second order (m = 2) angular separation is

∆θ =

0.6 × 10 −9 m 0.6945 × 10 −12 m 2 − 0.3476 × 10 −12 m 2

= 1.02 × 10 −3 rad = 0.058°


22.70. Model: The intensity in a double-slit interference pattern is determined by diffraction effects from the slits. Visualize: Please refer to Figure CP22.70. Solve: (a) For the two wavelengths λ and λ + ∆λ passing simultaneously through the grating, their first-order peaks are at λL (λ + ∆λ ) L y1′ = y1 = d d Subtracting the two equations gives an expression for the separation of the peaks: ∆λL ∆y = y1′ − y1 = d

(b) For a double-slit, the intensity pattern is πd  Idouble = 4 I1 cos 2  y  λL  The intensity oscillates between zero and 4I1, so the maximum intensity is 4I1. The width is measured at the point where the intensity is half of its maximum value. For the intensity to be 12 I max = 2 I1 for the m = 1 peak:

1 πd πd πd π λL 2 I1 = 4 I1 cos 2  y  ⇒ cos 2  y = ⇒ yhalf = ⇒ yhalf =  λL half   λL half  2 4 4d λL The width of the fringe is twice yhalf. This means λL w = 2 yhalf = 2d But the location of the m = 1 peak is y1 = λL d , so we get w = 12 y1 . (c) We can extend the result obtained in (b) for two slits to w = y1 N . The condition for barely resolving two diffraction fringes or peaks is w = ∆ymin. From part (a) we have an expression for the separation of the first-order peaks and from part (b) we have an expression for the width. Thus, combining these two pieces of information,

∆λ min L y1 y d  λL  d λ = ∆ymin = ⇒ ∆λ min = 1 =   = N d LN  d  NL N (d) Using the result of part (c),

N=

λ 656.27 × 10 −9 m = = 3646 lines 0.18 × 10 −9 m ∆λ min


22.71. Model: A diffraction grating produces an interference pattern like the one shown in Figure 22.8, when the incident light is normal to the grating. The equation for constructive interference will change when the light is incident at a nonzero angle. Visualize: Please refer to Figure CP22.71.

Solve: (a) The path difference between waves 1 and 2 on the incident side is ∆r1 = d sin φ . The path difference between the waves 1 and 2 on the diffracted side, however, is ∆r2 = d sin θ . The total path difference between waves 1 and 2 is thus ∆r1 + ∆r2 = d (sin θ + sin φ ) . Because the path difference for constructive interference must be equal to mλ,

d (sin θ + sin φ ) = mλ

m = 0, ±1, ±2, …

(b) For φ = 30° the angles of diffraction are

sin θ1 =

(1)(500 × 10 −9 m)

− sin 30° = −0.20 ⇒ θ1 = −11.5°

(1.0 × 10 m) 600 (−1)(500 × 10 m) = − sin 30° = −0.80 ⇒ θ (1.0 × 10 m) 600 −3

−9

sin θ −1

−3

−1

= −53.1°


22.72. Visualize: Please refer to Figure CP22.72. Solve:

(a)

We have two incoming and two diffracted light rays at angles φ and θ and two wave fronts perpendicular to the rays. We can see from the figure that the wave 1 travels an extra distance ∆r = d sin φ to reach the reflection spot. Wave 2 travels an extra distance ∆r = d sin θ from the reflection spot to the outgoing wave front. The path difference between the two waves is

∆r = ∆r1 − ∆r2 = d (sin θ − sin φ ) (b) The condition for diffraction, with all the waves in phase, is still ∆r = mλ. Using the results from part (a), the diffraction condition is

d sin θ m = mλ + d sin φ

m = … −2, −1, 0, 1, 2, …

Negative values of m will give a different diffraction angle than the corresponding positive values. (c) The “zeroth order” diffraction from the reflection grating is m = 0. From the diffraction condition of part (b), this implies d sin θ 0 = d sin φ and hence θ 0 = φ . That is, the zeroth order diffraction obeys the law of reflection—the angle of reflection equals the angle of incidence. 1 (d) A 700 lines per millimeter grating has spacing d = 700 mm = 1.429 × 10 −6 m = 1429 nm . The diffraction angles are given by

 m(500 nm )   mλ  θ m = sin −1  + sin 40° + sin φ  = sin −1   d   1429 nm  M ≤−5 −4 −3 −2 −1 0 1 ≥2 (e)

θm not defined −49.2° −24.0° −3.3° 17.0° 40.0° 83.1° not defined


22.73. Model: Diffraction patterns from two objects can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. Visualize: Please refer to Figure CP22.73. Solve: (a) Using Equation 22.24 with the width be equal to the aperture diameter,

w= D=

2.44λL ⇒ D = 2.44λL = D

(2.44)(550 × 10 −7 m)(0.20 m) = 0.52 mm

(b) We can now use Equation 22.23 to find the angle between two distant sources that can be resolved. The angle is −9 λ 1.22(550 × 10 m ) = = 1.29 × 10 −3 rad = 0.074° D 0.52 × 10 −3 m (c) The distance that can be resolved is

α = 1.22

(1000 m)α = (1000 m)(1.29 × 10 −3 rad) = 1.3 m


Chapter 22