22.SOLUTIONS TO CONCEPTS

Page 1

SOLUTIONS TO CONCEPTS CHAPTER 22 1. 2.

3. 4.

5.

6.

Total energy emitted 45 = = 3W Time 15s To get equally intense lines on the photographic plate, the radiant flux (energy) should be same. S0, 10W × 12sec = 12W × t 10 W  12 sec t= = 10 sec. 12W it can be found out from the graph by the student. Lu min ous flux of a source of given wavelength Relative luminousity = Lu min ous flux of a source of 555 nm of same power Let the radiant flux needed be P watt. Lu min ous flux of source ' P' watt Ao, 0.6 = 685 P  Luminous flux of the source = (685 P)× 0.6 = 120 × 685 120 P= = 200W 0 .6 The luminous flux of the given source of 1W is 450 lumen/watt Lu min ous flux of the source of given wavelength 450  Relative luminosity = = = 66% Lu min ous flux of 555 nm source of same power 685 [ Since, luminous flux of 555nm source of 1W = 685 lumen] The radiant flux of 555nm part is 40W and of the 600nm part is 30W (a) Total radiant flux = 40W + 30W = 70W (b) Luminous flux = (L.Fllux)555nm + (L.Flux)600nm = 1 × 40× 685 + 0.6 × 30 × 685 = 39730 lumen Total lu min ous flux 39730 = = 567.6 lumen/W (c) Luminous efficiency = Total radiant flux 70 Radiant Flux =

Total lu min ous flux 35  685 = = 239.75 lumen/W Power input 100 8. Radiant flux = 31.4W, Solid angle = 4 Luminous efficiency = 60 lumen/W So, Luminous flux = 60 × 31.4 lumen Lu min ous Flux 60  31.4 = = 150 candela And luminous intensity = 4 4 628 9. I = luminous intensity = = 50 Candela 4 r = 1m,  = 37° Source I cos  50  cos 37 = = 40 lux  So, illuminance, E = r2 12 10. Let, I = Luminous intensity of source EA = 900 lumen/m2 2 EB = 400 lumen/m I cos  I cos  Now, Ea = and EB = x2 ( x  10)2 7.

Overall luminous efficiency =

E x2 E ( x  10)2 So, I = A = B cos  cos 

O

x 2  900x = 400(x + 10)  =  3x = 2x + 20  x = 20 cm x  10 3 So, The distance between the source and the original position is 20cm. 2

2

22.1

x

Normal 37° 1m

Area

B

A 10cm


Chapter 22 11. Given that, Ea = 15 lux =

I0

O

60 2 2  I0 = 15 × (0.6) = 5.4 candela



0.6m 3 5 .4    I0 cos  5  = = 3.24 lux So, EB = (OB)2 12 A 12. The illuminance will not change. 13. Let the height of the source is ‘h’ and the luminous intensity in the normal direction is I0. So, illuminance at the book is given by, Ih I cos  I h = 03 = 2 0 2 3 / 2 E= 0 2 r r (r  h ) 1/ 2 3   I0 (R 2  h 2 )3 / 2  h  (R 2  h 2)  2h dE 2  For maximum E, =0  dh (R 2  h 2 )3

2

2 1/2

2

2

2

 (R + h ) [R + h – 3h ] = 0 R 2 2  R – 2h = 0  h = 2

♠♠♠♠♠

22.2

1m

B

0.8m

h

 R

r


Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.