Page 1 22.1 SOLUTIONS TO CONCEPTS CHAPTER 22 1. Radiant Flux = Time emitted energy Total = s 15 45 = 3W 2. To get equally intense lines on the photographic plate, the radiant flux (energy) should be same. S0, 10W × 12sec = 12W × t ? t = W 12 sec 12 W 10 ? = 10 sec. ? 3. it can be found out from the graph by the student. 4. Relative luminousity = power same of nm 555 of source a of flux ous min Lu wavelength given of source a of flux ous min Lu Let the radiant flux needed be P watt. Ao, 0.6 = P 685 watt ' P ' source of flux ous min Lu ? Luminous flux of the source = (685 P)× 0.6 = 120 × 685 ? P = 6 . 0 120 = 200W 5. The luminous flux of the given source of 1W is 450 lumen/watt ? Relative luminosity = power same of source nm 555 of flux ous min Lu wavelength given of source the of flux ous min Lu = 685 450 = 66% [ ? Since, luminous flux of 555nm source of 1W = 685 lumen] 6. 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 (c) Luminous efficiency = flux radiant Total flux ous min lu Total = 70 39730 = 567.6 lumen/W 7. Overall luminous efficiency = input Power flux ous min lu Total = 100 685 35 ? = 239.75 lumen/W 8. Radiant flux = 31.4W, Solid angle = 4 ? Luminous efficiency = 60 lumen/W So, Luminous flux = 60 × 31.4 lumen And luminous intensity = ? 4 Flux ous min Lu = ? ? 4 4 . 31 60 = 150 candela 9. I = luminous intensity = ? 4 628 = 50 Candela r = 1m, ? = 37° So, illuminance, E = 2 r cos I ? = 2 1 37 cos 50 ? ? = 40 lux ? 10. Let, I = Luminous intensity of source E A = 900 lumen/m 2 E B = 400 lumen/m 2 Now, E a = 2 x cos I ? and E B = 2 ) 10 x ( cos I ? ? So, I = ? cos x E 2 A = ? ? cos ) 10 x ( E 2 B ? 900x 2 = 400(x + 10) 2 ? 10 x x ? = 3 2 ? 3x = 2x + 20 ? x = 20 cm So, The distance between the source and the original position is 20cm. 1m Source 37° Area Normal 10cm x O B A Page 2 22.1 SOLUTIONS TO CONCEPTS CHAPTER 22 1. Radiant Flux = Time emitted energy Total = s 15 45 = 3W 2. To get equally intense lines on the photographic plate, the radiant flux (energy) should be same. S0, 10W × 12sec = 12W × t ? t = W 12 sec 12 W 10 ? = 10 sec. ? 3. it can be found out from the graph by the student. 4. Relative luminousity = power same of nm 555 of source a of flux ous min Lu wavelength given of source a of flux ous min Lu Let the radiant flux needed be P watt. Ao, 0.6 = P 685 watt ' P ' source of flux ous min Lu ? Luminous flux of the source = (685 P)× 0.6 = 120 × 685 ? P = 6 . 0 120 = 200W 5. The luminous flux of the given source of 1W is 450 lumen/watt ? Relative luminosity = power same of source nm 555 of flux ous min Lu wavelength given of source the of flux ous min Lu = 685 450 = 66% [ ? Since, luminous flux of 555nm source of 1W = 685 lumen] 6. 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 (c) Luminous efficiency = flux radiant Total flux ous min lu Total = 70 39730 = 567.6 lumen/W 7. Overall luminous efficiency = input Power flux ous min lu Total = 100 685 35 ? = 239.75 lumen/W 8. Radiant flux = 31.4W, Solid angle = 4 ? Luminous efficiency = 60 lumen/W So, Luminous flux = 60 × 31.4 lumen And luminous intensity = ? 4 Flux ous min Lu = ? ? 4 4 . 31 60 = 150 candela 9. I = luminous intensity = ? 4 628 = 50 Candela r = 1m, ? = 37° So, illuminance, E = 2 r cos I ? = 2 1 37 cos 50 ? ? = 40 lux ? 10. Let, I = Luminous intensity of source E A = 900 lumen/m 2 E B = 400 lumen/m 2 Now, E a = 2 x cos I ? and E B = 2 ) 10 x ( cos I ? ? So, I = ? cos x E 2 A = ? ? cos ) 10 x ( E 2 B ? 900x 2 = 400(x + 10) 2 ? 10 x x ? = 3 2 ? 3x = 2x + 20 ? x = 20 cm So, The distance between the source and the original position is 20cm. 1m Source 37° Area Normal 10cm x O B A Chapter 22 22.2 11. Given that, E a = 15 lux = 2 0 60 I ? I 0 = 15 × (0.6) 2 = 5.4 candela So, E B = 2 0 ) OB ( cos I ? = 2 1 5 3 4 . 5 ? ? ? ? ? ? ? = 3.24 lux 12. The illuminance will not change. 13. Let the height of the source is ‘h’ and the luminous intensity in the normal direction is I 0 . So, illuminance at the book is given by, E = 2 0 r cos I ? = 3 0 r h I = 2 / 3 2 2 0 ) h r ( h I ? For maximum E, dh dE = 0 ? 3 2 2 ) 2 2 2 / 3 2 2 0 ) h R ( h 2 h R ( h 2 3 ) h R ( I 2 / 1 ? ? ? ? ? ? ? ? ? ? ? ? ? (R 2 + h 2 ) 1/2 [R 2 + h 2 – 3h 2 ] = 0 ? R 2 – 2h 2 = 0 ? h = 2 R ? ? ? ? ? ? ? 0.6m A 1m O B 0.8m r h ? RRead More

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