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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 ?
R
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FAQs on HC Verma Solutions: Chapter 22 - Photometry - Physics Class 11 - NEET

1. What is photometry and why is it important in physics?
Ans. Photometry is the branch of physics that deals with the measurement of light and its properties, such as intensity and color. It is important in physics because light plays a crucial role in many physical phenomena and understanding its behavior is essential for various applications in fields such as optics, astronomy, and electronics.
2. How is photometry used in astronomy?
Ans. Photometry is extensively used in astronomy to study celestial objects and their properties. By measuring the intensity of light emitted by stars or other astronomical objects, photometry helps astronomers determine their distance, size, temperature, and composition. It also provides valuable information about the evolution and behavior of these objects.
3. What are the units used in photometry to measure light intensity?
Ans. The units commonly used in photometry to measure light intensity are the candela (cd) and the lumen (lm). The candela is the unit of luminous intensity, which measures the power of light emitted in a particular direction. The lumen is the unit of luminous flux, which measures the total amount of light emitted by a source.
4. How can photometry be used to measure the color of light?
Ans. Photometry can be used to measure the color of light by analyzing its spectral distribution. Different colors of light have different wavelengths, and by using filters or spectrometers, photometry can separate the light into its different wavelengths and measure their intensity. This information can then be used to determine the color of the light source.
5. What are some practical applications of photometry outside of physics?
Ans. Photometry finds applications in various fields outside of physics. In lighting design, photometry is used to measure and evaluate the performance of different light sources, ensuring optimal lighting conditions in buildings and outdoor spaces. In photography, photometry helps determine the correct exposure settings to capture well-balanced and properly lit images. Additionally, in environmental monitoring, photometry is used to measure light pollution levels and assess its impact on ecosystems and human health.
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