Chapter 22 : Photometry - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 12

JEE : Chapter 22 : Photometry - HC Verma Solution, Physics Class 11 Notes | EduRev

 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 ?
R
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of JEE

JEE

Dynamic Test

Content Category

Related Searches

practice quizzes

,

Exam

,

past year papers

,

Physics Class 11 Notes | EduRev

,

MCQs

,

mock tests for examination

,

video lectures

,

Chapter 22 : Photometry - HC Verma Solution

,

Summary

,

Physics Class 11 Notes | EduRev

,

Sample Paper

,

Semester Notes

,

Viva Questions

,

study material

,

Extra Questions

,

Important questions

,

pdf

,

Chapter 22 : Photometry - HC Verma Solution

,

Physics Class 11 Notes | EduRev

,

Previous Year Questions with Solutions

,

shortcuts and tricks

,

ppt

,

Chapter 22 : Photometry - HC Verma Solution

,

Objective type Questions

,

Free

;