HC Verma Solutions: Chapter 18 - Geometrical Optics | Physics Class 11 - NEET PDF Download

 Page 1


18.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 18
SIGN CONVENTION :
1) The direction of incident ray (from object to the mirror or lens) is taken as positive direction.
2) All measurements are taken from pole (mirror) or optical centre (lens) as the case may be.
1. u = –30 cm, R = – 40 cm
From the mirror equation, 
1 1 2
v u R
? ?
?
1 2 1 2 1
v R u 40 30
? ? ? ?
? ?
= 
1
60
?
or, v = –60 cm
So, the image will be formed at a distance of 60 cm in front of the mirror.
2. Given that,
H
1
= 20 cm, v = –5 m = –500 cm, h
2
= 50 cm
Since, 
2
1
h v
u h
?
?
or 
500 50
u 20
? ? (because the image in inverted)
or u = 
500 2
5
?
? = –200 cm = – 2 m
1 1 1
v u f
? ? or 
1 1 1
5 2 f
? ?
? ?
or f = 
10
7
?
= –1.44 m
So, the focal length is 1.44 m.
3. For the concave mirror, f = –20 cm, M = –v/u = 2
? v = –2u
1
st
case 2
nd
case
1 1 1
v u f
? ?
1 1 1
2u u f
?
? ? ?
?
1 1 1
2u u f
? ? ? ?
3 1
2u f
?
? u = f/2 = 10 cm ? u = 3f/2 = 30 cm
? The positions are 10 cm or 30 cm from the concave mirror.
4. m = –v/u = 0.6 and f = 7.5 cm = 15/2 cm
From mirror equation,
?
1 1 1
v u f
? ? ?
1 1 1
0.6u u f
? ?
? u = 5 cm
5. Height of the object AB = 1.6 cm
Diameter of the ball bearing = d = 0.4 cm
? R = 0.2 cm
Given, u = 20 cm
We know, 
1 1 2
u v R
? ?
30cm ?
P ?
C ? S
– Sign convertion
+ve
? ?
40cm ?
P ?
A ?
B ?
F
– Sign convertion
+ve
? ?
h 1 ?
A ? ?
B ? ?
500cm ?
h 2 ?
Case II(Real) ?
P ?
A ?
B ?
A ? ?
B ? ?
A ?
B ?
A ? ?
B ? ?
Case I (Virtual) ?
C ?
P ?
– Sign convertion
+ve
? ?
20cm ?
0.2cm ?
Page 2


18.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 18
SIGN CONVENTION :
1) The direction of incident ray (from object to the mirror or lens) is taken as positive direction.
2) All measurements are taken from pole (mirror) or optical centre (lens) as the case may be.
1. u = –30 cm, R = – 40 cm
From the mirror equation, 
1 1 2
v u R
? ?
?
1 2 1 2 1
v R u 40 30
? ? ? ?
? ?
= 
1
60
?
or, v = –60 cm
So, the image will be formed at a distance of 60 cm in front of the mirror.
2. Given that,
H
1
= 20 cm, v = –5 m = –500 cm, h
2
= 50 cm
Since, 
2
1
h v
u h
?
?
or 
500 50
u 20
? ? (because the image in inverted)
or u = 
500 2
5
?
? = –200 cm = – 2 m
1 1 1
v u f
? ? or 
1 1 1
5 2 f
? ?
? ?
or f = 
10
7
?
= –1.44 m
So, the focal length is 1.44 m.
3. For the concave mirror, f = –20 cm, M = –v/u = 2
? v = –2u
1
st
case 2
nd
case
1 1 1
v u f
? ?
1 1 1
2u u f
?
? ? ?
?
1 1 1
2u u f
? ? ? ?
3 1
2u f
?
? u = f/2 = 10 cm ? u = 3f/2 = 30 cm
? The positions are 10 cm or 30 cm from the concave mirror.
4. m = –v/u = 0.6 and f = 7.5 cm = 15/2 cm
From mirror equation,
?
1 1 1
v u f
? ? ?
1 1 1
0.6u u f
? ?
? u = 5 cm
5. Height of the object AB = 1.6 cm
Diameter of the ball bearing = d = 0.4 cm
? R = 0.2 cm
Given, u = 20 cm
We know, 
1 1 2
u v R
? ?
30cm ?
P ?
C ? S
– Sign convertion
+ve
? ?
40cm ?
P ?
A ?
B ?
F
– Sign convertion
+ve
? ?
h 1 ?
A ? ?
B ? ?
500cm ?
h 2 ?
Case II(Real) ?
P ?
A ?
B ?
A ? ?
B ? ?
A ?
B ?
A ? ?
B ? ?
Case I (Virtual) ?
C ?
P ?
– Sign convertion
+ve
? ?
20cm ?
0.2cm ?
Chapter 18
18.2
Putting the values according to sign conventions 
1 1 2
20 v 0.2
? ?
?
?
1 1 201
10
v 20 20
? ? ? ? v = 0.1 cm = 1 mm inside the ball bearing.
Magnification = m = 
A B v 0.1 1
AB u 20 200
? ?
? ? ? ? ?
?
? A ?B ? = 
AB 16
200 200
? = +0.008 cm = +0.8 mm.
6. Given AB = 3 cm, u = –7.5 cm, f = 6 cm.
Using 
1 1 1
v u f
? ? ?
1 1 1
v f u
? ?
Putting values according to sign conventions, 
1 1 1 3
v 6 7.5 10
? ? ?
?
? v = 10/3 cm
? magnification = m = 
v 10
u 7.5 3
? ?
?
?
A B 10 100 4
A B 1.33
AB 7.5 3 72 3
? ?
? ? ? ? ? ? ?
?
cm.
? Image will form at a distance of 10/3 cm. From the pole and image is 1.33 cm (virtual and erect).
7. R = 20 cm, f = R/2 = –10 cm
For part AB, PB = 30 + 10 = 40 cm
So, u = –40 cm ?
1 1 1 1 1 3
v f u 10 40 40
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
40
3
? = –13.3 cm.
So, PB ? = 13.3 cm
m = 
A B v 13.3 1
AB u 40 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
?
? ? ? ?
? A ?B ? = –10/3 = –3.33 cm
For part CD, PC = 30, So, u = –30 cm
1 1 1 1 1 1
v f u 10 30 15
? ?
? ? ? ? ? ? ? ?
? ?
? ?
? v = –15 cm = PC ?
So, m = 
C D v 15 1
CD u 30 2
? ? ? ? ?
? ? ? ? ? ?
? ?
?
? ?
? C D ? ? = 5 cm
B ?C ? = PC ? – PB ? = 15 – 13.3 = 17 cm
So, total length A ?B ? + B ?C ? + C ?D ? = 3.3 + 1.7 + 5 = 10 cm.
8. u = –25 cm
m = 
A B v v 14 v
1.4
AB u 25 10 25
? ?
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
25 14
10
?
= 35 cm.
Now, 
1 1 1
v u f
? ?
?
1 1 1 5 7 2
f 35 25 175 175
? ? ?
? ? ? ? ?
? ?
?
? ?
? f = –87.5 cm.
So, focal length of the concave mirror is 87.5 cm.
6cm ?
F
+ve
? ?
A ?
B ?
C
7.5cm ?
10cm ?
A ? ?
B ? ?
C ?
P ?
A ?
B ?
D
30cm ?
D ? ?
C ? ?
Page 3


18.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 18
SIGN CONVENTION :
1) The direction of incident ray (from object to the mirror or lens) is taken as positive direction.
2) All measurements are taken from pole (mirror) or optical centre (lens) as the case may be.
1. u = –30 cm, R = – 40 cm
From the mirror equation, 
1 1 2
v u R
? ?
?
1 2 1 2 1
v R u 40 30
? ? ? ?
? ?
= 
1
60
?
or, v = –60 cm
So, the image will be formed at a distance of 60 cm in front of the mirror.
2. Given that,
H
1
= 20 cm, v = –5 m = –500 cm, h
2
= 50 cm
Since, 
2
1
h v
u h
?
?
or 
500 50
u 20
? ? (because the image in inverted)
or u = 
500 2
5
?
? = –200 cm = – 2 m
1 1 1
v u f
? ? or 
1 1 1
5 2 f
? ?
? ?
or f = 
10
7
?
= –1.44 m
So, the focal length is 1.44 m.
3. For the concave mirror, f = –20 cm, M = –v/u = 2
? v = –2u
1
st
case 2
nd
case
1 1 1
v u f
? ?
1 1 1
2u u f
?
? ? ?
?
1 1 1
2u u f
? ? ? ?
3 1
2u f
?
? u = f/2 = 10 cm ? u = 3f/2 = 30 cm
? The positions are 10 cm or 30 cm from the concave mirror.
4. m = –v/u = 0.6 and f = 7.5 cm = 15/2 cm
From mirror equation,
?
1 1 1
v u f
? ? ?
1 1 1
0.6u u f
? ?
? u = 5 cm
5. Height of the object AB = 1.6 cm
Diameter of the ball bearing = d = 0.4 cm
? R = 0.2 cm
Given, u = 20 cm
We know, 
1 1 2
u v R
? ?
30cm ?
P ?
C ? S
– Sign convertion
+ve
? ?
40cm ?
P ?
A ?
B ?
F
– Sign convertion
+ve
? ?
h 1 ?
A ? ?
B ? ?
500cm ?
h 2 ?
Case II(Real) ?
P ?
A ?
B ?
A ? ?
B ? ?
A ?
B ?
A ? ?
B ? ?
Case I (Virtual) ?
C ?
P ?
– Sign convertion
+ve
? ?
20cm ?
0.2cm ?
Chapter 18
18.2
Putting the values according to sign conventions 
1 1 2
20 v 0.2
? ?
?
?
1 1 201
10
v 20 20
? ? ? ? v = 0.1 cm = 1 mm inside the ball bearing.
Magnification = m = 
A B v 0.1 1
AB u 20 200
? ?
? ? ? ? ?
?
? A ?B ? = 
AB 16
200 200
? = +0.008 cm = +0.8 mm.
6. Given AB = 3 cm, u = –7.5 cm, f = 6 cm.
Using 
1 1 1
v u f
? ? ?
1 1 1
v f u
? ?
Putting values according to sign conventions, 
1 1 1 3
v 6 7.5 10
? ? ?
?
? v = 10/3 cm
? magnification = m = 
v 10
u 7.5 3
? ?
?
?
A B 10 100 4
A B 1.33
AB 7.5 3 72 3
? ?
? ? ? ? ? ? ?
?
cm.
? Image will form at a distance of 10/3 cm. From the pole and image is 1.33 cm (virtual and erect).
7. R = 20 cm, f = R/2 = –10 cm
For part AB, PB = 30 + 10 = 40 cm
So, u = –40 cm ?
1 1 1 1 1 3
v f u 10 40 40
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
40
3
? = –13.3 cm.
So, PB ? = 13.3 cm
m = 
A B v 13.3 1
AB u 40 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
?
? ? ? ?
? A ?B ? = –10/3 = –3.33 cm
For part CD, PC = 30, So, u = –30 cm
1 1 1 1 1 1
v f u 10 30 15
? ?
? ? ? ? ? ? ? ?
? ?
? ?
? v = –15 cm = PC ?
So, m = 
C D v 15 1
CD u 30 2
? ? ? ? ?
? ? ? ? ? ?
? ?
?
? ?
? C D ? ? = 5 cm
B ?C ? = PC ? – PB ? = 15 – 13.3 = 17 cm
So, total length A ?B ? + B ?C ? + C ?D ? = 3.3 + 1.7 + 5 = 10 cm.
8. u = –25 cm
m = 
A B v v 14 v
1.4
AB u 25 10 25
? ?
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
25 14
10
?
= 35 cm.
Now, 
1 1 1
v u f
? ?
?
1 1 1 5 7 2
f 35 25 175 175
? ? ?
? ? ? ? ?
? ?
?
? ?
? f = –87.5 cm.
So, focal length of the concave mirror is 87.5 cm.
6cm ?
F
+ve
? ?
A ?
B ?
C
7.5cm ?
10cm ?
A ? ?
B ? ?
C ?
P ?
A ?
B ?
D
30cm ?
D ? ?
C ? ?
Chapter 18
18.3
9. u = –3.8 ? 10
5
km
diameter of moon = 3450 km ; f = –7.6 m
?
1 1 1
v u f
? ? ?
5
1 1 1
v 7.6 3.8 10
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
Since, distance of moon from earth is very large as compared to focal 
length it can be taken as ?.
? Image will be formed at focus, which is inverted.
?
1 1
v 7.6
v 7.6
? ?
? ? ? ? ?
? ?
? ?
m.
m = 
image
object
d
v
u d
? ? ?
image
8 3
d
( 7.6)
( 3.8 10 ) 3450 10
? ?
?
? ? ?
d
image 
= 
3
8
3450 7.6 10
3.8 10
? ?
?
= 0.069 m = 6.9 cm.
10. u = –30 cm, f = –20 cm
We know, 
1 1 1
v u f
? ?
?
1 1 1
v 30 20
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? v = –60 cm.
Image of the circle is formed at a distance 60 cm in front of the mirror.
? m = 
image
object
R
v
u R
? ? ?
image
R
60
30 2
?
? ?
?
? R
image
= 4 cm
Radius of image of the circle is 4 cm.
11. Let the object be placed at a height x above the surface of the water.
The apparent position of the object with respect to mirror should be at the centre 
of curvature so that the image is formed at the same position.
Since, 
Real depth 1
Apparent depth
?
?
(with respect to mirror)
Now, 
x 1 R h
x
R h
?
? ? ?
? ? ?
.
12. Both the mirrors have equal focal length f.
They will produce one image under two conditions.
Case I : When the source is at distance ‘2f’ from each mirror i.e. the source is at 
centre of curvature of the mirrors, the image will be produced at the same point 
S. So, d = 2f + 2f = 4f.
Case II : When the source S is at distance ‘f’ from each mirror, the rays from the 
source after reflecting from one mirror will become parallel and so these parallel 
rays after the reflection from the other mirror the object itself. So, only sine 
image is formed.
Here, d = f + f = 2f.
13. As shown in figure, for 1
st
reflection in M
1
, u = –30 cm, f = –20 cm
?
1 1 1
v 30 20
? ? ?
?
? v = –60 cm.
So, for 2
nd
reflection in M
2
u = 60 – (30 + x) = 30 – x 
v = –x ; f = 20 cm
?
2
1 1 1
x 10x 600 0
30 x x 20
? ? ? ? ? ?
?
7.6cm ?
F
+ve
? ?
F
R
object
20cm ?
30cm ?
R
Image
C ?
R ?
O ?
h ?
x
(R–h) ?
S ?
2f ? 2f ?
S ?
f ? f ?
– for M 1 ?
30cm ?
S
+ve
? ?
+ve
? ?
– for M 2 ?
x ?
Page 4


18.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 18
SIGN CONVENTION :
1) The direction of incident ray (from object to the mirror or lens) is taken as positive direction.
2) All measurements are taken from pole (mirror) or optical centre (lens) as the case may be.
1. u = –30 cm, R = – 40 cm
From the mirror equation, 
1 1 2
v u R
? ?
?
1 2 1 2 1
v R u 40 30
? ? ? ?
? ?
= 
1
60
?
or, v = –60 cm
So, the image will be formed at a distance of 60 cm in front of the mirror.
2. Given that,
H
1
= 20 cm, v = –5 m = –500 cm, h
2
= 50 cm
Since, 
2
1
h v
u h
?
?
or 
500 50
u 20
? ? (because the image in inverted)
or u = 
500 2
5
?
? = –200 cm = – 2 m
1 1 1
v u f
? ? or 
1 1 1
5 2 f
? ?
? ?
or f = 
10
7
?
= –1.44 m
So, the focal length is 1.44 m.
3. For the concave mirror, f = –20 cm, M = –v/u = 2
? v = –2u
1
st
case 2
nd
case
1 1 1
v u f
? ?
1 1 1
2u u f
?
? ? ?
?
1 1 1
2u u f
? ? ? ?
3 1
2u f
?
? u = f/2 = 10 cm ? u = 3f/2 = 30 cm
? The positions are 10 cm or 30 cm from the concave mirror.
4. m = –v/u = 0.6 and f = 7.5 cm = 15/2 cm
From mirror equation,
?
1 1 1
v u f
? ? ?
1 1 1
0.6u u f
? ?
? u = 5 cm
5. Height of the object AB = 1.6 cm
Diameter of the ball bearing = d = 0.4 cm
? R = 0.2 cm
Given, u = 20 cm
We know, 
1 1 2
u v R
? ?
30cm ?
P ?
C ? S
– Sign convertion
+ve
? ?
40cm ?
P ?
A ?
B ?
F
– Sign convertion
+ve
? ?
h 1 ?
A ? ?
B ? ?
500cm ?
h 2 ?
Case II(Real) ?
P ?
A ?
B ?
A ? ?
B ? ?
A ?
B ?
A ? ?
B ? ?
Case I (Virtual) ?
C ?
P ?
– Sign convertion
+ve
? ?
20cm ?
0.2cm ?
Chapter 18
18.2
Putting the values according to sign conventions 
1 1 2
20 v 0.2
? ?
?
?
1 1 201
10
v 20 20
? ? ? ? v = 0.1 cm = 1 mm inside the ball bearing.
Magnification = m = 
A B v 0.1 1
AB u 20 200
? ?
? ? ? ? ?
?
? A ?B ? = 
AB 16
200 200
? = +0.008 cm = +0.8 mm.
6. Given AB = 3 cm, u = –7.5 cm, f = 6 cm.
Using 
1 1 1
v u f
? ? ?
1 1 1
v f u
? ?
Putting values according to sign conventions, 
1 1 1 3
v 6 7.5 10
? ? ?
?
? v = 10/3 cm
? magnification = m = 
v 10
u 7.5 3
? ?
?
?
A B 10 100 4
A B 1.33
AB 7.5 3 72 3
? ?
? ? ? ? ? ? ?
?
cm.
? Image will form at a distance of 10/3 cm. From the pole and image is 1.33 cm (virtual and erect).
7. R = 20 cm, f = R/2 = –10 cm
For part AB, PB = 30 + 10 = 40 cm
So, u = –40 cm ?
1 1 1 1 1 3
v f u 10 40 40
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
40
3
? = –13.3 cm.
So, PB ? = 13.3 cm
m = 
A B v 13.3 1
AB u 40 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
?
? ? ? ?
? A ?B ? = –10/3 = –3.33 cm
For part CD, PC = 30, So, u = –30 cm
1 1 1 1 1 1
v f u 10 30 15
? ?
? ? ? ? ? ? ? ?
? ?
? ?
? v = –15 cm = PC ?
So, m = 
C D v 15 1
CD u 30 2
? ? ? ? ?
? ? ? ? ? ?
? ?
?
? ?
? C D ? ? = 5 cm
B ?C ? = PC ? – PB ? = 15 – 13.3 = 17 cm
So, total length A ?B ? + B ?C ? + C ?D ? = 3.3 + 1.7 + 5 = 10 cm.
8. u = –25 cm
m = 
A B v v 14 v
1.4
AB u 25 10 25
? ?
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
25 14
10
?
= 35 cm.
Now, 
1 1 1
v u f
? ?
?
1 1 1 5 7 2
f 35 25 175 175
? ? ?
? ? ? ? ?
? ?
?
? ?
? f = –87.5 cm.
So, focal length of the concave mirror is 87.5 cm.
6cm ?
F
+ve
? ?
A ?
B ?
C
7.5cm ?
10cm ?
A ? ?
B ? ?
C ?
P ?
A ?
B ?
D
30cm ?
D ? ?
C ? ?
Chapter 18
18.3
9. u = –3.8 ? 10
5
km
diameter of moon = 3450 km ; f = –7.6 m
?
1 1 1
v u f
? ? ?
5
1 1 1
v 7.6 3.8 10
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
Since, distance of moon from earth is very large as compared to focal 
length it can be taken as ?.
? Image will be formed at focus, which is inverted.
?
1 1
v 7.6
v 7.6
? ?
? ? ? ? ?
? ?
? ?
m.
m = 
image
object
d
v
u d
? ? ?
image
8 3
d
( 7.6)
( 3.8 10 ) 3450 10
? ?
?
? ? ?
d
image 
= 
3
8
3450 7.6 10
3.8 10
? ?
?
= 0.069 m = 6.9 cm.
10. u = –30 cm, f = –20 cm
We know, 
1 1 1
v u f
? ?
?
1 1 1
v 30 20
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? v = –60 cm.
Image of the circle is formed at a distance 60 cm in front of the mirror.
? m = 
image
object
R
v
u R
? ? ?
image
R
60
30 2
?
? ?
?
? R
image
= 4 cm
Radius of image of the circle is 4 cm.
11. Let the object be placed at a height x above the surface of the water.
The apparent position of the object with respect to mirror should be at the centre 
of curvature so that the image is formed at the same position.
Since, 
Real depth 1
Apparent depth
?
?
(with respect to mirror)
Now, 
x 1 R h
x
R h
?
? ? ?
? ? ?
.
12. Both the mirrors have equal focal length f.
They will produce one image under two conditions.
Case I : When the source is at distance ‘2f’ from each mirror i.e. the source is at 
centre of curvature of the mirrors, the image will be produced at the same point 
S. So, d = 2f + 2f = 4f.
Case II : When the source S is at distance ‘f’ from each mirror, the rays from the 
source after reflecting from one mirror will become parallel and so these parallel 
rays after the reflection from the other mirror the object itself. So, only sine 
image is formed.
Here, d = f + f = 2f.
13. As shown in figure, for 1
st
reflection in M
1
, u = –30 cm, f = –20 cm
?
1 1 1
v 30 20
? ? ?
?
? v = –60 cm.
So, for 2
nd
reflection in M
2
u = 60 – (30 + x) = 30 – x 
v = –x ; f = 20 cm
?
2
1 1 1
x 10x 600 0
30 x x 20
? ? ? ? ? ?
?
7.6cm ?
F
+ve
? ?
F
R
object
20cm ?
30cm ?
R
Image
C ?
R ?
O ?
h ?
x
(R–h) ?
S ?
2f ? 2f ?
S ?
f ? f ?
– for M 1 ?
30cm ?
S
+ve
? ?
+ve
? ?
– for M 2 ?
x ?
Chapter 18
18.4
? x = 
10 50 40
2 2
?
? = 20 cm or –30 cm
? Total distance between the two lines is 20 + 30 = 50 cm.
14. We know, 
8
sin i 3 10 sin45
2
sin r v sin30
? ?
? ? ?
?
? v = 
8
3 10
2
?
m/sec.
Distance travelled by light in the slab is, 
x = 
1m 2
cos30 3
?
?
m
So, time taken = 
8
2 2
3 3 10
?
? ?
= 0.54 ? 10
–8
= 5.4 ? 10
–9
sec.
15. Shadow length = BA ? = BD + A ?D = 0.5 + 0.5 tan r
Now, 1.33 = 
sin45
sin r
?
? sin r = 0.53.
? cos r = 
2 2
1 sin r 1 (0.53) ? ? ? = 0.85
So, tan r = 0.6235
So, shadow length = (0.5) (1 + 0.6235) = 81.2 cm.
16. Height of the lake = 2.5 m
When the sun is just setting, ? is approximately = 90°
?
2
1
sin i
sin r
?
?
?
?
1 4 / 3 3
sinr
sin r 1 4
? ? ? ? r = 49°
As shown in the figure, x/2.5 = tan r = 1.15
? x = 2.5 ? 1.15 = 2.8 m. ?
17. The thickness of the glass is d = 2.1 cm and ? =1.5
Shift due to the glass slab 
?T = 
1 1
1 d 1 2.1
1.5
? ? ? ?
? ? ?
? ? ? ?
?
? ? ? ?
= 0.7 CM
So, the microscope should be shifted 0.70 cm to focus the object again. ?
18. Shift due to water ?t
w
= 
1 1
1 d 1 20
1.33
? ? ? ?
? ? ?
? ? ? ?
?
? ? ? ?
= 5 cm
Shift due to oil, ?t
o
= 
1
1 20
1.3
? ?
?
? ?
? ?
= 4.6 cm
Total shift ?t = 5 + 4.6 = 9.6 cm
Apparent depth = 40 – (9.6) = 30.4 cm below the surface. ?
19. The presence of air medium in between the sheets does not affect the shift.
The shift will be due to 3 sheets of different refractive index other than air.
= 
1 1 1
1 (0.2) 1 (0.3) 1 (0.4)
1.2 13 14
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
= 0.2 cm above point P.
20. Total no. of slabs = k, thickness = t
1
, t
2
, t
3
… t
k
Refractive index = ?
1
, ??
2
, ??
3
, ??
4
,… ??
k
? The shift ?t = 
1 2 k
1 2 k
1 1 1
1 t 1 t ...... 1 t
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
…(1)
If, ? ? refractive index of combination of slabs and image is formed at same place,
?t = 
1 2 k
1
1 (t t ... t )
? ?
? ? ? ?
? ?
?
? ?
…(2)
45° ?
x ?
30° ?
1m ?
45° ?
A ?
r ?
0.5m
0.5m ?
0.5m ?
A ? ? B ? D ?
45° ?
x ?
O
i=90° ?
2.5m ?
w ?
w ?
Oil
20cm ?
Water
20cm ?
t = 0.4 cm
P ?
? ?= 1.4 ?
1 cm ?
??= 1.2 ?
??= 1.3 ?
t = 0.3 cm
t = 0.2 cm
1 cm ?
? ?
Page 5


18.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 18
SIGN CONVENTION :
1) The direction of incident ray (from object to the mirror or lens) is taken as positive direction.
2) All measurements are taken from pole (mirror) or optical centre (lens) as the case may be.
1. u = –30 cm, R = – 40 cm
From the mirror equation, 
1 1 2
v u R
? ?
?
1 2 1 2 1
v R u 40 30
? ? ? ?
? ?
= 
1
60
?
or, v = –60 cm
So, the image will be formed at a distance of 60 cm in front of the mirror.
2. Given that,
H
1
= 20 cm, v = –5 m = –500 cm, h
2
= 50 cm
Since, 
2
1
h v
u h
?
?
or 
500 50
u 20
? ? (because the image in inverted)
or u = 
500 2
5
?
? = –200 cm = – 2 m
1 1 1
v u f
? ? or 
1 1 1
5 2 f
? ?
? ?
or f = 
10
7
?
= –1.44 m
So, the focal length is 1.44 m.
3. For the concave mirror, f = –20 cm, M = –v/u = 2
? v = –2u
1
st
case 2
nd
case
1 1 1
v u f
? ?
1 1 1
2u u f
?
? ? ?
?
1 1 1
2u u f
? ? ? ?
3 1
2u f
?
? u = f/2 = 10 cm ? u = 3f/2 = 30 cm
? The positions are 10 cm or 30 cm from the concave mirror.
4. m = –v/u = 0.6 and f = 7.5 cm = 15/2 cm
From mirror equation,
?
1 1 1
v u f
? ? ?
1 1 1
0.6u u f
? ?
? u = 5 cm
5. Height of the object AB = 1.6 cm
Diameter of the ball bearing = d = 0.4 cm
? R = 0.2 cm
Given, u = 20 cm
We know, 
1 1 2
u v R
? ?
30cm ?
P ?
C ? S
– Sign convertion
+ve
? ?
40cm ?
P ?
A ?
B ?
F
– Sign convertion
+ve
? ?
h 1 ?
A ? ?
B ? ?
500cm ?
h 2 ?
Case II(Real) ?
P ?
A ?
B ?
A ? ?
B ? ?
A ?
B ?
A ? ?
B ? ?
Case I (Virtual) ?
C ?
P ?
– Sign convertion
+ve
? ?
20cm ?
0.2cm ?
Chapter 18
18.2
Putting the values according to sign conventions 
1 1 2
20 v 0.2
? ?
?
?
1 1 201
10
v 20 20
? ? ? ? v = 0.1 cm = 1 mm inside the ball bearing.
Magnification = m = 
A B v 0.1 1
AB u 20 200
? ?
? ? ? ? ?
?
? A ?B ? = 
AB 16
200 200
? = +0.008 cm = +0.8 mm.
6. Given AB = 3 cm, u = –7.5 cm, f = 6 cm.
Using 
1 1 1
v u f
? ? ?
1 1 1
v f u
? ?
Putting values according to sign conventions, 
1 1 1 3
v 6 7.5 10
? ? ?
?
? v = 10/3 cm
? magnification = m = 
v 10
u 7.5 3
? ?
?
?
A B 10 100 4
A B 1.33
AB 7.5 3 72 3
? ?
? ? ? ? ? ? ?
?
cm.
? Image will form at a distance of 10/3 cm. From the pole and image is 1.33 cm (virtual and erect).
7. R = 20 cm, f = R/2 = –10 cm
For part AB, PB = 30 + 10 = 40 cm
So, u = –40 cm ?
1 1 1 1 1 3
v f u 10 40 40
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
40
3
? = –13.3 cm.
So, PB ? = 13.3 cm
m = 
A B v 13.3 1
AB u 40 3
? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
?
? ? ? ?
? A ?B ? = –10/3 = –3.33 cm
For part CD, PC = 30, So, u = –30 cm
1 1 1 1 1 1
v f u 10 30 15
? ?
? ? ? ? ? ? ? ?
? ?
? ?
? v = –15 cm = PC ?
So, m = 
C D v 15 1
CD u 30 2
? ? ? ? ?
? ? ? ? ? ?
? ?
?
? ?
? C D ? ? = 5 cm
B ?C ? = PC ? – PB ? = 15 – 13.3 = 17 cm
So, total length A ?B ? + B ?C ? + C ?D ? = 3.3 + 1.7 + 5 = 10 cm.
8. u = –25 cm
m = 
A B v v 14 v
1.4
AB u 25 10 25
? ?
? ?
? ? ? ? ? ? ?
? ?
?
? ?
? v = 
25 14
10
?
= 35 cm.
Now, 
1 1 1
v u f
? ?
?
1 1 1 5 7 2
f 35 25 175 175
? ? ?
? ? ? ? ?
? ?
?
? ?
? f = –87.5 cm.
So, focal length of the concave mirror is 87.5 cm.
6cm ?
F
+ve
? ?
A ?
B ?
C
7.5cm ?
10cm ?
A ? ?
B ? ?
C ?
P ?
A ?
B ?
D
30cm ?
D ? ?
C ? ?
Chapter 18
18.3
9. u = –3.8 ? 10
5
km
diameter of moon = 3450 km ; f = –7.6 m
?
1 1 1
v u f
? ? ?
5
1 1 1
v 7.6 3.8 10
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ?
Since, distance of moon from earth is very large as compared to focal 
length it can be taken as ?.
? Image will be formed at focus, which is inverted.
?
1 1
v 7.6
v 7.6
? ?
? ? ? ? ?
? ?
? ?
m.
m = 
image
object
d
v
u d
? ? ?
image
8 3
d
( 7.6)
( 3.8 10 ) 3450 10
? ?
?
? ? ?
d
image 
= 
3
8
3450 7.6 10
3.8 10
? ?
?
= 0.069 m = 6.9 cm.
10. u = –30 cm, f = –20 cm
We know, 
1 1 1
v u f
? ?
?
1 1 1
v 30 20
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? v = –60 cm.
Image of the circle is formed at a distance 60 cm in front of the mirror.
? m = 
image
object
R
v
u R
? ? ?
image
R
60
30 2
?
? ?
?
? R
image
= 4 cm
Radius of image of the circle is 4 cm.
11. Let the object be placed at a height x above the surface of the water.
The apparent position of the object with respect to mirror should be at the centre 
of curvature so that the image is formed at the same position.
Since, 
Real depth 1
Apparent depth
?
?
(with respect to mirror)
Now, 
x 1 R h
x
R h
?
? ? ?
? ? ?
.
12. Both the mirrors have equal focal length f.
They will produce one image under two conditions.
Case I : When the source is at distance ‘2f’ from each mirror i.e. the source is at 
centre of curvature of the mirrors, the image will be produced at the same point 
S. So, d = 2f + 2f = 4f.
Case II : When the source S is at distance ‘f’ from each mirror, the rays from the 
source after reflecting from one mirror will become parallel and so these parallel 
rays after the reflection from the other mirror the object itself. So, only sine 
image is formed.
Here, d = f + f = 2f.
13. As shown in figure, for 1
st
reflection in M
1
, u = –30 cm, f = –20 cm
?
1 1 1
v 30 20
? ? ?
?
? v = –60 cm.
So, for 2
nd
reflection in M
2
u = 60 – (30 + x) = 30 – x 
v = –x ; f = 20 cm
?
2
1 1 1
x 10x 600 0
30 x x 20
? ? ? ? ? ?
?
7.6cm ?
F
+ve
? ?
F
R
object
20cm ?
30cm ?
R
Image
C ?
R ?
O ?
h ?
x
(R–h) ?
S ?
2f ? 2f ?
S ?
f ? f ?
– for M 1 ?
30cm ?
S
+ve
? ?
+ve
? ?
– for M 2 ?
x ?
Chapter 18
18.4
? x = 
10 50 40
2 2
?
? = 20 cm or –30 cm
? Total distance between the two lines is 20 + 30 = 50 cm.
14. We know, 
8
sin i 3 10 sin45
2
sin r v sin30
? ?
? ? ?
?
? v = 
8
3 10
2
?
m/sec.
Distance travelled by light in the slab is, 
x = 
1m 2
cos30 3
?
?
m
So, time taken = 
8
2 2
3 3 10
?
? ?
= 0.54 ? 10
–8
= 5.4 ? 10
–9
sec.
15. Shadow length = BA ? = BD + A ?D = 0.5 + 0.5 tan r
Now, 1.33 = 
sin45
sin r
?
? sin r = 0.53.
? cos r = 
2 2
1 sin r 1 (0.53) ? ? ? = 0.85
So, tan r = 0.6235
So, shadow length = (0.5) (1 + 0.6235) = 81.2 cm.
16. Height of the lake = 2.5 m
When the sun is just setting, ? is approximately = 90°
?
2
1
sin i
sin r
?
?
?
?
1 4 / 3 3
sinr
sin r 1 4
? ? ? ? r = 49°
As shown in the figure, x/2.5 = tan r = 1.15
? x = 2.5 ? 1.15 = 2.8 m. ?
17. The thickness of the glass is d = 2.1 cm and ? =1.5
Shift due to the glass slab 
?T = 
1 1
1 d 1 2.1
1.5
? ? ? ?
? ? ?
? ? ? ?
?
? ? ? ?
= 0.7 CM
So, the microscope should be shifted 0.70 cm to focus the object again. ?
18. Shift due to water ?t
w
= 
1 1
1 d 1 20
1.33
? ? ? ?
? ? ?
? ? ? ?
?
? ? ? ?
= 5 cm
Shift due to oil, ?t
o
= 
1
1 20
1.3
? ?
?
? ?
? ?
= 4.6 cm
Total shift ?t = 5 + 4.6 = 9.6 cm
Apparent depth = 40 – (9.6) = 30.4 cm below the surface. ?
19. The presence of air medium in between the sheets does not affect the shift.
The shift will be due to 3 sheets of different refractive index other than air.
= 
1 1 1
1 (0.2) 1 (0.3) 1 (0.4)
1.2 13 14
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
= 0.2 cm above point P.
20. Total no. of slabs = k, thickness = t
1
, t
2
, t
3
… t
k
Refractive index = ?
1
, ??
2
, ??
3
, ??
4
,… ??
k
? The shift ?t = 
1 2 k
1 2 k
1 1 1
1 t 1 t ...... 1 t
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
…(1)
If, ? ? refractive index of combination of slabs and image is formed at same place,
?t = 
1 2 k
1
1 (t t ... t )
? ?
? ? ? ?
? ?
?
? ?
…(2)
45° ?
x ?
30° ?
1m ?
45° ?
A ?
r ?
0.5m
0.5m ?
0.5m ?
A ? ? B ? D ?
45° ?
x ?
O
i=90° ?
2.5m ?
w ?
w ?
Oil
20cm ?
Water
20cm ?
t = 0.4 cm
P ?
? ?= 1.4 ?
1 cm ?
??= 1.2 ?
??= 1.3 ?
t = 0.3 cm
t = 0.2 cm
1 cm ?
? ?
Chapter 18
18.5
Equation (1) and (2), we get,
?
1 2 k
1
1 (t t ... t )
? ?
? ? ? ?
? ?
?
? ?
???
1 2 k
1 2 k
1 1 1
1 t 1 t ...... 1 t
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
= 
1 2 k
1 2 k
1 2 k
t t t
(t t ... t ) ...
? ?
? ? ? ? ? ? ?
? ?
? ? ?
? ?
= 
k
i
k k
1 i 1
1
k
1 i 1 i 1
1 1
i 1
t
t 1
t
(t / )
?
? ?
?
? ?
? ? ? ? ? ?
? ?
? ?
? ?
?
?
? ?
?
. ?
21. Given r = 6 cm, r
1
= 4 cm, h
1
= 8 cm
Let, h = final height of water column.
The volume of the cylindrical water column after the glass piece is put will be,
?r
2
h = 800 ? + ?r
1
2
h
1
or r
2
h = 800 + r
1
2
h
1
or 6
2
h = 800 + 4
2
? 8 = 25.7 cm
There are two shifts due to glass block as well as water.
So, ?t
1
= 
0
0
1
1 t
? ?
?
? ?
?
? ?
= 
1
1 8
3 / 2
? ?
?
? ?
? ?
= 2.26 cm
And, ?t
2
= 
w
w
1 1
1 t 1 (25.7 8)
4 / 3
? ? ? ?
? ? ? ?
? ?
? ?
? ? ?
? ?
= 4.44 cm.
Total shift = (2.66 + 4.44) cm = 7.1 cm above the bottom. ?
22. a) Let x = distance of the image of the eye formed above the surface as seen by the fish
So, 
H Real depth 1
x Apparent depth
? ?
?
or x = ?H
So, distance of the direct image = 
H 1
H H( )
2 2
? ? ? ? ?
Similarly, image through mirror = 
H 3H 3
(H x) H H( )
2 2 2
? ? ? ? ? ? ? ?
b) Here, 
H/ 2
y
= ?, so, y = 
H
2 ?
Where, y = distance of the image of fish below the surface as seen by eye.
So, Direct image = H + y = 
H 1
H H 1
2 2
? ?
? ? ?
? ?
? ?
? ?
Again another image of fish will be formed H/2 below the mirror.
So, the real depth for that image of fish becomes H + H/2 = 3H/2
So, Apparent depth from the surface of water = 3H/2 ?
So, distance of the image from the eye = 
3H 3
H H(1 )
2 2
? ? ?
? ?
. ?
23. According to the figure, x/3 = cot r …(1)
Again, 
sini 1 3
sinr 1.33 4
? ?
? sin r = 
4 4 3 4
sini
3 3 5 5
? ? ? (because sin i = 
BC 3
AC 5
? )
? cot r = 3/4 …(2)
From (1) and (2) ? x/3 = ¾
? x = 9/4 = 2.25 cm.
? Ratio of real and apparent depth = 4 : (2.25) = 1.78.
h ?
h–h 1 ?
Glass ?
h 1
8cm 
Water ?
8cm 
12cm ?
? ?
y
x
?
S ?
H
? ?
?
? ?
? ?
H/2
H/2
90° ?
r ?
4cm ?
A ?
? ?
? ?
r ?
B ?
D ?
3cm ?
i ?
C ?