DPP for NEET: Daily Practice Problems, Ch 51: Ray Optics- 3 (Solutions) | Physics Class 12 PDF Download

 Page 1


1. (a).  According to sign convention, it is given that
u = – 0.6 m, R = 0.25 m, µ
1
 = 1 (air), µ
2
 = 1.5
Therefore, using
1 2 21
µ µ µµ
u vR
-
- +=
, we get
1.5
v
= 
1 1.51
( 0.6) 0.25
-
+
-
 = – 
1 0.5
0.6 0.25
+
= –
 
51
2
33
+=
  Þ v = 4.5 m
The image is formed on the other side of the object (i.e.
inside the refracting surface).
2. (a). On viewing from the closer surface A (near to object)
:
The final image is formed at I.
C O A I
·
C
B A
O
1cm
Nearing
surface
Far
surface
From sign convention
u = OA = – 4 cm, v = ?
R = AC = – 5cm
µ = 
2
1
µ 2
µ3
=
1 µ1
v uR
m-
-= Þ 
2/3 1 2/3 1
v 45
-
+=
-
Þ 
2 1 1 4 15 11
3v 15 4 60 60
--
=-==
v = 
2 60 40
3 11 11
-
´ =-  = – 3.63 cm
\ AI = 3.63 cm
3. (c).
u v 
f f x
1
x
2
F
1
F
2
O I
• • • • • 
As in case of thin lens the distance of either foci from the
optical centre is f ,
| u | = ( f + x
1 
) and | v | = ( f + x
2 
)
Substituting thses values of u and v in lens formula with
proper sign
( ) ( )
21
1 11
fx fx f
-=
+ -+
or  
( )( )
12
12
x x 2f 1
fx fx f
++
=
++
i.e.,  fx
1
 + fx
2
 + 2f 
2
 = f 
2
 + fx
1
+ fx
2
 + x
1
 x
2
or,  x
1
 x
2
 = f 
2
4. (b). According to Lens-maker’s formula :
1
f
 = (m – 1) 
12
11
RR
éù
-
êú
ëû
 with  m = 
L
M
m
m
Here f = 30 cm and R
1
 =10 cm and  R
2
  = ¥
So,  
1
30
 = ( m – 1 ) 
11
10
éù
-
êú
¥
ëû
3m – 3  = 1 or, m = ( 4 / 3 )
5. (a). For combination of lenses
 
12
1 1 1 1 1 251
f f f 10 15 150 6
= + = + ==
Therefore, f = 6 cm.
Page 2


1. (a).  According to sign convention, it is given that
u = – 0.6 m, R = 0.25 m, µ
1
 = 1 (air), µ
2
 = 1.5
Therefore, using
1 2 21
µ µ µµ
u vR
-
- +=
, we get
1.5
v
= 
1 1.51
( 0.6) 0.25
-
+
-
 = – 
1 0.5
0.6 0.25
+
= –
 
51
2
33
+=
  Þ v = 4.5 m
The image is formed on the other side of the object (i.e.
inside the refracting surface).
2. (a). On viewing from the closer surface A (near to object)
:
The final image is formed at I.
C O A I
·
C
B A
O
1cm
Nearing
surface
Far
surface
From sign convention
u = OA = – 4 cm, v = ?
R = AC = – 5cm
µ = 
2
1
µ 2
µ3
=
1 µ1
v uR
m-
-= Þ 
2/3 1 2/3 1
v 45
-
+=
-
Þ 
2 1 1 4 15 11
3v 15 4 60 60
--
=-==
v = 
2 60 40
3 11 11
-
´ =-  = – 3.63 cm
\ AI = 3.63 cm
3. (c).
u v 
f f x
1
x
2
F
1
F
2
O I
• • • • • 
As in case of thin lens the distance of either foci from the
optical centre is f ,
| u | = ( f + x
1 
) and | v | = ( f + x
2 
)
Substituting thses values of u and v in lens formula with
proper sign
( ) ( )
21
1 11
fx fx f
-=
+ -+
or  
( )( )
12
12
x x 2f 1
fx fx f
++
=
++
i.e.,  fx
1
 + fx
2
 + 2f 
2
 = f 
2
 + fx
1
+ fx
2
 + x
1
 x
2
or,  x
1
 x
2
 = f 
2
4. (b). According to Lens-maker’s formula :
1
f
 = (m – 1) 
12
11
RR
éù
-
êú
ëû
 with  m = 
L
M
m
m
Here f = 30 cm and R
1
 =10 cm and  R
2
  = ¥
So,  
1
30
 = ( m – 1 ) 
11
10
éù
-
êú
¥
ëû
3m – 3  = 1 or, m = ( 4 / 3 )
5. (a). For combination of lenses
 
12
1 1 1 1 1 251
f f f 10 15 150 6
= + = + ==
Therefore, f = 6 cm.
DPP/ P 51
142
6. (c). Let f
2
 is the focal length of the diverging lens. Then ,
12
1 11
f ff
=+
It is given that f
1
 = +20 cm, f = 30 cm
2
1 11
30 20 f
=+
or 
2
1 1 1 231
f 30 20 60 60
-
= - = =-
Thus f
2
 = – 60 cm
7. (a). As radius of curvature of silvered surface is 22 cm, so,
f
M
 = 
R 22
22
-
= = –11 cm = – 0.11 m
and hence, P
M
 = – 
M
1 11
D
f 0.11 0.11
==
-
Further as the focal length of lens is 20 cm, i.e. 0.20 m, its
power will be given by :
P
L
 = 
L
11
D
f 0.20
=
Now as in image formation, light after passing through the
lens will be reflected back by the curved mirror the lens
again  P = P
L
 + P
M
 + P
L
 = 2P
L
 + P
M
i.e.   P = 
2 1 210
D
0.20 0.11 11
+=
So the focal length of equivalent mirror
F = – 
1 11 110
m cm
P 210 21
=- =-
i.e. the silvered lens behaves as a concave mirror of focal
length (110/21) cm. So for object at a distance 10 cm in
front of it,
1 1 21
v 10 110
+ =-
-
i.e. v = –11 cm
i.e. image will be 11 cm in front of the silvered lens
8. (b). On covering the lens half by a black paper will reduce
the intensity of image and not the part of image. So full
image is formed.
9. (b). The focal length of lens in water is given by
ag
a
ag
a
1
ff
1
m-
=
m
-
m
l
l
 = 
1.21
1.2
1
1.33
-
-
f
a
 
a
0.2 1.33
ff
0.13
´
=-
l
Hence f is negative and as such it behaves as a divergent
lens.
10. (a). The focal length of an equiconvex lens is given by
1 2 ( 1)
fR
m-
=
It is given that 
1
f
 = + 5 and m = 1.5
Therefore, 5 = 
2 (1.5 1)
R
-
or  R = 
1
5
 metre = 20 cm
11. (b). The question is based on the conventional method of
measurement of focal length by displacement method.
According to this method where D is the distance between
object and the image, and x is the displacement given to
the object.
From the data x = 25 cm and D = 75 cm .
Thus
22
(75) (25)
f
4 75
-
=
´
   = 
(75 25) (75 25)
4 75
-+
´
   = 
50 100 50
4 753
´
=
´
= 16.7 cm
12. (c). m
1
 = 
1
A
O
 and m
2
 
2
A
O
Þ m
1
m
2
 = 
12
2
AA
O
Also it can be proved that m
1
m
2
 =  1
So O = 
12
AA
13. (c). Effective power P = P
1
 + P
2
 = 4 – 3 = 1D
14. (d). One part of combination will behave as converging
lens and the other as diverging lens of same focal length.
As such total power will be zero.
15. (c). Let the image of an object at O is formed at the same
point as shown in figure. The distance of O from the plane
surface is x. The rays suffer refraction at first surface
(curved) as they reach lens. After wards become parallel
and gets reflected form plane surface and so retrace the
path and image is formed at O itself.
Page 3


1. (a).  According to sign convention, it is given that
u = – 0.6 m, R = 0.25 m, µ
1
 = 1 (air), µ
2
 = 1.5
Therefore, using
1 2 21
µ µ µµ
u vR
-
- +=
, we get
1.5
v
= 
1 1.51
( 0.6) 0.25
-
+
-
 = – 
1 0.5
0.6 0.25
+
= –
 
51
2
33
+=
  Þ v = 4.5 m
The image is formed on the other side of the object (i.e.
inside the refracting surface).
2. (a). On viewing from the closer surface A (near to object)
:
The final image is formed at I.
C O A I
·
C
B A
O
1cm
Nearing
surface
Far
surface
From sign convention
u = OA = – 4 cm, v = ?
R = AC = – 5cm
µ = 
2
1
µ 2
µ3
=
1 µ1
v uR
m-
-= Þ 
2/3 1 2/3 1
v 45
-
+=
-
Þ 
2 1 1 4 15 11
3v 15 4 60 60
--
=-==
v = 
2 60 40
3 11 11
-
´ =-  = – 3.63 cm
\ AI = 3.63 cm
3. (c).
u v 
f f x
1
x
2
F
1
F
2
O I
• • • • • 
As in case of thin lens the distance of either foci from the
optical centre is f ,
| u | = ( f + x
1 
) and | v | = ( f + x
2 
)
Substituting thses values of u and v in lens formula with
proper sign
( ) ( )
21
1 11
fx fx f
-=
+ -+
or  
( )( )
12
12
x x 2f 1
fx fx f
++
=
++
i.e.,  fx
1
 + fx
2
 + 2f 
2
 = f 
2
 + fx
1
+ fx
2
 + x
1
 x
2
or,  x
1
 x
2
 = f 
2
4. (b). According to Lens-maker’s formula :
1
f
 = (m – 1) 
12
11
RR
éù
-
êú
ëû
 with  m = 
L
M
m
m
Here f = 30 cm and R
1
 =10 cm and  R
2
  = ¥
So,  
1
30
 = ( m – 1 ) 
11
10
éù
-
êú
¥
ëû
3m – 3  = 1 or, m = ( 4 / 3 )
5. (a). For combination of lenses
 
12
1 1 1 1 1 251
f f f 10 15 150 6
= + = + ==
Therefore, f = 6 cm.
DPP/ P 51
142
6. (c). Let f
2
 is the focal length of the diverging lens. Then ,
12
1 11
f ff
=+
It is given that f
1
 = +20 cm, f = 30 cm
2
1 11
30 20 f
=+
or 
2
1 1 1 231
f 30 20 60 60
-
= - = =-
Thus f
2
 = – 60 cm
7. (a). As radius of curvature of silvered surface is 22 cm, so,
f
M
 = 
R 22
22
-
= = –11 cm = – 0.11 m
and hence, P
M
 = – 
M
1 11
D
f 0.11 0.11
==
-
Further as the focal length of lens is 20 cm, i.e. 0.20 m, its
power will be given by :
P
L
 = 
L
11
D
f 0.20
=
Now as in image formation, light after passing through the
lens will be reflected back by the curved mirror the lens
again  P = P
L
 + P
M
 + P
L
 = 2P
L
 + P
M
i.e.   P = 
2 1 210
D
0.20 0.11 11
+=
So the focal length of equivalent mirror
F = – 
1 11 110
m cm
P 210 21
=- =-
i.e. the silvered lens behaves as a concave mirror of focal
length (110/21) cm. So for object at a distance 10 cm in
front of it,
1 1 21
v 10 110
+ =-
-
i.e. v = –11 cm
i.e. image will be 11 cm in front of the silvered lens
8. (b). On covering the lens half by a black paper will reduce
the intensity of image and not the part of image. So full
image is formed.
9. (b). The focal length of lens in water is given by
ag
a
ag
a
1
ff
1
m-
=
m
-
m
l
l
 = 
1.21
1.2
1
1.33
-
-
f
a
 
a
0.2 1.33
ff
0.13
´
=-
l
Hence f is negative and as such it behaves as a divergent
lens.
10. (a). The focal length of an equiconvex lens is given by
1 2 ( 1)
fR
m-
=
It is given that 
1
f
 = + 5 and m = 1.5
Therefore, 5 = 
2 (1.5 1)
R
-
or  R = 
1
5
 metre = 20 cm
11. (b). The question is based on the conventional method of
measurement of focal length by displacement method.
According to this method where D is the distance between
object and the image, and x is the displacement given to
the object.
From the data x = 25 cm and D = 75 cm .
Thus
22
(75) (25)
f
4 75
-
=
´
   = 
(75 25) (75 25)
4 75
-+
´
   = 
50 100 50
4 753
´
=
´
= 16.7 cm
12. (c). m
1
 = 
1
A
O
 and m
2
 
2
A
O
Þ m
1
m
2
 = 
12
2
AA
O
Also it can be proved that m
1
m
2
 =  1
So O = 
12
AA
13. (c). Effective power P = P
1
 + P
2
 = 4 – 3 = 1D
14. (d). One part of combination will behave as converging
lens and the other as diverging lens of same focal length.
As such total power will be zero.
15. (c). Let the image of an object at O is formed at the same
point as shown in figure. The distance of O from the plane
surface is x. The rays suffer refraction at first surface
(curved) as they reach lens. After wards become parallel
and gets reflected form plane surface and so retrace the
path and image is formed at O itself.
DPP/ P 51
143
·
O
x
11
v uR
m m-
-=
u = – x, v = ¥
11
xR
m m-
+=
¥
x = 
R
1 m-
As such O behaves as equivalent to centre of curvature of
equivalent concave mirror.
\ Radius = x = 
R
1 m-
16. (a). As here f = 10 cm and v = 5m = 500 cm
So from lens formula 
1 11
v uf
-= , we have
1 11
500 u 10
-= i.e., u = – 
500
49
éù
êú
ëû
cm
So that m = 
v 500
u 500 / 49
=
-
= – 49
Here negative sign means the image is inverted with respect
to object. Now as here object is (2 cm × 2cm ) so the size
of picture on the screen
A
i
 = ( 2 × 49 cm) × ( 2 × 49cm) = ( 98 × 98) cm
2
17. (a). As power of a lens is reciprocal of focal length in m,
P = 
2
1
5 10m
-
´
 = 
1
0.05
 diopter = 20 D
18. (b). For relaxed eye, MP is minimum and will be
MP = 
D 25
f5
= = 5×
While for strained eye, MP is maximum and will be
MP = 1 + 
D
f
 = 1 + 5 = 6×
19. (a). As in case of projector, m = 
Iv
Ou
=
So  – 
20 100 cm
3.5cm
´
 = 
40 100
u
´
i.e., u = – 7 cm
i.e., film is at a distance of 7 cm in front of projection lens.
And from lens formula 
1 11
v uf
-=
, here we have
1 11
4000 7 f
-=
-
 or f @ 7 cm = 70 mm
[as ( 1/4000) << (1/7) ]
i.e., focal length of projection lens is 70 mm.
20. (a). As final image is at 25 cm in front of eye piece
e
1 11
25 u5
-=
-
 i.e., u
e
 = – 
25
6
And so,  m
e
 = 
e
e
v 25
u ( 25 / 6)
-
=
-
 = 6 ...(1)
Now for objective,
v = L – u
e
 = 20 – (25/6) = (95/6)
So if object is at a distance u from the objective,
6 11
95 u 0.95
-= i.e., u = – 
95
94
cm
i.e. object is at a distance (95/94) cm in front of field lens.
Also, m = 
v (95 / 6) 94
u ( 95 / 94) 6
éù
= =-
êú
- ëû
...(2)
So total magnification,
M = m xm
e
 = – 
94
6
éù
êú
ëû
 × (6) = – 94
i.e., final image is inverted, virtual and 94 times that of
object.
21. (a). As object is distant, i.e., u = – 
¥
, so
0
1 11
vf
-=
-¥
 i.e. v = f
0
 = 12 cm
i.e. objective will form the image I
M
 at its focus which is
at a distance of 12 cm from O. Now as eye- piece of focal
length – 4 cm forms image I at a distance of 24 cm from it,
e
1 11
24u4
-=
--
   Þ u
e
 = 
24
5
 = 4.8 cm.
i.e, the distance of I
M
 from eye lens EA is 4.8 cm. So the
length of tube L = OA – EA = 12 – 4.8 = 7.2 cm.
Page 4


1. (a).  According to sign convention, it is given that
u = – 0.6 m, R = 0.25 m, µ
1
 = 1 (air), µ
2
 = 1.5
Therefore, using
1 2 21
µ µ µµ
u vR
-
- +=
, we get
1.5
v
= 
1 1.51
( 0.6) 0.25
-
+
-
 = – 
1 0.5
0.6 0.25
+
= –
 
51
2
33
+=
  Þ v = 4.5 m
The image is formed on the other side of the object (i.e.
inside the refracting surface).
2. (a). On viewing from the closer surface A (near to object)
:
The final image is formed at I.
C O A I
·
C
B A
O
1cm
Nearing
surface
Far
surface
From sign convention
u = OA = – 4 cm, v = ?
R = AC = – 5cm
µ = 
2
1
µ 2
µ3
=
1 µ1
v uR
m-
-= Þ 
2/3 1 2/3 1
v 45
-
+=
-
Þ 
2 1 1 4 15 11
3v 15 4 60 60
--
=-==
v = 
2 60 40
3 11 11
-
´ =-  = – 3.63 cm
\ AI = 3.63 cm
3. (c).
u v 
f f x
1
x
2
F
1
F
2
O I
• • • • • 
As in case of thin lens the distance of either foci from the
optical centre is f ,
| u | = ( f + x
1 
) and | v | = ( f + x
2 
)
Substituting thses values of u and v in lens formula with
proper sign
( ) ( )
21
1 11
fx fx f
-=
+ -+
or  
( )( )
12
12
x x 2f 1
fx fx f
++
=
++
i.e.,  fx
1
 + fx
2
 + 2f 
2
 = f 
2
 + fx
1
+ fx
2
 + x
1
 x
2
or,  x
1
 x
2
 = f 
2
4. (b). According to Lens-maker’s formula :
1
f
 = (m – 1) 
12
11
RR
éù
-
êú
ëû
 with  m = 
L
M
m
m
Here f = 30 cm and R
1
 =10 cm and  R
2
  = ¥
So,  
1
30
 = ( m – 1 ) 
11
10
éù
-
êú
¥
ëû
3m – 3  = 1 or, m = ( 4 / 3 )
5. (a). For combination of lenses
 
12
1 1 1 1 1 251
f f f 10 15 150 6
= + = + ==
Therefore, f = 6 cm.
DPP/ P 51
142
6. (c). Let f
2
 is the focal length of the diverging lens. Then ,
12
1 11
f ff
=+
It is given that f
1
 = +20 cm, f = 30 cm
2
1 11
30 20 f
=+
or 
2
1 1 1 231
f 30 20 60 60
-
= - = =-
Thus f
2
 = – 60 cm
7. (a). As radius of curvature of silvered surface is 22 cm, so,
f
M
 = 
R 22
22
-
= = –11 cm = – 0.11 m
and hence, P
M
 = – 
M
1 11
D
f 0.11 0.11
==
-
Further as the focal length of lens is 20 cm, i.e. 0.20 m, its
power will be given by :
P
L
 = 
L
11
D
f 0.20
=
Now as in image formation, light after passing through the
lens will be reflected back by the curved mirror the lens
again  P = P
L
 + P
M
 + P
L
 = 2P
L
 + P
M
i.e.   P = 
2 1 210
D
0.20 0.11 11
+=
So the focal length of equivalent mirror
F = – 
1 11 110
m cm
P 210 21
=- =-
i.e. the silvered lens behaves as a concave mirror of focal
length (110/21) cm. So for object at a distance 10 cm in
front of it,
1 1 21
v 10 110
+ =-
-
i.e. v = –11 cm
i.e. image will be 11 cm in front of the silvered lens
8. (b). On covering the lens half by a black paper will reduce
the intensity of image and not the part of image. So full
image is formed.
9. (b). The focal length of lens in water is given by
ag
a
ag
a
1
ff
1
m-
=
m
-
m
l
l
 = 
1.21
1.2
1
1.33
-
-
f
a
 
a
0.2 1.33
ff
0.13
´
=-
l
Hence f is negative and as such it behaves as a divergent
lens.
10. (a). The focal length of an equiconvex lens is given by
1 2 ( 1)
fR
m-
=
It is given that 
1
f
 = + 5 and m = 1.5
Therefore, 5 = 
2 (1.5 1)
R
-
or  R = 
1
5
 metre = 20 cm
11. (b). The question is based on the conventional method of
measurement of focal length by displacement method.
According to this method where D is the distance between
object and the image, and x is the displacement given to
the object.
From the data x = 25 cm and D = 75 cm .
Thus
22
(75) (25)
f
4 75
-
=
´
   = 
(75 25) (75 25)
4 75
-+
´
   = 
50 100 50
4 753
´
=
´
= 16.7 cm
12. (c). m
1
 = 
1
A
O
 and m
2
 
2
A
O
Þ m
1
m
2
 = 
12
2
AA
O
Also it can be proved that m
1
m
2
 =  1
So O = 
12
AA
13. (c). Effective power P = P
1
 + P
2
 = 4 – 3 = 1D
14. (d). One part of combination will behave as converging
lens and the other as diverging lens of same focal length.
As such total power will be zero.
15. (c). Let the image of an object at O is formed at the same
point as shown in figure. The distance of O from the plane
surface is x. The rays suffer refraction at first surface
(curved) as they reach lens. After wards become parallel
and gets reflected form plane surface and so retrace the
path and image is formed at O itself.
DPP/ P 51
143
·
O
x
11
v uR
m m-
-=
u = – x, v = ¥
11
xR
m m-
+=
¥
x = 
R
1 m-
As such O behaves as equivalent to centre of curvature of
equivalent concave mirror.
\ Radius = x = 
R
1 m-
16. (a). As here f = 10 cm and v = 5m = 500 cm
So from lens formula 
1 11
v uf
-= , we have
1 11
500 u 10
-= i.e., u = – 
500
49
éù
êú
ëû
cm
So that m = 
v 500
u 500 / 49
=
-
= – 49
Here negative sign means the image is inverted with respect
to object. Now as here object is (2 cm × 2cm ) so the size
of picture on the screen
A
i
 = ( 2 × 49 cm) × ( 2 × 49cm) = ( 98 × 98) cm
2
17. (a). As power of a lens is reciprocal of focal length in m,
P = 
2
1
5 10m
-
´
 = 
1
0.05
 diopter = 20 D
18. (b). For relaxed eye, MP is minimum and will be
MP = 
D 25
f5
= = 5×
While for strained eye, MP is maximum and will be
MP = 1 + 
D
f
 = 1 + 5 = 6×
19. (a). As in case of projector, m = 
Iv
Ou
=
So  – 
20 100 cm
3.5cm
´
 = 
40 100
u
´
i.e., u = – 7 cm
i.e., film is at a distance of 7 cm in front of projection lens.
And from lens formula 
1 11
v uf
-=
, here we have
1 11
4000 7 f
-=
-
 or f @ 7 cm = 70 mm
[as ( 1/4000) << (1/7) ]
i.e., focal length of projection lens is 70 mm.
20. (a). As final image is at 25 cm in front of eye piece
e
1 11
25 u5
-=
-
 i.e., u
e
 = – 
25
6
And so,  m
e
 = 
e
e
v 25
u ( 25 / 6)
-
=
-
 = 6 ...(1)
Now for objective,
v = L – u
e
 = 20 – (25/6) = (95/6)
So if object is at a distance u from the objective,
6 11
95 u 0.95
-= i.e., u = – 
95
94
cm
i.e. object is at a distance (95/94) cm in front of field lens.
Also, m = 
v (95 / 6) 94
u ( 95 / 94) 6
éù
= =-
êú
- ëû
...(2)
So total magnification,
M = m xm
e
 = – 
94
6
éù
êú
ëû
 × (6) = – 94
i.e., final image is inverted, virtual and 94 times that of
object.
21. (a). As object is distant, i.e., u = – 
¥
, so
0
1 11
vf
-=
-¥
 i.e. v = f
0
 = 12 cm
i.e. objective will form the image I
M
 at its focus which is
at a distance of 12 cm from O. Now as eye- piece of focal
length – 4 cm forms image I at a distance of 24 cm from it,
e
1 11
24u4
-=
--
   Þ u
e
 = 
24
5
 = 4.8 cm.
i.e, the distance of I
M
 from eye lens EA is 4.8 cm. So the
length of tube L = OA – EA = 12 – 4.8 = 7.2 cm.
DPP/ P 51
144
>
4.8 cm
u
e
A
I
M
B
L= 7.2 cm
24 cm
f
0
I
O
q
q
0
q
0
q
>
>>
>>
>
>>
f
e
f
0
=12 cm
E
)
)
( (
22. (d)  In case of astronomical telescope if object and final
image both are at infinity .
MP = – (f
0
/f
e
) and L = f
0
 + f
e
So here –(f
0
/f
e
) = – 5 and f
0
 + f
e
 = 36
Solving these for f
0
 and f
e, 
we get
f
0
 = 30 cm and f
e
 = 6 cm
23. (a) Resolving power of microscope 
1
µ
l
24. (a) Light gathering power µ Area of lens aperture or d
2
25. (b) For lens L
1
, ray must move parallel to the axis after
refraction
1 1
1
–
10
µ µµ µ
x cm
xR
ww
+ = Þ=
¥
26. (a) For lens L
2
, image must form at centre of curvature of
the curved surface after refraction through plane part
2
2
0
–'
µ µ
Rx
w
+=
Þ x¢ = 8 cm
27. (b) Length of tube = x + x¢ = 18 cm
28. (a) Focal length of lens immersed in water is four times
the focal length of lens in air. It means
f
w
 = 4f
a
 = 4×10 =40 cm
29. (d) Focal length of the lens depends upon its refractive
index as 
( )
1
1
f
µ m-
.
Since m
b
 > m
r 
so f
b
 < f
r
Therefore, the focal length of a lens decreases when red
light is replaced by blue light.
30. (b) The light gathering power (or brightness) of a telescope
µ
 (diameter)
2
. So by increasing the objective diameter even
far off stars may produce images of optimum brightness.