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 Page 1


19.1
SOLUTIONS TO CONCEPTS 
CHAPTER 19
1. The visual angles made by the tree with the eyes can be 
calculated be below.
? = 
A
Height of the tree AB 2
0.04
Distance from the eye OB 50
? ? ? ? ?
similarly, ?
B
= 2.5 / 80 = 0.03125
?
C
= 1.8 / 70 = 0.02571
?
D
= 2.8 / 100 = 0.028
Since, ?
A
> ??
B
> ??
D
> ??
C
, the arrangement in decreasing order is given by A, B, D and C. ?
2. For the given simple microscope, 
f = 12 cm and D = 25 cm
For maximum angular magnification, the image should be produced 
at least distance of clear vision.
So, v = – D = –25 cm
Now, 
1 1 1
v u f
? ?
?
1 1 1 1 1 37
u v f 25 12 300
? ? ? ? ? ?
?
? u = –8.1 cm
So, the object should be placed 8.1 cm away from the lens.
3. The simple microscope has, m = 3, when image is formed at D = 25 cm
a) m = 
D
1
f
? ?
25
3 1
f
? ?
? f = 25/2 = 12.5 cm
b) When the image is formed at infinity (normal adjustment)
Magnifying power = 
D 25
f 12.5
? = 2.0
4. The child has D = 10 cm and f = 10 cm
The maximum angular magnification is obtained when the image is formed at near point.
m = 
D 10
1 1
f 10
? ? ? = 1 + 1 = 2
5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm).
Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, m = 5 = 
D 25
f f
? ? f = 5 cm
For the relaxed farsighted eye, D = 40 cm
So, m = 
D 40
f 5
? = 8
So, its magnifying power is 8X.
A ?
A
+ve
B
B ?
D=25cm
(Simple Microscope)
Distance 
A
height
B 
? ?
Page 2


19.1
SOLUTIONS TO CONCEPTS 
CHAPTER 19
1. The visual angles made by the tree with the eyes can be 
calculated be below.
? = 
A
Height of the tree AB 2
0.04
Distance from the eye OB 50
? ? ? ? ?
similarly, ?
B
= 2.5 / 80 = 0.03125
?
C
= 1.8 / 70 = 0.02571
?
D
= 2.8 / 100 = 0.028
Since, ?
A
> ??
B
> ??
D
> ??
C
, the arrangement in decreasing order is given by A, B, D and C. ?
2. For the given simple microscope, 
f = 12 cm and D = 25 cm
For maximum angular magnification, the image should be produced 
at least distance of clear vision.
So, v = – D = –25 cm
Now, 
1 1 1
v u f
? ?
?
1 1 1 1 1 37
u v f 25 12 300
? ? ? ? ? ?
?
? u = –8.1 cm
So, the object should be placed 8.1 cm away from the lens.
3. The simple microscope has, m = 3, when image is formed at D = 25 cm
a) m = 
D
1
f
? ?
25
3 1
f
? ?
? f = 25/2 = 12.5 cm
b) When the image is formed at infinity (normal adjustment)
Magnifying power = 
D 25
f 12.5
? = 2.0
4. The child has D = 10 cm and f = 10 cm
The maximum angular magnification is obtained when the image is formed at near point.
m = 
D 10
1 1
f 10
? ? ? = 1 + 1 = 2
5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm).
Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, m = 5 = 
D 25
f f
? ? f = 5 cm
For the relaxed farsighted eye, D = 40 cm
So, m = 
D 40
f 5
? = 8
So, its magnifying power is 8X.
A ?
A
+ve
B
B ?
D=25cm
(Simple Microscope)
Distance 
A
height
B 
? ?
Chapter 19
19.2
6. For the given compound microscope
f
0
= 
1
25 diopter
= 0.04 m = 4 cm, f
e
= 
1
5 diopter
= 0.2 m = 20 cm
D = 25 cm, separation between objective and eyepiece = 30 cm
The magnifying power is maximum when the image is formed by 
the eye piece at least distance of clear vision i.e. D = 25 cm
for the eye piece, v
e
= –25 cm, f
e
= 20 cm
For lens formula, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1
u v f 25 20
? ? ? ?
?
? u
e
= 11.11 cm
So, for the objective lens, the image distance should be 
v
0
= 30 – (11.11) = 18.89 cm
Now, for the objective lens,
v
0
= +18.89 cm (because real image is produced)
f
0 
= 4 cm
So, 
o o o
1 1 1 1 1
u v f 18.89 4
? ? ? ? = 0.053 – 0.25 = –0.197
? u
o
= –5.07 cm
So, the maximum magnificent power is given by 
m = 
o
o e
v D 18.89 25
1 1
u f 5.07 20
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ?
? ?
= 3.7225 ? 2.25 = 8.376  
7. For the given compound microscope
f
o
= 1 cm, f
e
= 6 cm, D = 24 cm
For the eye piece, v
e
= –24 cm, f
e
= 6 cm
Now, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1 5
u v f 24 6 24
? ?
? ? ? ? ? ? ?
? ?
? ?
? u
e
= –4.8 cm
a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective 
lens must be (9.8) – (4.8) = 5.0 cm
Now,  
0 0
u
1
v
1
? = 
0
f
1
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
5
1
? = –
5
4
? u
0
= –
4
5
= – 1.25 cm
So, the magnifying power is given by,
m = 
?
?
?
?
?
?
?
f
D
1
u
v
o
0
= 
?
?
?
?
?
?
?
?
?
6
24
1
25 . 1
5
= 4 × 5 = 20
(b) When the separation is 11.8 cm,
v
0
= 11.8 – 4.8 = 7.0 cm, f
0
= 1 cm 
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
7
1
? = 
7
6
?
f o = 0.04m 
objective
A ?
B
A
B ? B ? ?
11.11
cm
25cm
30cm
A ??
f e = 0.2m 
objective
Fig-A
5 cm ?
9.8cm ?
4.8cm ?
24 cm ?
Fig-B
7 cm ?
11.8cm ?
4.8cm ?
24 cm ?
Page 3


19.1
SOLUTIONS TO CONCEPTS 
CHAPTER 19
1. The visual angles made by the tree with the eyes can be 
calculated be below.
? = 
A
Height of the tree AB 2
0.04
Distance from the eye OB 50
? ? ? ? ?
similarly, ?
B
= 2.5 / 80 = 0.03125
?
C
= 1.8 / 70 = 0.02571
?
D
= 2.8 / 100 = 0.028
Since, ?
A
> ??
B
> ??
D
> ??
C
, the arrangement in decreasing order is given by A, B, D and C. ?
2. For the given simple microscope, 
f = 12 cm and D = 25 cm
For maximum angular magnification, the image should be produced 
at least distance of clear vision.
So, v = – D = –25 cm
Now, 
1 1 1
v u f
? ?
?
1 1 1 1 1 37
u v f 25 12 300
? ? ? ? ? ?
?
? u = –8.1 cm
So, the object should be placed 8.1 cm away from the lens.
3. The simple microscope has, m = 3, when image is formed at D = 25 cm
a) m = 
D
1
f
? ?
25
3 1
f
? ?
? f = 25/2 = 12.5 cm
b) When the image is formed at infinity (normal adjustment)
Magnifying power = 
D 25
f 12.5
? = 2.0
4. The child has D = 10 cm and f = 10 cm
The maximum angular magnification is obtained when the image is formed at near point.
m = 
D 10
1 1
f 10
? ? ? = 1 + 1 = 2
5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm).
Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, m = 5 = 
D 25
f f
? ? f = 5 cm
For the relaxed farsighted eye, D = 40 cm
So, m = 
D 40
f 5
? = 8
So, its magnifying power is 8X.
A ?
A
+ve
B
B ?
D=25cm
(Simple Microscope)
Distance 
A
height
B 
? ?
Chapter 19
19.2
6. For the given compound microscope
f
0
= 
1
25 diopter
= 0.04 m = 4 cm, f
e
= 
1
5 diopter
= 0.2 m = 20 cm
D = 25 cm, separation between objective and eyepiece = 30 cm
The magnifying power is maximum when the image is formed by 
the eye piece at least distance of clear vision i.e. D = 25 cm
for the eye piece, v
e
= –25 cm, f
e
= 20 cm
For lens formula, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1
u v f 25 20
? ? ? ?
?
? u
e
= 11.11 cm
So, for the objective lens, the image distance should be 
v
0
= 30 – (11.11) = 18.89 cm
Now, for the objective lens,
v
0
= +18.89 cm (because real image is produced)
f
0 
= 4 cm
So, 
o o o
1 1 1 1 1
u v f 18.89 4
? ? ? ? = 0.053 – 0.25 = –0.197
? u
o
= –5.07 cm
So, the maximum magnificent power is given by 
m = 
o
o e
v D 18.89 25
1 1
u f 5.07 20
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ?
? ?
= 3.7225 ? 2.25 = 8.376  
7. For the given compound microscope
f
o
= 1 cm, f
e
= 6 cm, D = 24 cm
For the eye piece, v
e
= –24 cm, f
e
= 6 cm
Now, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1 5
u v f 24 6 24
? ?
? ? ? ? ? ? ?
? ?
? ?
? u
e
= –4.8 cm
a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective 
lens must be (9.8) – (4.8) = 5.0 cm
Now,  
0 0
u
1
v
1
? = 
0
f
1
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
5
1
? = –
5
4
? u
0
= –
4
5
= – 1.25 cm
So, the magnifying power is given by,
m = 
?
?
?
?
?
?
?
f
D
1
u
v
o
0
= 
?
?
?
?
?
?
?
?
?
6
24
1
25 . 1
5
= 4 × 5 = 20
(b) When the separation is 11.8 cm,
v
0
= 11.8 – 4.8 = 7.0 cm, f
0
= 1 cm 
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
7
1
? = 
7
6
?
f o = 0.04m 
objective
A ?
B
A
B ? B ? ?
11.11
cm
25cm
30cm
A ??
f e = 0.2m 
objective
Fig-A
5 cm ?
9.8cm ?
4.8cm ?
24 cm ?
Fig-B
7 cm ?
11.8cm ?
4.8cm ?
24 cm ?
Chapter 19
19.3
So, m = –
?
?
?
?
?
?
?
f
D
1
u
v
o
0
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
24
1
6
7
7
= 6 × 5 = 30
So, the range of magnifying power will be 20 to 30.
8. For the given compound microscope.
f
0
= 
D 20
1
= 0.05 m = 5 cm, f
e
= 
D 10
1
= 0.1 m = 10 cm.
D = 25 cm, separation between objective & eyepiece= 20 cm
For the minimum separation between two points which can be distinguished by eye using the 
microscope, the magnifying power should be maximum.
For the eyepiece, v
0
= –25 cm, f
e
= 10 cm
So, 
e
u
1
= 
e e
f
1
v
1
? =
10
1
25
1
?
?
= –
?
?
?
?
?
? ?
50
5 2
? u
e
= –
7
50
cm
So, the image distance for the objective lens should be,
V
0
= 20 –
7
50
=  
7
90
cm
Now, for the objective lens,
0 0 0
f
1
v
1
u
1
? ? = 
5
1
90
7
? = –
90
11
? u
0
= –
11
90
cm
So, the maximum magnifying power is given by,
m = 
?
?
?
?
?
?
?
?
e 0
0
f
D
1
u
v
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
10
25
1
11
90
7
90
= 5 . 3
7
11
? = 5.5
Thus, minimum separation eye can distinguish = 
5 . 5
22 . 0
mm = 0.04 mm
9. For the give compound microscope,
f
0
= 0.5cm, tube length = 6.5cm
magnifying power = 100 (normal adjustment)
Since, the image is formed at infinity, the real image produced by the objective lens should lie on the 
focus of the eye piece.
So, v
0
+ f
e
= 6.5 cm …(1)
Again, magnifying power= 
e 0
0
f
D
u
v
? [for normal adjustment]
? m = –
e 0
0
f
D
f
v
1
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
0
0
0
0
f
v
1
u
v
?
? 100 = –
e
0
f
25
5 . 0
v
1 ?
?
?
?
?
?
?
? [Taking D = 25 cm]
? 100 f
e
= –(1 – 2v
0
) × 25
? 2v
0
– 4f
e
= 1 …(2)
? ?
v 0 ?
Eye piece ?
f e ?
F e ?
Objective ?
Page 4


19.1
SOLUTIONS TO CONCEPTS 
CHAPTER 19
1. The visual angles made by the tree with the eyes can be 
calculated be below.
? = 
A
Height of the tree AB 2
0.04
Distance from the eye OB 50
? ? ? ? ?
similarly, ?
B
= 2.5 / 80 = 0.03125
?
C
= 1.8 / 70 = 0.02571
?
D
= 2.8 / 100 = 0.028
Since, ?
A
> ??
B
> ??
D
> ??
C
, the arrangement in decreasing order is given by A, B, D and C. ?
2. For the given simple microscope, 
f = 12 cm and D = 25 cm
For maximum angular magnification, the image should be produced 
at least distance of clear vision.
So, v = – D = –25 cm
Now, 
1 1 1
v u f
? ?
?
1 1 1 1 1 37
u v f 25 12 300
? ? ? ? ? ?
?
? u = –8.1 cm
So, the object should be placed 8.1 cm away from the lens.
3. The simple microscope has, m = 3, when image is formed at D = 25 cm
a) m = 
D
1
f
? ?
25
3 1
f
? ?
? f = 25/2 = 12.5 cm
b) When the image is formed at infinity (normal adjustment)
Magnifying power = 
D 25
f 12.5
? = 2.0
4. The child has D = 10 cm and f = 10 cm
The maximum angular magnification is obtained when the image is formed at near point.
m = 
D 10
1 1
f 10
? ? ? = 1 + 1 = 2
5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm).
Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, m = 5 = 
D 25
f f
? ? f = 5 cm
For the relaxed farsighted eye, D = 40 cm
So, m = 
D 40
f 5
? = 8
So, its magnifying power is 8X.
A ?
A
+ve
B
B ?
D=25cm
(Simple Microscope)
Distance 
A
height
B 
? ?
Chapter 19
19.2
6. For the given compound microscope
f
0
= 
1
25 diopter
= 0.04 m = 4 cm, f
e
= 
1
5 diopter
= 0.2 m = 20 cm
D = 25 cm, separation between objective and eyepiece = 30 cm
The magnifying power is maximum when the image is formed by 
the eye piece at least distance of clear vision i.e. D = 25 cm
for the eye piece, v
e
= –25 cm, f
e
= 20 cm
For lens formula, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1
u v f 25 20
? ? ? ?
?
? u
e
= 11.11 cm
So, for the objective lens, the image distance should be 
v
0
= 30 – (11.11) = 18.89 cm
Now, for the objective lens,
v
0
= +18.89 cm (because real image is produced)
f
0 
= 4 cm
So, 
o o o
1 1 1 1 1
u v f 18.89 4
? ? ? ? = 0.053 – 0.25 = –0.197
? u
o
= –5.07 cm
So, the maximum magnificent power is given by 
m = 
o
o e
v D 18.89 25
1 1
u f 5.07 20
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ?
? ?
= 3.7225 ? 2.25 = 8.376  
7. For the given compound microscope
f
o
= 1 cm, f
e
= 6 cm, D = 24 cm
For the eye piece, v
e
= –24 cm, f
e
= 6 cm
Now, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1 5
u v f 24 6 24
? ?
? ? ? ? ? ? ?
? ?
? ?
? u
e
= –4.8 cm
a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective 
lens must be (9.8) – (4.8) = 5.0 cm
Now,  
0 0
u
1
v
1
? = 
0
f
1
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
5
1
? = –
5
4
? u
0
= –
4
5
= – 1.25 cm
So, the magnifying power is given by,
m = 
?
?
?
?
?
?
?
f
D
1
u
v
o
0
= 
?
?
?
?
?
?
?
?
?
6
24
1
25 . 1
5
= 4 × 5 = 20
(b) When the separation is 11.8 cm,
v
0
= 11.8 – 4.8 = 7.0 cm, f
0
= 1 cm 
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
7
1
? = 
7
6
?
f o = 0.04m 
objective
A ?
B
A
B ? B ? ?
11.11
cm
25cm
30cm
A ??
f e = 0.2m 
objective
Fig-A
5 cm ?
9.8cm ?
4.8cm ?
24 cm ?
Fig-B
7 cm ?
11.8cm ?
4.8cm ?
24 cm ?
Chapter 19
19.3
So, m = –
?
?
?
?
?
?
?
f
D
1
u
v
o
0
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
24
1
6
7
7
= 6 × 5 = 30
So, the range of magnifying power will be 20 to 30.
8. For the given compound microscope.
f
0
= 
D 20
1
= 0.05 m = 5 cm, f
e
= 
D 10
1
= 0.1 m = 10 cm.
D = 25 cm, separation between objective & eyepiece= 20 cm
For the minimum separation between two points which can be distinguished by eye using the 
microscope, the magnifying power should be maximum.
For the eyepiece, v
0
= –25 cm, f
e
= 10 cm
So, 
e
u
1
= 
e e
f
1
v
1
? =
10
1
25
1
?
?
= –
?
?
?
?
?
? ?
50
5 2
? u
e
= –
7
50
cm
So, the image distance for the objective lens should be,
V
0
= 20 –
7
50
=  
7
90
cm
Now, for the objective lens,
0 0 0
f
1
v
1
u
1
? ? = 
5
1
90
7
? = –
90
11
? u
0
= –
11
90
cm
So, the maximum magnifying power is given by,
m = 
?
?
?
?
?
?
?
?
e 0
0
f
D
1
u
v
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
10
25
1
11
90
7
90
= 5 . 3
7
11
? = 5.5
Thus, minimum separation eye can distinguish = 
5 . 5
22 . 0
mm = 0.04 mm
9. For the give compound microscope,
f
0
= 0.5cm, tube length = 6.5cm
magnifying power = 100 (normal adjustment)
Since, the image is formed at infinity, the real image produced by the objective lens should lie on the 
focus of the eye piece.
So, v
0
+ f
e
= 6.5 cm …(1)
Again, magnifying power= 
e 0
0
f
D
u
v
? [for normal adjustment]
? m = –
e 0
0
f
D
f
v
1
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
0
0
0
0
f
v
1
u
v
?
? 100 = –
e
0
f
25
5 . 0
v
1 ?
?
?
?
?
?
?
? [Taking D = 25 cm]
? 100 f
e
= –(1 – 2v
0
) × 25
? 2v
0
– 4f
e
= 1 …(2)
? ?
v 0 ?
Eye piece ?
f e ?
F e ?
Objective ?
Chapter 19
19.4
Solving equation (1) and (2) we can get,
V
0
= 4.5 cm and f
e
= 2 cm
So, the focal length of the eye piece is 2cm.
10. Given that, 
f
o
  = = 1 cm, f
e
= 5 cm, u
0
= 0.5 cm, v
e
= 30 cm
For the objective lens, u
0
= – 0.5 cm, f
0
= 1 cm.
From lens formula,
0 0 0
f
1
u
1
v
1
? ? ?
0 0 0
f
1
u
1
v
1
? ? = 
1
1
5 . 0
1
?
?
= – 1
? v
0
= – 1 cm
So, a virtual image is formed by the objective on the same side as that of the object at a distance of 1 
cm from the objective lens. This image acts as a virtual object for the eyepiece.
For the eyepiece,
0 0 0
f
1
u
1
v
1
? ? ?
0 0 0
f
1
v
1
u
1
? ? = 
5
1
30
1
? = 
30
5 ?
= 
6
1 ?
? u
0
= – 6 cm
So, as shown in figure,
Separation between the lenses = u
0
– v
0
= 6 – 1 = 5 cm
11. The optical instrument has
f
0
= 
D 25
1
= 0.04 m = 4 cm
f
e
= 
D 20
1
= 0.05 m = 5 cm
tube length = 25 cm (normal adjustment)
(a) The instrument must be a microscope as f
0
< f
e
(b) Since the final image is formed at infinity, the image produced 
by the objective should lie on the focal plane of the eye piece.
So, image distance for objective = v
0
= 25 – 5 = 20 cm
Now, using lens formula.
0 0 0
f
1
u
1
v
1
? ? ?
0 0 0
f
1
v
1
u
1
? ? = 
4
1
20
1
? = 
20
4 ?
= 
5
1 ?
? u
0
= – 5 cm
So, angular magnification = m = –
e 0
0
f
D
u
v
? [Taking D = 25 cm]
= –
5
25
5
20
?
?
= 20 
12. For the astronomical telescope in normal adjustment.
Magnifying power = m = 50, length of the tube = L = 102 cm
Let f
0
and f
e
be the focal length of objective and eye piece respectively.
m = 
e
0
f
f
= 50 ? f
0
= 50 f
e
…(1)
and, L = f
0
+ f
e
= 102 cm …(2)
Putting the value of f
0
from equation (1) in (2), we get, 
f
0
+ f
e
= 102 ? 51f
e
= 102 ? f
e
= 2 cm = 0.02 m
So, f
0
= 100 cm = 1 m
? Power of the objective lens = 
0
f
1
= 1D
And Power of the eye piece lens = 
e
f
1
= 
02 . 0
1
= 50D
B ?
A ?
? ?
20cm
?
5cm
?
F e ?
? ?
f 0 ?
Eye piece ?
f e ?
F e ?
Objective ?
A ?
A ? ?
0.5cm
?
Eye piece ? 1cm
?
Objective ?
30cm
?
60cm
?
B ? ?
A ? ?
B ? ?
B ?
Page 5


19.1
SOLUTIONS TO CONCEPTS 
CHAPTER 19
1. The visual angles made by the tree with the eyes can be 
calculated be below.
? = 
A
Height of the tree AB 2
0.04
Distance from the eye OB 50
? ? ? ? ?
similarly, ?
B
= 2.5 / 80 = 0.03125
?
C
= 1.8 / 70 = 0.02571
?
D
= 2.8 / 100 = 0.028
Since, ?
A
> ??
B
> ??
D
> ??
C
, the arrangement in decreasing order is given by A, B, D and C. ?
2. For the given simple microscope, 
f = 12 cm and D = 25 cm
For maximum angular magnification, the image should be produced 
at least distance of clear vision.
So, v = – D = –25 cm
Now, 
1 1 1
v u f
? ?
?
1 1 1 1 1 37
u v f 25 12 300
? ? ? ? ? ?
?
? u = –8.1 cm
So, the object should be placed 8.1 cm away from the lens.
3. The simple microscope has, m = 3, when image is formed at D = 25 cm
a) m = 
D
1
f
? ?
25
3 1
f
? ?
? f = 25/2 = 12.5 cm
b) When the image is formed at infinity (normal adjustment)
Magnifying power = 
D 25
f 12.5
? = 2.0
4. The child has D = 10 cm and f = 10 cm
The maximum angular magnification is obtained when the image is formed at near point.
m = 
D 10
1 1
f 10
? ? ? = 1 + 1 = 2
5. The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm).
Because, the eye is relaxed the image is formed at infinity (normal adjustment)
So, m = 5 = 
D 25
f f
? ? f = 5 cm
For the relaxed farsighted eye, D = 40 cm
So, m = 
D 40
f 5
? = 8
So, its magnifying power is 8X.
A ?
A
+ve
B
B ?
D=25cm
(Simple Microscope)
Distance 
A
height
B 
? ?
Chapter 19
19.2
6. For the given compound microscope
f
0
= 
1
25 diopter
= 0.04 m = 4 cm, f
e
= 
1
5 diopter
= 0.2 m = 20 cm
D = 25 cm, separation between objective and eyepiece = 30 cm
The magnifying power is maximum when the image is formed by 
the eye piece at least distance of clear vision i.e. D = 25 cm
for the eye piece, v
e
= –25 cm, f
e
= 20 cm
For lens formula, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1
u v f 25 20
? ? ? ?
?
? u
e
= 11.11 cm
So, for the objective lens, the image distance should be 
v
0
= 30 – (11.11) = 18.89 cm
Now, for the objective lens,
v
0
= +18.89 cm (because real image is produced)
f
0 
= 4 cm
So, 
o o o
1 1 1 1 1
u v f 18.89 4
? ? ? ? = 0.053 – 0.25 = –0.197
? u
o
= –5.07 cm
So, the maximum magnificent power is given by 
m = 
o
o e
v D 18.89 25
1 1
u f 5.07 20
? ? ? ?
? ? ? ? ?
? ? ? ?
? ? ?
? ?
= 3.7225 ? 2.25 = 8.376  
7. For the given compound microscope
f
o
= 1 cm, f
e
= 6 cm, D = 24 cm
For the eye piece, v
e
= –24 cm, f
e
= 6 cm
Now, 
e e e
1 1 1
v u f
? ?
?
e e e
1 1 1 1 1 5
u v f 24 6 24
? ?
? ? ? ? ? ? ?
? ?
? ?
? u
e
= –4.8 cm
a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective 
lens must be (9.8) – (4.8) = 5.0 cm
Now,  
0 0
u
1
v
1
? = 
0
f
1
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
5
1
? = –
5
4
? u
0
= –
4
5
= – 1.25 cm
So, the magnifying power is given by,
m = 
?
?
?
?
?
?
?
f
D
1
u
v
o
0
= 
?
?
?
?
?
?
?
?
?
6
24
1
25 . 1
5
= 4 × 5 = 20
(b) When the separation is 11.8 cm,
v
0
= 11.8 – 4.8 = 7.0 cm, f
0
= 1 cm 
?
0
u
1
= 
0 0
f
1
v
1
? = 
1
1
7
1
? = 
7
6
?
f o = 0.04m 
objective
A ?
B
A
B ? B ? ?
11.11
cm
25cm
30cm
A ??
f e = 0.2m 
objective
Fig-A
5 cm ?
9.8cm ?
4.8cm ?
24 cm ?
Fig-B
7 cm ?
11.8cm ?
4.8cm ?
24 cm ?
Chapter 19
19.3
So, m = –
?
?
?
?
?
?
?
f
D
1
u
v
o
0
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
6
24
1
6
7
7
= 6 × 5 = 30
So, the range of magnifying power will be 20 to 30.
8. For the given compound microscope.
f
0
= 
D 20
1
= 0.05 m = 5 cm, f
e
= 
D 10
1
= 0.1 m = 10 cm.
D = 25 cm, separation between objective & eyepiece= 20 cm
For the minimum separation between two points which can be distinguished by eye using the 
microscope, the magnifying power should be maximum.
For the eyepiece, v
0
= –25 cm, f
e
= 10 cm
So, 
e
u
1
= 
e e
f
1
v
1
? =
10
1
25
1
?
?
= –
?
?
?
?
?
? ?
50
5 2
? u
e
= –
7
50
cm
So, the image distance for the objective lens should be,
V
0
= 20 –
7
50
=  
7
90
cm
Now, for the objective lens,
0 0 0
f
1
v
1
u
1
? ? = 
5
1
90
7
? = –
90
11
? u
0
= –
11
90
cm
So, the maximum magnifying power is given by,
m = 
?
?
?
?
?
?
?
?
e 0
0
f
D
1
u
v
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
10
25
1
11
90
7
90
= 5 . 3
7
11
? = 5.5
Thus, minimum separation eye can distinguish = 
5 . 5
22 . 0
mm = 0.04 mm
9. For the give compound microscope,
f
0
= 0.5cm, tube length = 6.5cm
magnifying power = 100 (normal adjustment)
Since, the image is formed at infinity, the real image produced by the objective lens should lie on the 
focus of the eye piece.
So, v
0
+ f
e
= 6.5 cm …(1)
Again, magnifying power= 
e 0
0
f
D
u
v
? [for normal adjustment]
? m = –
e 0
0
f
D
f
v
1
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
0
0
0
0
f
v
1
u
v
?
? 100 = –
e
0
f
25
5 . 0
v
1 ?
?
?
?
?
?
?
? [Taking D = 25 cm]
? 100 f
e
= –(1 – 2v
0
) × 25
? 2v
0
– 4f
e
= 1 …(2)
? ?
v 0 ?
Eye piece ?
f e ?
F e ?
Objective ?
Chapter 19
19.4
Solving equation (1) and (2) we can get,
V
0
= 4.5 cm and f
e
= 2 cm
So, the focal length of the eye piece is 2cm.
10. Given that, 
f
o
  = = 1 cm, f
e
= 5 cm, u
0
= 0.5 cm, v
e
= 30 cm
For the objective lens, u
0
= – 0.5 cm, f
0
= 1 cm.
From lens formula,
0 0 0
f
1
u
1
v
1
? ? ?
0 0 0
f
1
u
1
v
1
? ? = 
1
1
5 . 0
1
?
?
= – 1
? v
0
= – 1 cm
So, a virtual image is formed by the objective on the same side as that of the object at a distance of 1 
cm from the objective lens. This image acts as a virtual object for the eyepiece.
For the eyepiece,
0 0 0
f
1
u
1
v
1
? ? ?
0 0 0
f
1
v
1
u
1
? ? = 
5
1
30
1
? = 
30
5 ?
= 
6
1 ?
? u
0
= – 6 cm
So, as shown in figure,
Separation between the lenses = u
0
– v
0
= 6 – 1 = 5 cm
11. The optical instrument has
f
0
= 
D 25
1
= 0.04 m = 4 cm
f
e
= 
D 20
1
= 0.05 m = 5 cm
tube length = 25 cm (normal adjustment)
(a) The instrument must be a microscope as f
0
< f
e
(b) Since the final image is formed at infinity, the image produced 
by the objective should lie on the focal plane of the eye piece.
So, image distance for objective = v
0
= 25 – 5 = 20 cm
Now, using lens formula.
0 0 0
f
1
u
1
v
1
? ? ?
0 0 0
f
1
v
1
u
1
? ? = 
4
1
20
1
? = 
20
4 ?
= 
5
1 ?
? u
0
= – 5 cm
So, angular magnification = m = –
e 0
0
f
D
u
v
? [Taking D = 25 cm]
= –
5
25
5
20
?
?
= 20 
12. For the astronomical telescope in normal adjustment.
Magnifying power = m = 50, length of the tube = L = 102 cm
Let f
0
and f
e
be the focal length of objective and eye piece respectively.
m = 
e
0
f
f
= 50 ? f
0
= 50 f
e
…(1)
and, L = f
0
+ f
e
= 102 cm …(2)
Putting the value of f
0
from equation (1) in (2), we get, 
f
0
+ f
e
= 102 ? 51f
e
= 102 ? f
e
= 2 cm = 0.02 m
So, f
0
= 100 cm = 1 m
? Power of the objective lens = 
0
f
1
= 1D
And Power of the eye piece lens = 
e
f
1
= 
02 . 0
1
= 50D
B ?
A ?
? ?
20cm
?
5cm
?
F e ?
? ?
f 0 ?
Eye piece ?
f e ?
F e ?
Objective ?
A ?
A ? ?
0.5cm
?
Eye piece ? 1cm
?
Objective ?
30cm
?
60cm
?
B ? ?
A ? ?
B ? ?
B ?
Chapter 19
19.5
13. For the given astronomical telescope in normal adjustment,
F
e
= 10 cm, L = 1 m = 100cm
S0, f
0
= L – f
e
= 100 – 10 = 90 cm
and, magnifying power = 
e
0
f
f
= 
10
90
= 9
14. For the given Galilean telescope, (When the image is formed at infinity)
f
0
= 30 cm, L = 27 cm
Since L = f
0
–
e
f
[Since, concave eyepiece lens is used in Galilean Telescope]
? f
e
= f
0
– L = 30 – 27 = 3 cm 
15. For the far sighted person,
u = – 20 cm, v = – 50 cm
from lens formula  
f
1
u
1
v
1
? ?
f
1
= 
20
1
50
1
?
?
?
= 
50
1
20
1
? = 
100
3
? f = 
3
100
cm = 
3
1
m
So, power of the lens = 
f
1
= 3 Diopter
16. For the near sighted person,
u = ? and v = – 200 cm = – 2m
So, 
u
1
v
1
f
1
? ? = 
?
?
?
1
2
1
= –
2
1
= – 0.5 
So, power of the lens is –0.5D
17. The person wears glasses of power –2.5D
So, the person must be near sighted.
u = ?, v = far point, f= 
5 . 2
1
?
= – 0.4m  = – 40 cm
Now, 
f
1
u
1
v
1
? ?
?
f
1
u
1
v
1
? ? =  
40
1
0
?
? ? v = – 40 cm
So, the far point of the person is 40 cm
18. On the 50
th
birthday, he reads the card at a distance 25cm using a glass of +2.5D.
Ten years later, his near point must have changed.
So after ten years,
u = – 50 cm, f = 
D 5 . 2
1
= 0.4m = 40 cm v = near point
Now, 
f
1
u
1
v
1
? ? ?
f
1
u
1
v
1
? ? =  
40
1
50
1
?
?
= 
200
1
So, near point = v = 200cm
To read the farewell letter at a distance of 25 cm,
U = – 25 cm
For lens formula,
f
1
u
1
v
1
? ? ?
f
1
= 
25 200
1
?
?
? = 
25
1
200
1
? = 
200
9
? f = 
9
200
cm = 
9
2
m
? Power of the lens = 
f
1
= 
2
9
= 4.5D
?He has to use a lens of power +4.5D.
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FAQs on HC Verma Solutions: Chapter 19 - Optical Instruments - Physics Class 11 - NEET

1. How do optical instruments work?
Ans. Optical instruments work by manipulating light to enhance our ability to observe and measure objects. They typically consist of lenses, mirrors, and other optical components that bend, focus, and magnify light to form an image. By using these instruments, we can see distant objects more clearly (telescopes), examine tiny objects (microscopes), and correct vision problems (eyeglasses).
2. What is the difference between a telescope and a microscope?
Ans. Telescopes and microscopes are both optical instruments, but they have different purposes. Telescopes are designed to magnify distant objects, such as stars and planets, allowing us to observe them in more detail. Microscopes, on the other hand, are used to magnify small objects, such as cells or microorganisms, enabling us to see them clearly. While telescopes have a longer focal length, microscopes have a shorter focal length to focus on nearby objects.
3. How does a microscope magnify objects?
Ans. A microscope magnifies objects by using two lenses: the objective lens and the eyepiece. The objective lens, located close to the object being observed, forms a magnified real image of the object. This image is then further magnified by the eyepiece, which acts as a magnifying glass for our eyes. The combination of these lenses allows us to see tiny details that would otherwise be invisible to the naked eye.
4. What is the purpose of a lens in optical instruments?
Ans. Lenses play a crucial role in optical instruments. They are used to refract (bend) light and focus it to form an image. In telescopes and microscopes, lenses help magnify distant or small objects, respectively. In cameras and projectors, lenses focus light onto a photosensitive surface or a screen. Lenses also correct vision problems by refracting light in a way that compensates for the eye's deficiencies.
5. Can optical instruments correct vision problems?
Ans. Yes, optical instruments can correct vision problems. Eyeglasses and contact lenses, for example, use specially designed lenses to compensate for refractive errors in the eye. These errors can cause nearsightedness, farsightedness, or astigmatism, making objects appear blurry. By using lenses that alter the path of light as it enters the eye, these instruments can help focus the light correctly onto the retina, resulting in clearer vision.
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