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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