CBSE Class 12 Physics 2023 Delhi Set 1 Paper

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Question : 33
Total: 35
(a) (i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point.
(ii) In a compound microscope an object is placed at a distance of 1.5cm from the objective of focal length 1.25cm. If the eye-piece has a focal length of 5cm and the final image is formed at the near point, find the magnifying power of the microscope.
OR
(b) (i) Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.
(ii) The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150cm. Find the focal lengths of the two lenses.
(a) (i) Ray diagram of compound microscope:

In a compound microscope there are two lenses - objective (O) and Eyepiece (E).

Object PQ is placed in front of the objective at a distance more than the focal length of the objective.

An inverted, magnified, real image P1Q1 is formed in front of eyepiece within the optical centre and the focus of the eyepiece. This acts as the object of the eyepiece. An (erect with respect to P1Q1, inverted with respect to PQ ), magnified, virtual image P2Q2 is formed at a distance D (minimum distance of distinct vision) from the eyepiece.

Magnification:

For objective:

Object distance =u
Image distance =v
Magnification =mo=
v
u
=
P1Q1
PQ

Applying the lens formula,
1
v
1
u
=
1
f0

Or, 1+
v
u
=
v
f0

Or,
v
u
=
v
f0
1

For eyepiece:

Magnification =me=1+
D
fe

Magnification of the combination of objective and eyepiece =m=mO×me
Or, m=
v
u
×(1+
D
fe
)

Or, m=(
v
f0
1
)
(1+
D
fe
)

P1Q1 image is formed very close to the eyepiece, hence v can be approximated as the distance between the two lenses i.e., the length of the tube(L).
m=(
L
f0
1
)
(1+
D
fe
)

Since, feD and f0L, hence the above expression may be approximated as,
m=
L
f0
×
D
fe

(ii) Given, u=1.5cm,f0=1.25cm,fe=5cm,D =25cm
Here, all alphabets are in their usual meanings Applying lens formula for objective lens,
1
v
1
u
=
1
f0

Or,
1
v
1
1.5
=
1
1.25

v=7.5cm
Magnification =m=
v
u
×(1+
D
fe
)

Or, m=
7.5
1.5
×(1+
25
5
)

m=30
OR
(b) (i) Astronomical telescope in normal adjustment:

In an astronomical telescope there are two lenses - objective (O) and Eyepiece (E).
The two lenses are so placed during focussing that the foci of the lenses meet at a point.
Objective is directed towards the object at infinity.
Parallel rays coming from the object meet at the focus of the objective and forms an inverted, real image P1Q1 in front of eyepiece. This point is the focus of eyepiece too.
This acts as the object of the eyepiece. An (inverted with respect to P1Q1, erect with respect to original object), highly magnified, real image is formed at infinity.
Magnification =m
=
Angle subtended at eye by the final image
Angle subtended at eye by the object

Or, m=
Angle subtended at eyepiece by the final image
Angle subtended at objective by the object

Or, m=
β
α

Or, m=
Q1EP1
Q1OP1

Or, m=
tanQ1EP1
tanQ1OP1

[ α and β being very small, tanα=α and tanβ=β ]
Or, m=
Q1P1
Q1E
Q1P1
Q1O

Or, m=
Q1O
Q1E

m=
fo
fe

(ii) Since, m=
f0
fe

Or, 2.9=
f0
fe

fo=2.9fe
Also, fo+fe=150
Or, 2.9fe+fe=150
fe=
150
3.9

=38.46cm
fO=2.9fc
=2.9×38.46
=111.54cm
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