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Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye-piece is found to be 20 cm. The focal lengths of the lenses are

A
18 cm, 2 cm
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B
11 cm, 9 cm
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C
10 cm, 10 cm
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D
15 cm, 5 cm
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Solution

The correct option is A 18 cm, 2 cm
Given,

m=9, L=20 cm

Let, the focal lengths of the objective and eye-piece are f0 and fe respectively.

We know,
m=f0fe=9

f0=9fe ....(i)

and L=f0+fe=20 ....(ii)

Solving equations (i) and (ii), we get

f0=18 cm and fe=2 cm

Hence, (A) is the correct answer.
Key Concept: Magnifying power of a telescope is the ration of focal length of the objective to that of the eye piece, when the final image formed at infinity.

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