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NEET 2012

Exam: NEET
Year: 2012
Subject: Physics
Chapter: Ray Optics

NEET 2012 · Physics · Ray Optics

The magnifying power of a telescope is 9. When it is adjusted for...

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is

20 \mathrm{~cm}

. The focal length of lenses are

Options

A $10 \mathrm{~cm}, 10 \mathrm{~cm}$
B $15 \mathrm{~cm}, 5 \mathrm{~cm}$
C $18 \mathrm{~cm}, 2 \mathrm{~cm}$
D $11 \mathrm{~cm}, 9 \mathrm{~cm}$

Correct answer

C. $18 \mathrm{~cm}, 2 \mathrm{~cm}$

Step-by-step solution

Given,

M=\frac{f_0}{f_e}=9

and f_0+f_e=20 f_o=9 f_e So,

9 f_e+f_e=20

\begin{aligned} & \therefore f_e=2 \mathrm{~cm} \\ & f_o=9 \times 2 \\ & f_0=18 \mathrm{~cm} \end{aligned}

$$

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