SATHEE: Optics Question 835

Optics Question 835

Question: The focal length of the objective and the eyepiece of a telescope are 50 cm and 5 cm respectively. If the telescope is focused for distinct vision on a scale distant 2m from its objective, then its magnifying power will be:

Options:

A) - 4

B) - 8

C) +8

D) - 2

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Given: $f_{0} = 50 c m, f_{e} = 5 c m$

$d = 25 c m, v_{0} = - 200 c m$

Magnification M=? As $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$

$\Rightarrow \frac{1}{v_{0}} = \frac{1}{f_{0}} + \frac{1}{u_{0}} = \frac{1}{50} - \frac{1}{200} = \frac{4 - 1}{200} = \frac{3}{200}$

$or v_{0} = \frac{200}{3} Now v_{e} = d = - 25 c m$

From, $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$

$- \frac{1}{u_{e}} = \frac{1}{f_{e}} - \frac{1}{v_{e}} = \frac{1}{5} + \frac{1}{25} = \frac{6}{25} or, v_{e} = \frac{- 25}{6} c m$

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Optics Question 835

Question: The focal length of the objective and the eyepiece of a telescope are 50 cm and 5 cm respectively. If the telescope is focused for distinct vision on a scale distant 2m from its objective, then its magnifying power will be:

Options:

Answer:

Solution:

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