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}=50cm,f _{e}=5cm $

$ d=25cm,v _{0}=-200cm $

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} $

$ \text{or }v _{0}=\frac{200}{3}\text{ Now }v _{e}=d=-25cm $

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}\text{ or, }v _{e}=\frac{-25}{6}cm $



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