Optics Question 835
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 an object distant 2m from its objective, then its magnifying power will be:
Options:
-
4
-
8
+8
- 2
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given: $ f _{0}=50cm,f _{e}=5cm $
$ d=25,cm,v _{0}=-200,cm $
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=-25\text{ cm}$
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{2}{25}\text{ or, }v _{e}=\frac{-25}{2}cm $
BETA
