Question

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

• A. $-4$
• B. $-8$
• C. $+8$
• D. $-2$

Answer 1

Answer: (D)

Given: $f_0 = 50\, cm, f_e = 5\, cm$
$d= 25\, cm, u_0=-200\, cm$
Magnification $M = ?$
As $\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$
$\Rightarrow \frac{1}{v_{0}}-\frac{1}{f_{0}}=\frac{1}{u_{0}}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}$
or $v_{0}=\frac{200}{3}cm$
$v_{e}=d=-25\,cm$
From, $\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$
$-\frac{1}{u_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}$
$=\frac{1}{5}+\frac{1}{25}=\frac{6}{25}$
or, $v_{e}=\frac{-25}{6}cm$
Magnification $M=M_{0}\times M_{e}$
$=\frac{v_{0}}{u_{0}}\times\frac{v_{e}}{u_{e}}=\frac{-200/3}{200}\times\frac{-25}{-25/6}$
$=-\frac{1}{3}\times6=-2$