Question

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

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

Answer 1

Answer: (D)

Given: $f_0=50cm,f_e=5cm$
$d=25cm,u_0=-200cm$
Magnification $M=?$
As $\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$
$\Rightarrow \frac{1}{v_0}-\frac{1}{f_0}=\frac{1}{u_0}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}$
or $v_0=\frac{200}{3}cm$
$v_e=d=-25cm$
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 $M=M_0\times M_e$
$=\frac{v_0}{u_0}\times \frac{v_e}{u_e}=\frac{-200/3}{200}\times \frac{-25}{-25/6}$
$=-\frac{1}{3}\times 6=-2$