Question

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

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

Answer 1

Answer: (A)

$-2$

Given $f_0=50\text{cm},f_e=5\text{cm}$ and $u_0=-2\text{m}=-200\text{cm}$

The telescope is used for distinct vision. Hence, $d=25\text{cm}$

Using the lens equation, $\frac{1}{v_0}=\frac{1}{f_0}+\frac{1}{u_0}=\frac{1}{50}-\frac{1}{200}=\frac{3}{200}\Rightarrow v_0=\frac{200}{3}\text{cm}$

Using the lens equation for the eye piece, $v_e=d=-25\text{cm}$

Hence $\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{5}-\frac{1}{25}=-\frac{6}{25}\Rightarrow u_e=-\frac{25}{6}\text{cm}$

Magnification is $M=M_0\times M_e=\frac{v_0}{u_0}\times \frac{v_e}{u_e}=\frac{200/3}{-200}\times \frac{-25}{-25/6}=-2$