Question

The focal length of objective of a telescope is $50cm$ and eye-piece is $5cm$ . The telescope is focused from a distinct vision on a scale $2m$ away. The separation between the objective and the eye-piece is, if final image is formed at the least distance of distinct vision,

• A. $75cm$
• B. $60cm$
• C. $71cm$
• D. $74cm$

Answer 1

Answer: (C)

$L=VO+\left|U_e\right|$
$\frac{1}{V_0}-\frac{1}{u_0}=\frac{1}{f_0}\text{,}\frac{1}{V_0}-\frac{1}{u_e}=\frac{1}{f_e}$
$\frac{1}{V_0}=\frac{1}{50}-\frac{1}{206}\text{, }\frac{1}{-25}-\frac{1}{u_e}=\frac{1}{5}$
$V_0=\frac{200}{3}cm\text{,}{u}_e=\frac{-25}{6}cm$
$L=\frac{200}{3}+\frac{25}{6}\Rightarrow L=70.83=71cm.$