A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distant vision on a scale 200 cm away from the objective. Calculate the separation between objective and eyepiece:

70.8 cm

The correct answer is Option (1) → 70.8 cm

for objective lens,

$μ_0$ (Object distance) = -20 cm

lens formula,

$\frac{1}{f}=\frac{1}{v_0}-\frac{1}{u_0}$

$⇒v_0=\frac{1}{\frac{1}{50}+\frac{1}{200}}=\frac{200}{3}cm$

for eye piece,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{-25}-(\frac{1}{-u_e})=\frac{1}{5}$

$u_e=\frac{25}{6}cm$

∴ L (Seperation) = $v_0+u_e$

$=\frac{200}{3}+\frac{25}{6}≃71cm$