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:
Answer & explanation
Correct answer: option 1
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) = $v_0+u_e$
$=\frac{200}{3}+\frac{25}{6}≃71cm$