A telescope has an objective of focal length 30 cm and an eyepiece of focal length 3.0 cm. It is focused on a scale distant 2.0 m. For seeing with relaxed eye, the separation between the objective and eyepiece would be:

38.3 cm

The correct answer is Option (2) → 38.3 cm

Using the lens formula,

$\frac{1}{f_o}=\frac{1}{u}+\frac{1}{v_o}$

$\frac{1}{30}=\frac{1}{200}+\frac{1}{v_o}$

$⇒\frac{1}{v_o}=\frac{20-3}{600}=\frac{17}{600}$

$⇒v_o=35.29cm$

Seperation (L) = $f_e+v_o$

$=3.35.29=38.29cm$