Question

A telescope has an objective of focal length $30\text{cm}$ and an eyepiece of focal length $3.0\text{cm}.$ It is focussed on a scale present at a distance of $2.0\text{m}$. For seeing with relaxed eye, calculate the separation between the objective and the eyepiece.

• A. $38.3\text{cm}$
• B. $42.3\text{cm}$
• C. $35.3\text{cm}$
• D. $49\text{cm}$

Answer 1

The correct option is A $38.3\text{cm}$

For seeing with relaxed eye, the final image should be formed at infinity.

For this, the image formed by the objective should fall at the focus of the eyepiece.

Let $v_0$ be the distance of the image formed by the objective.

Then, from the lens formula,

$\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$

Here $u_0=-2.0\text{m}=-200\text{cm},f_0=30\text{cm}$

$\Rightarrow \frac{1}{v_0}-\frac{1}{-200}=\frac{1}{30}$

$\Rightarrow \frac{1}{v_0}=\frac{1}{30}-\frac{1}{200}=\frac{20-3}{600}=\frac{17}{600}$

$\Rightarrow v_0=\frac{600}{17}=35.3\text{cm}$

$\therefore$ distance between the objective and the eyepiece,

$d=v_0+f_e=35.3+3.0=38.3\text{cm}$

Hence, option $\left(A\right)$ is correct.

Why this question? Key Note:For seeing with relaxed eye, the final image should be formed at infinity. For this, the image formed by the objective should fall at the focus of the eyepiece.