Question

A terrestrial telescope is constructed by using two convex lenses of focal length $100\text{cm}$ and $4\text{cm}$. If the final image obtained is $50\text{cm}$ away from the eyepiece. Find the length of the telescope (approximately).

• A. $101\text{cm}$
• B. $104\text{cm}$
• C. $107\text{cm}$
• D. $112\text{cm}$

Answer 1

The correct option is B $104\text{cm}$

We know, the focal length of the objective is larger than the focal length of the eyepiece in a telescope.

$\therefore f_0=100;\text{and}{f}_e=4\text{cm}$



For the eyepiece,

$f_e=4\text{cm};v_e=-50\text{cm}$

Using lens formula,

$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$

$\Rightarrow \frac{1}{4}=\frac{1}{-50}-\frac{1}{u_e}$

$\Rightarrow \frac{1}{u_e}=\frac{-1}{50}-\frac{1}{4}=\frac{-2-25}{100}$

$\therefore u_e=\frac{-100}{27}\text{cm}$

The length of the telescope is,

$L=f_0+\left|\begin{array}{c}{u}_e\end{array}\right|$

$L=100+\frac{100}{27}=103.7\text{cm}\text{(or)}$

$L\approx 104\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question? Caution: Length of the telescope is not always, $L=f_0+f_e$ (or) $L=f_0+\frac{D{f}_e}{D+f_e}$ The length of the astronomical telescope is always equal to the sum of the distances of the first image from the objective lens and eyepiece of the telescope.