Question

An astronomical telescope has objective and eyepiece of focal lengths $40\text{cm}$ and $4\text{cm}$ respectively. To view an object $200\text{cm}$ away from the objective, the lenses must be separated by a distance of-

• A. $37.3\text{cm}$
• B. $46.0\text{cm}$
• C. $50.0\text{cm}$
• D. $53.4\text{cm}$

Answer 1

The correct option is D $53.4\text{cm}$

Given, $f_0=40\text{cm};f_e=4\text{cm}{u}_0=-200\text{cm}L=\left|\begin{array}{c}{v}_0\end{array}\right|+\left|\begin{array}{c}{u}_e\end{array}\right|=?$

Using lens formula, for the objective lens,

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

$\Rightarrow \frac{1}{40}=\frac{1}{v_0}-\frac{1}{-200}$

$\Rightarrow v_0=50\text{cm}$

Using lens formula, for the eyepiece lens,

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

$\Rightarrow \frac{1}{4}=\frac{1}{-25}-\frac{1}{u_e}[\because v_e=-25\text{cm}]$

$\frac{1}{u_e}=-(\frac{1}{25}+\frac{1}{4})=-\frac{29}{100}$

$\Rightarrow u_e=-\frac{100}{29}=-3.4\text{cm}$

$\therefore$ The separation between lenses,

$L=\left|\begin{array}{c}{v}_0\end{array}\right|+\left|\begin{array}{c}{u}_e\end{array}\right|=50+3.4=53.4\text{cm}$

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