Question

Magnifying power of a terrestrial telescope of objective lens having focal length $200\text{cm}$ is $40$. Find the length of the telescope when the final image is formed at least distance of distinct vision, Focal length of erecting lens is $3\text{cm}$.

• A. $205\text{cm}$
• B. $216\text{cm}$
• C. $212\text{cm}$
• D. $208\text{cm}$

Answer 1

The correct option is B $216\text{cm}$

Given,

$m=40;f_0=200\text{cm};f=3\text{cm}$

Where, $f=\text{Focal length of the erecting lens}{f}_0=\text{Focal length of the objective}{f}_e=\text{Focal length of the eyepiece}$

We know, magnifying power of the terrestrial telescope is,

$m=\frac{f_0}{f_e}$

$\Rightarrow 40=\frac{200}{f_e}$

$\therefore f_e=5\text{cm}$

When the final image is formed, at least distance of distinct vision, then



$v_e=-D=-25\text{cm};f_e=5\text{cm};u=u_e=?$

From the lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

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

$\Rightarrow \frac{1}{5}=\frac{1}{-25}-\frac{1}{u_e}$

$\Rightarrow \frac{1}{u_e}=\frac{-1}{25}-\frac{1}{5}=\frac{-6}{25}$

$\therefore u_e=\frac{-25}{6}\text{cm}$

Now, length of the terrestrial telescope is,

$L=f_0+4f+\left|\begin{array}{c}{u}_e\end{array}\right|$

$=200+4\times 3+\frac{25}{6}=216.1\text{cm}\text{(or)}$

$L\approx 216\text{cm}$

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