Question

Magnifying power of a terrestrial telescope is $50$. Find the length of the telescope if focal length of eyepiece and erecting lens is $4\text{cm}$ and $6\text{cm}$ respectively.

• A. $200\text{cm}$
• B. $224\text{cm}$
• C. $226\text{cm}$
• D. $228\text{cm}$

Answer 1

The correct option is D $228\text{cm}$

Given,

$m=50;f_e=4\text{cm};f=6\text{cm}$

We know, magnifying power of the terrestrial telescope is,

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

$\Rightarrow 50=\frac{f_0}{4}$

$\therefore f_0=200\text{cm}$

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

For normal adjustment, the length of the telescope is,

$L=f_0+4f+f_e$

$\therefore L=200+4\times 6+4=228\text{cm}$

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

Why this question? To teach the following concepts: $\left(i\right)$ Astronomical telescopes are used for viewing distant stars and planets, whereas terrestrial telescopes are used for viewing distant objects on earth. $\left(ii\right)L_{\text{Astronomical}}=f_o+f_e\text{and}{L}_{\text{Terrestrial}}=f_o+4f+f_e$