Question

In an astronomical telescope in normal adjustment, a straight black line of length $L$ is drawn on the inside part of the objective lens. The eyepiece forms a real image of this line. The length of this image is $I$. The magnification of the telescope is -

• A. $\frac{L}{I}+1$
• B. $\frac{L}{I}-1$
• C. $\frac{L+1}{L-1}$
• D. $\frac{L}{I}$

Answer 1

The correct option is D $\frac{L}{I}$



We know that, in normal adjustment, the final image is formed at infinity.

The angular magnification is given by,

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

Here, for the image formation of the straight black line by the eyepiece,

$u=-(f_o+f_e)$

So, using lens formula,

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

$\Rightarrow \frac{1}{v}-\frac{1}{-(f_o+f_e)}=\frac{1}{f_e}$

$\Rightarrow v=\frac{f_e(f_o+f_e)}{f_o}$

Further,

$\frac{h_i}{h_o}=\frac{v}{u}$

$\Rightarrow \frac{-I}{L}=\frac{\frac{f_e(f_o+f_e)}{f_o}}{-(f_o+f_e)}$

$\Rightarrow \frac{f_o}{f_e}=\frac{L}{I}$

$\Rightarrow m=\frac{L}{I}$

​​​​​​​Hence, option $\left(D\right)$ is the correct answer.