Question

The magnifying power of a telescope is $9$ in normal adjustment. When it is adjusted for parallel rays, the distance between the objective lens and the eyepiece is $20\text{cm}$. The focal length of the lenses are -

• A. $10\text{cm},10\text{cm}$
• B. $15\text{cm},5\text{cm}$
• C. $18\text{cm},2\text{cm}$
• D. $11\text{cm},9\text{cm}$

Answer 1

The correct option is C $18\text{cm},2\text{cm}$

Given:
$m=9$
$L=20\text{cm}$

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

The angular magnification is given by,

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

$\Rightarrow 9=\frac{f_o}{f_e}$

$\Rightarrow f_o=9f_e$

Also, tube length,

$L=f_o+f_e$

$\Rightarrow 20=9f_e+f_e$

​​​​​​​$\Rightarrow f_e=2\text{cm}$

So, $f_o=9f_e=9\times 2=18\text{cm}$

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