Question

An astronomical telescope, in normal adjustment position, has magnifying power $5$. The distance between the objective and the eye-piece is $120$ cm. Calculate the focal lengths of the objective and of the eye-piece.

• A. $100$cm,$20$m
• B. $20$m and $100$cm
• C. $100$cm and $100$cm
• D. $20$m and $20$m

Answer 1

The correct option is A $100$cm,$20$m

Here, Magnifying power ,$m=5$

$f_0=$Focal length of the objective

$f_e=$Focal length of the eye-piece

$f_0+f_e=120$cm

$m=\frac{f_0}{f_e}=5$

$\Rightarrow f_0=5f_e$

$\Rightarrow f_0+f_e=120$cm

$\Rightarrow 5f_e+f_e=120$cm

$\Rightarrow 6f_e=120$cm

$\Rightarrow f_e=\frac{120}{6}=20$cm

And $f_0=100$cm