Question

If an astronomical telescope has objective and eye-pieces of focal length 200 cm and 4 cm respectively,then the magnifying power of the telescope for the normal vision is:

• A. 42
• B. 50
• C. 58
• D. 204

Answer 1

The correct option is C 58

It is given that,

Focal length of eye- piece f_e = 4 cm

Focal length of object is f_o = 200 cm

Least distance of distinct vision is d = 25 cm

So, magnifying power of microscope is

$M=\frac{-f_0}{f_e}(1+\frac{f_e}{d})$

$M=\frac{-200}{4}(1+\frac{4}{25})$

$=-58cm$