Question

The magnifying power of a telescope is $9$. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be $20cm$. The focal length of lenses are

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

Answer 1

Answer: (A)

$18cm,2cm$

Lets consider, $f_o=$ focal length of objective lens

$f_e=$ focal length of eyepiece

Given,

$m=9$ the magnifying power of a telescope.

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

$f_o=9f_e$. . . . . .(1)

The distance between the objective and eyepiece is $20cm$

$f_e+f_o=20cm$

$9f_e+f_e=20cm$ (from equation 1)

$10f_e=20$

$f_e=2cm$

From equation (1),

$f_o=9\times 2=18cm$

The correct option is A.