Question

The objective lens of a compound microscope produces magnification of $10$. In order to get an overall magnification of $100$ when image is formed at $25cm$ from the eye, the focal length of the eye lens should be :(in $cm$)

• A. $4$
• B. $10$
• C. $\frac{25}{9}$
• D. $9$

Answer 1

The correct option is C $\frac{25}{9}$

If $M_0=$ magnification of objective lens of compound microscope.

$M=$ magnification of compound microscope

$\Rightarrow M=M_0(1+\frac{D}{fe})$............(1)

where $D=$ distance of nearest vision $=25cm$

$fe=$ focal length of eye lens

Given,

$m=100$

$m_0=10$

$D=25$

substituting in (1)

$=100-10(1+\frac{25}{fe})$

$fe=\frac{25}{9}cm$