Question

The magnifying power of a compound microscope is $20$ and the distance between its two lenses is $30cm$ when the final image is at the near point of the eye. If the focal length of eyepiece is $6.25cm$, the focal length of objective is :

• A. $2.5cm$
• B. $3.5cm$
• C. $4.5cm$
• D. $5.0cm$

Answer 1

The correct option is D $5.0cm$

$M=20$
$L=30cm$
$v_e=D=-25cm$
$f_e=6.25$
$f_o=?$

Using lens formula for eyepiece:

$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$

$\frac{1}{-25}-\frac{1}{u_e}=\frac{1}{6.25}$

$\frac{1}{-25}+\frac{1}{-6.25}=\frac{1}{u_e}$

$u_e=5cm$

From diagram length between lenses is:

$L=v_o+u_e$

$30=v_o+5$

$\Rightarrow v_o=25cm$

In compound microscope the magnification is given by:

$M=\frac{v_o}{u_o}(1+\frac{D}{f_e})$

$20=\frac{25}{u_o}(1+\frac{25}{6.25})\Rightarrow u_o=\frac{25}{4}$

Using lens formula for objective lens:

$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$

$\frac{1}{25}-\frac{4}{-25}=\frac{1}{f_o}$

$\Rightarrow f_o=5cm$