In a compound microscope the focal length of objective lens and eye lens are 4 cm and 10 cm respectively. An object is placed at 6 cm from the objective lens. If the final image is formed at the near point of eye, then the magnifying power of microscope is

7

The correct answer is Option (1) → 7

Using lens formula,

$\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$

$\frac{1}{4}=\frac{1}{v_0}-\frac{1}{-6}$

$⇒\frac{1}{v_0}=\frac{1}{4}-\frac{1}{-6}=\frac{1}{12}$

$v_0=12cm$

Magnify power of objective lens $(M_0)$ is -

$M_0=\frac{v_0}{u_0}=\frac{12}{-6}=-2$

Magnifying power of the eyepiece is,

$M_e=1+\frac{D}{f_e}=1+\frac{25}{10}$

$=3.5$

Total magnifying power of microscope $(M)$ is -

$M=M_0×M_e$

$=-2×3.5=7$