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
Answer & explanation
Correct answer: option 1
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 -
$M=M_0×M_e$
$=-2×3.5=7$