Q. If the focal length of objective and eye lens are 1.2 $cm$ and $3\ cm$ respectively and the object is put $1.25\ cm$ away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is
A) 150
B) 200
C) 250
D) 400
Explanation
(b) $m_{\infty} = - \frac{v_{o}}{u_{o}} \times \frac{D}{f_{e}}$
From $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$
$\Rightarrow \frac{1}{( + 1.2)} = \frac{1}{v_{o}} - \frac{1}{( - 1.25)} \Rightarrow v_{o} = 30\ cm$
$\therefore\left| m_{\infty} \right| = \frac{30}{1.25} \times \frac{25}{3} = 200\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.