Question

A microscope has an objective of focal length $1.5cm$ and eye piece of focal length $2.5cm$. If the distance between objective and eyepiece is $25cm$. what is the approximate value of magnification produced for relaxed eye?

• A. 75
• B. 110
• C. 140
• D. 25

Answer 1

Answer: (C)

Length of the tube is $L=v_0+f_e$
$v_0=L-f_e$
$=25-2.5=22.5cm$
Now applying $\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$
we have $\frac{1}{22.5}-\frac{1}{u_0}=\frac{1}{1.5}$
$\therefore \left|u_0\right|\approx 1.6cm$
$\therefore |M|=\frac{v_0}{u_0}\times \frac{D}{f_e}$
$=\left(\frac{22.5}{1.6}\right)\left(\frac{25}{2.5}\right)\approx 140\approx 140$