Question

A microscope has an objective of focal length $1.5 \,cm$ and eye piece of focal length $2.5 \,cm$. If the distance between objective and eyepiece is $25 \,cm$. 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 $L=v_{0}+f_{e}$
$v_{0}=L-f_{e}$
$=25-2.5=22.5\, cm$
Now applying $\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$
we have $\frac{1}{22.5}-\frac{1}{u_{0}}=\frac{1}{1.5}$
$\therefore \left|u_{0}\right| \approx 1.6 \,cm$
$\therefore |M|=\frac{v_{0}}{u_{0}} \times \frac{D}{f_{e}}$
$=\left(\frac{22.5}{1.6}\right)\left(\frac{25}{2.5}\right) \approx 140 \approx 140$