Question

A compound microscope has an objective of focal length 4mm and an eyepiece of focal length $25$ mm. The objective produces a real image at a distance of $180$ mm. If the eye-piece is in normal adjustment, then the magnification is

• A. $45$
• B. $90$
• C. $225$
• D. $440$

Answer 1

The correct option is D $440$

$f_0=4mm$

$f_e=25mm$

$v_0=180mm$

$M_D=?$

$M_D=\frac{v_0}{v_0}[1+\frac{D}{fe}]$

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

$\frac{1}{180}-\frac{1}{v_0}=\frac{1}{4}$

$\frac{1}{v_0}=\frac{11}{45}$

$M_D=\frac{180\times 11}{45}\left[10\right]=\frac{180\times 11}{45}\times 10$ $=440$