Question

The separation L between the objective $(f=0.5cm)$ and the eye piece $(f=5cm)$ of a compound microscope is $7$cm. Where should a small object be placed, so that the eye is least strained to see the image? Also, find the angular magnification produced by the microscope.

• A. $-\frac{2}{3}cm,-15$
• B. $-\frac{1}{3}cm,-10$
• C. $-\frac{4}{3}cm,-15$
• D. $-\frac{2}{3}cm,-10$

Answer 1

Answer: (A)

$-\frac{2}{3}cm,-15$

Given, $f_0=0.5cm,f_e=5cm,L=7cm$
The eye is least strained, i.e., final image is formed at infinity.
$L=v_o+f_e$
$7=v_o+5$
$\Rightarrow v_o=2cm$
For objective, $u_o=-x$
$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$
$\frac{1}{2}-\frac{1}{(-x)}=\frac{1}{0.5}$
$\frac{1}{x}=2-\frac{1}{2}=\frac{3}{2}$
$x=\frac{2}{3}cm,u_o=-x=-\frac{2}{3}$cm
Angular magnification,
$m=\frac{v_o}{u_o}\cdot \frac{D}{f_e}$
$=\frac{2}{-2/3}\cdot \frac{25}{5}$
$=-3\times 5=-15$.