Question

The separation L between the objective (f${}_o$ = 0.5 cm) and the eyepiece (f${}_o$ = 5 cm) 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 and what is the angular magnification produced?

• A. $u=\frac{2}{3}cm,m=15$
• B. $u=\frac{1}{3}cm,m=17$
• C. $u=\frac{2}{5}cm,m=24$
• D. $u=\frac{2}{7}cm,m=18$

Answer 1

The correct option is A $u=\frac{2}{3}cm,m=15$

$L=7cm$ $f_0=0.5cm$ $f_e=5cm$ $7=v_0+f_e$
$\Rightarrow v_0=2cm$

$\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$

$\frac{1}{2}-\frac{1}{u}=+2$

$\frac{1}{2}-2=\frac{1}{u}$

$u=-\frac{2}{3}cm$

For relaxed eye
$u_e=f_e$

$M=\frac{v_0}{u_0}\times \frac{D}{f_e}$

$L=v_0+f_e$

$M=\frac{2\times 3}{2}\times \frac{25}{5}$

$=15$