Question

The separation L between the objective ($f_o=$ 0.5 cm) and the eye piece($f_e=5cm$)of a compound microscope is 7.0 cm. Where should a small object be placed so that the eye is least strained ?

• A. $\frac{1}{2}cm$
• B. $\frac{3}{2}cm$
• C. $\frac{2}{3}cm$
• D. $\frac{1}{3}cm$

Answer 1

The correct option is C $\frac{2}{3}cm$

$f_o=0.5cm$

$L={\gamma }_o+fe$

$f_e=5cm$

$U_e=f_e$

$L=7cm$

$M=\frac{v_o}{u_o}\frac{D}{f_e}$

$7=v_o+5$

${\gamma }_o=2cm$

$\frac{1}{2}-\frac{1}{u_o}=\frac{+2}{0}$

$\frac{1}{u_o}=\frac{1}{2}\mp 2$

$\frac{1}{u_o}=\frac{1}{2}\mp \frac{4}{2}=\frac{-3}{2}$

$\frac{1}{u_o}=\frac{-3}{2}$

$u_o=\frac{2}{3}cm$