Question

A compound nmicroscope has an objective and eye-piece as thin lenses of focal lengths 1 cm and 5 cm respectively.The distance between the objective and the eye piece is 20 cm. The distance at which the object must be placed in front of the objective if the final image is located at 25 cm from the eye-piece, is numerically

• A. $\frac{95}{6}\text{cm}$
• B. $\frac{95}{89}\text{cm}$
• C. $\frac{5}{6}\text{cm}$
• D. $\frac{25}{6}\text{cm}$

Answer 1

The correct option is B $\frac{95}{89}\text{cm}$

Final image distance $L=v_0+u_e=20$

for objective $\frac{1}{f_0}=\frac{1}{v_0}+\frac{1}{u_p}$

for eye piece $\frac{1}{fe}=\frac{-1}{25}+\frac{1}{4}$

or $\frac{1}{5}+\frac{1}{25}=\frac{1}{u_e}(f_e=5cm)$

$\Rightarrow \frac{2S+S}{25\cdot 5}=\frac{1}{u_e}$

$\therefore u_e=\frac{25}{6}cm$

$V_0=20-4e$

$V_0=20-\frac{25}{6}=\frac{95}{6}cm$

$f_0=1cm$

$\therefore \frac{1}{v_0}+\frac{1}{4_0}=1$

$\therefore \frac{1}{95/6}+\frac{1}{40}=1$

$\Rightarrow \frac{1}{40}=1-\frac{9}{95}$

$\Rightarrow u_0=-\frac{95}{89}cm$ (magnitude $=\frac{95}{89})$

Hence option (B) is correct