Question

The focal length of the objective and eyepiece of a microscope are $2\text{cm}$ and $5\text{cm}$ respectively, and the distance between them is $20\text{cm}$. Find the distance of the object from the objective, when the final image seen by the eye is $25\text{cm}$ from the eyepiece.

• A. $3\text{cm}$
• B. $2.3\text{cm}$
• C. $3.2\text{cm}$
• D. $4\text{cm}$

Answer 1

The correct option is B $2.3\text{cm}$

Given:
$f_0=2\text{cm},f_e=5\text{cm}$

$|v_o|+|u_e|=20\text{cm}$

The final image is formed by the eyepiece,
$\therefore v_e=-25\text{cm}$

Applying lens formula for the eyepiece,
$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$

$\Rightarrow \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{5}$

$\Rightarrow u_e=-\frac{25}{6}\text{cm}$

$\therefore$ Distance of real image from objective
$v_o=20-|u_e|=20-\frac{25}{6}$

$\Rightarrow v_o=\frac{120-25}{6}=\frac{95}{6}\text{cm}$

Now, applying lens formula for objective,
$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$

$\Rightarrow \frac{1}{u_o}=\frac{1}{v_o}-\frac{1}{f_o}=\frac{1}{\left(95/6\right)}-\frac{1}{2}$

$\Rightarrow \frac{1}{u_o}=\frac{6}{95}-\frac{1}{2}=\frac{12-95}{190}$

$\Rightarrow \frac{1}{u_o}=-\frac{83}{190}$

$\Rightarrow u_o=-\frac{190}{83}=-2.3\text{cm}$

Hence, option $\left(B\right)$ is correct.