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 object from the objective, when the final image seen by the eye is $25\text{cm}$ from the eyepiece.

• A. $8\text{cm}$
• B. $4\text{cm}$
• C. $3.6\text{cm}$
• D. $2.3\text{cm}$

Answer 1

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

Given, $f_0=2\text{cm};f_e=5\text{cm}L=\left|\begin{array}{c}{v}_0\end{array}\right|+\left|\begin{array}{c}{u}_e\end{array}\right|=20\text{cm}{v}_e=-25\text{cm}$

From lens formula,

$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$

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

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

Distance of real image from objective,

$v_0=20-|u_e|=20-\frac{25}{6}$

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

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

$\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}=\frac{1}{\left(\frac{95}{6}\right)}-\frac{1}{2}$

$=\frac{6}{95}-\frac{1}{2}=\frac{12-95}{190}=-\frac{83}{190}$

$\therefore u_0=-\frac{190}{83}=-2.3\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.