Question

The length of the compound microscope is $18\text{cm}$. If the magnifying power for the relaxed eye is 28, find the object distance for the objective lens. Focal length of the eyepiece is $4\text{cm}$ and $D=25\text{cm}$.

• A. $2.75\text{cm}$
• B. $3.75\text{cm}$
• C. $2.50\text{cm}$
• D. $3.12\text{cm}$

Answer 1

The correct option is D $3.12\text{cm}$

Given,
Length of the compound microscope $L=18\text{cm}$
Magnifying power, $MP=28$ (for relaxed eye)
Focal length of the eyepiece, $f_e=4\text{cm}$

We know that for relaxed eye viewing, the final image is formed at infinity. So, for eyepiece, object distance will be equal to $f_e$

Now, length of microscope $L=v_0+f_e$
$18=v_0+4$
$\therefore v_0=14\text{cm}$

Now, $MP=\frac{v_0}{u_0}\times \frac{D}{f_e}$
$\Rightarrow 28=\frac{14}{u_0}\times \frac{25}{4}$
$\therefore u_0=3.125\text{cm}$

Hence, $\left(A\right)$ is the correct answer.