Question

The length of the compound microscope is $20\text{cm}$. The focal length of the eyepiece is $5\text{cm}$. If the magnifying power for the relaxed eye is $25$, then find the object distance from the objective lens.

• A. $6\text{cm}$
• B. $5\text{cm}$
• C. $4\text{cm}$
• D. $3\text{cm}$

Answer 1

The correct option is D $3\text{cm}$

We know that, for the relaxed eye, the final image is formed at infinity.

So, $|u_e|=|f_e|$

Length of the compound microscope,

$L=|v_o|+|u_e|=|v_o|+|f_e|$

$\Rightarrow 20=|v_o|+5$

$\Rightarrow |v_o|=15\text{cm}$

Now, magnifying power for the relaxed eye is,

$m=\frac{v_o}{|u_o|}\left(\frac{D}{f_e}\right)$

$\Rightarrow 25=\frac{15}{|u_o|}\left(\frac{25}{5}\right)$

$\Rightarrow |u_o|=3\text{cm}$

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