Question

When the object is placed at $4\text{cm}$ from the objective lens of the compound microscope, the final image is formed at the position of the object which is the least distance of distinct vision $\left(24\text{cm}\right)$. If the focal length of the eyepiece is $8\text{cm}$ then the focal length of the objective lens is -

Answer 1

Given:
$u_o=-4\text{cm}$
$v_e=-24\text{cm}$
$f_e=8\text{cm}$

Applying lens formula for eyepiece,

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

$\Rightarrow \frac{1}{8}=\frac{1}{(-24)}-\frac{1}{u_e}$

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

Since, the final image is formed at the position of the object, the length of the tube is given by,

$L=\left|v_e\right|-\left|u_o\right|$

$\Rightarrow L=24-4=20\text{cm}$

The length of the tube is also given by,

$L=\left|v_o\right|+\left|u_e\right|$

$\therefore |v_o|=20-6=14\text{cm}$

Further, applying lens formula for objective lens,

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

$\Rightarrow \frac{1}{f_0}=\frac{1}{14}-\frac{1}{(-4)}$

$\Rightarrow f_o\approx 3\text{cm}$

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