Question

The focal lengths of the objective and the eyepiece of a compound microscope are $2.0cm$ and $3.0cm$, respectively. The distance between the objective and the eyepiece is $15.0cm$. The final image formed by the eyepiece is at infinity. The two lenses are thin. The distance, in $cm$, of the object and the image produced by the objective, measure from the objective lens, are respectively.

• A. $2.4$ and $12.0$
• B. $2.4$ and $15.0$
• C. $2.0$ and $12.0$
• D. $2.0$ and $3.0$

Answer 1

The correct option is A $2.4$ and $12.0$

Here, $f_v=2cm$ and $f_e=3cm$.
Using lens formula for eyepiece, $-\frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_e}$
$\Rightarrow \frac{-1}{u}+\frac{1}{\propto }=\frac{1}{3}\Rightarrow u_1=-3cm[\because i=0]$
But the distance between objective and eyepiece is $15cm$ (given).
Therefore, distance of image formed by the objective, $v=15-3=12cm$. Let $u$ be the object distance from the objective, then for objective lens $-\frac{1}{u}+\frac{1}{v}=\frac{1}{f_0}$ or $\frac{-1}{u}+\frac{1}{12}=\frac{1}{2}$
$\Rightarrow \frac{-1}{u}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}u=-\frac{12}{5}=2.4cm$
a. is the correct option.