Question

Find the maximum magnifying power of a compound microscope having a $25\text{D}$ lens as the objective, a $5\text{D}$ lens as the eyepiece and a separation of $30\text{cm}$ between the two lenses. The least distance for clear vision is $25\text{cm}$.

• A. $3.3$
• B. $6.8$
• C. $8.5$
• D. $9.1$

Answer 1

The correct option is C $8.5$

Given:
Power of objective, $P_o=25\text{D}$
Power of eyepiece, $P_e=5\text{D}$
Length of compound microscope, $L=30\text{cm}$.
$D=25\text{cm}$
$f_o=\frac{1}{P_o}=\frac{1}{25}\text{m}=4\text{cm}$
$f_e=\frac{1}{P_e}=\frac{1}{5}\text{m}=20\text{cm}$

Magnifying power is maximum when the final image is formed at the least distance of clear vision (i.e. DDV).
$\Rightarrow v_e=D=25\text{cm}$
Now, for the eye - piece,
$v_e=-25\text{cm},f_e=20\text{cm}$

Using lens formula for the eyepiece,
$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$
$\frac{1}{20}=\frac{1}{-25}-\frac{1}{u_e}$
$\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{20}$
$\therefore u_e=-11.1\text{cm}$

The distance between the two lenses is, $L=v_o+|u_e|$
So, for the objective lens, the image distance should be,
$v_0=30-\left(11.1\right)=18.9\text{cm}$

Now, for the objective lens.
$v_0=18.9\text{cm},f_0=4\text{cm}$

Using lens formula for the objective lens,
$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$
$\Rightarrow \frac{1}{4}=\frac{1}{18.9}-\frac{1}{u_o}$
$\Rightarrow u_o\approx -5\text{cm}$

The maximum magnifying power is given by,
$m=-\frac{v_o}{u_o}(1+\frac{D}{f_e})$
$m=\frac{-18.9}{-5}(1+\frac{25}{20})=8.5$

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

Why this question? Concept : Maximum magnified image is obtained when the image is formed at least distance of clear vision from the eye - piece.