Question

The distance between two point sources of light is $24\text{cm}$. Where should be a converging lens of focal length $9\text{cm}$ must be placed, so that the images of both the point sources are formed at the same point?

• A. $6\text{cm}or18\text{cm}$ either of the sources
• B. $9\text{cm}$ from either of the sources
• C. $8\text{cm}or18\text{cm}$ from either of the sources
• D. $5\text{cm}$ from either of the sources

Answer 1

The correct option is A $6\text{cm}or18\text{cm}$ either of the sources


Let the lens is placed at a distance $x$ from the $s_1$ and image of the point source $s_1$ is formed at $v_1$.
For source $s_1$, applying the lens formula
$u=-x$
$f=+9$
$\Rightarrow \frac{1}{v_1}-\frac{1}{-x}=\frac{1}{9}$
or $\frac{1}{v_1}=\frac{1}{9}-\frac{1}{x}......$(i)
For source $s_2$, object distance
$u=-(24-x)$
$\Rightarrow \frac{1}{v_2}-\frac{1}{-(24-x)}=\frac{1}{9}$
or $\frac{1}{v_2}=\frac{1}{9}-\frac{1}{(24-x)}$....(ii)
The sign of image distance for $s_1$ and $s_2$ will be opposite.
$\Rightarrow v_1=-v_2$
or $\frac{1}{v_1}=\frac{-1}{v_2}$
$\Rightarrow \frac{1}{9}-\frac{1}{x}=-\frac{1}{9}+\frac{1}{(24-x)}$
or , $\frac{2}{9}=\frac{x+24-x}{x(24-x)}$
or, $-2x^2+48x-216=0$
or, $x^2-24x+108=0$
or, $x^2-18x-6x+108=0$
$x(x-18)-6(x-18)=0$
$\therefore x=6\text{cm}or18\text{cm}$

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: From the principle of reversibility of light if}\\ \text{position of both sources are interchanged then}\\ \text{it won't affect the final image formation.}\end{array}$