Question

A converging lens of focal length 15 cm and a converging mirror of focal length 20 cm are placed with their principal axes coinciding. A point source S is placed on the principal axis at a distance of 12 cm from the lens as shown in figure It is found that the final beam comes out parallel to the principal axis. Find the separation between the mirror and the lens.

• A. 20 cm
• B. 40 cm
• C. 60 cm
• D. 80 cm

Answer 1

The correct option is B

40 cm

Let us first locate the image of S formed by the lens L. Here u = -12 cm and f = 15 cm. We have,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
or, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$=\frac{1}{15cm}-\frac{1}{12cm}$
or, $v=-60cm.$
The negative sign shows that the image is formed to the left of the lens as suggested in the figure. The image $I_1$ acts as the source for the mirror. The mirror forms an image $I_2$ of the source $I_1$. This image $I_2$ then acts as the source for the lens and the final beam comes out parallel to the principal axis. Clearly $I_2$ must be at the focus of the lens. We have,
$I_1{I}_2=I_{1}L+L{I}_2$ = 60 cm + 15 cm = 75 cm.
Suppose the distance of the mirror from $I_2$ is x cm. For the reflection from the mirror,
$u=M{I}_1$ = -(75+x)cm, v=-x cm and f = - 20 cm.
Using, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{x}+\frac{1}{75+x}=\frac{1}{20}$
or, $\frac{75+2x}{(75+x)x}=\frac{1}{20}$
or, $x^2+35x-1500=0$
or, $x=\frac{-35\pm \sqrt{35\times 35+4\times 1500}}{2}$
This gives x = 25 or - 60.
As the negative sign has no physical meaning, only positive sign should be taken. Taking x = 25, the separation between the lens and the mirror is (15 + 25) cm = 40 cm.