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. Let the separation between the mirror and the lens be $10\times k$ . Find $k$ .

Answer 1

Let us first locate the image of $S$ formed by the lens $L$ . Let's name this first image be $I_{1}$ .
Here, $u=-12\,cm$ and $f=15\,cm$ .
We have,
$\frac{1}{ v } - \frac{1}{ u ⁡} = \frac{1}{f}$
or, $\frac{1}{ v } = \frac{1}{ f ⁡} + \frac{1}{ u ⁡}$
$\frac{1}{v}=\frac{1}{15}-\frac{1}{12}$
or, $v=-60\text{ cm}$ .

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 a source for the mirror. The mirror forms an image $I_{2}$ of the source $I_{1}$ . This image $I_{2}$ then acts as a 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,
$I_{1}I_{2}=I_{1}P+PI_{2}=60\,cm+15\,cm=75\,cm$ .
Suppose the distance of the mirror from $I_{2}$ is $x\,cm$ . For reflection from the mirror,
$u=MI_{1}=-\left(\right.75+x\left.\right)cm\text{,}v=-xcm\text{and}f=-20\,cm$ .
Using, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ ,
$\frac{1}{x}+\frac{1}{75 + x}=\frac{1}{20}$
or, $\frac{75 + 2 x}{\left(75 + x\right) x}=\frac{1}{20}$
or, $x^{2}+35x-1500=0$
or, $x=\frac{- 3 5 \pm \sqrt{3 5 \times 3 5 + 4 \times 1 5 0 0}}{2}$ .
This gives, $x=25\text{or}-60$ .
As the negative sign has no physical meaning, only the positive sign should be taken. Taking $x=25$ , the separation between the lens and the mirror is, $\left(\right.15+25\left.\right)cm=40\,cm$ .
Hence, $k=4$ .