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.

Explanation

Let us first locate the image of S formed byi the lens L. Here `u=-12cm and f=15 cm`. We have
`1/v-1/u=1/f`
`or 1/v=1/f+1/u`
`=1/(15cm)-1/(12cm)`
`or v=-60cm`
The negative sign shows that the image is formed to the left of the lens as suggeted in the ure. The image `I_1` acts as the source for the mirror. The mirror forms and image `I_2` of the source `I_1`. This image I_2 then acts as the source for the lens and the fiN/Al beam comes out parallel to the principal axis. Clearly `I_2` must be at the focus of the lens. We have
`I_1I_2=I_1L+L_2=60cm15cm=75cm`.
Suppose the distance of the mirror from `I_2` is x cm. For the reflection from the mirror.
`u=MI_1(=-(75+x)cm, v=-xcm =75cm`
Suppose the distance of the mirror from `I_2` is cm. For the reflection from the mirror.
`u=MI_1=-(75+x)cm,v=-xcm and f=-20cm`
Using `1/v+1/u=1/f`
`1/x+1/(75+x)=1/20`
`or (75+2x)/((75+x)x)=1/20`
`or x^2+35-1500=0`
or `x=(-35+-sqrt(35xx35+4x1500))/2`
This gives x=25 or -60.
As the negatives sign has no physical meaning only positive sigh should be taken. Taking `x=25` the separatioin between the lens and the mirror is `(15+25)cm=40cm.`