Question

A point source S is placed at a distance of $15cm$ from a converging lens of focal length $10cm$ on its principal axis. Where should a diverging mirror of focal length $12cm$ be placed so that a real image is formed on the source itself?

Answer 1

Formula for refraction through the lens is,

$u=-15cm$ and $f=\pm 10cm$

$\frac{1}{v}+\frac{1}{15}=\frac{1}{10}$

$v=30cm$

• As v is positive,the image $I_1$,is formed to the right of lens.
  
• The convex mirror is to be placed to the right of lens in such a way hat light rays fall on the mirror normally.
  
• Then only the light rays will retrace their path and form final image on the object itself.
• This shows that image $I_1$ formed by the lens should be at the center of curvature of the mirror i.e,$P{I}_1=R=2f=24cm$
• As $|v|=O{I}_1=30cm.$ ,Hence $OP=30-24=6cm$
• Thus,the convex mirror should be placed at a distance of $6cm$ to the right of lens