A point object O is placed in front of a concave mirror of focal
length `10 cm`. A glass slab of refractive index `mu=3/2` and thickness `6cm` is
inserted between object and mirror. Find the position of final image when the
distance x shown in figure is

(a) `5cm` , (b) `20cm`

Explanation

As we have read in the above article the normal shift produced by a glass slab is,
`Deltax=(1-1/mu)t=(1-2/3)(6)= 2 cm`
i.e., for the mirror the object is placed at a distance
(32-`Deltax`)=32-2=30 cm from it. Applying mirror formula,
`1/v+1/u=1/for1/v-1/30=-1/10orv=-15cm`
When x=5 cm The light falls on the slab on its return journey as shown. But the slab will again shift it by distance, `Deltax=2` cm. Hence, the final real image is formed at a distance (15+ 2)= 17 cm from the mirror.

When x=20 cm This time also the final image is at a distance 17 cm from the mirror but it is virtual as shown.