A point object O is placed in front of a concave mirror of focal length 10cm. A glass slab of refractive index `mu=3//2` and thickness 6cm is inserted between the object and mirror.
Find the position of final image when the distance x shown in Fig... is, (a) 5cm and (b) 20 cm.

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)=2cm`
i.e. for the mirror the object is placed at a distance `(32-Deltax)=30cm` from it. Applying
mirror formula
`1/v+1/u=1/f or 1/v-1/30=-1/10 or v=-15cm`
(a) When `x=5cm` The light falls on the slab on
its return journey as shown. But the slab will
again shift it by a distance `Delta x=2cm` Hence, the
final real image is formed at a distance
`(15+2)=17cm` from the mirror.

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