Question

What will be position of the final image?

An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and `mu=1.50` is then placed close to the mirror in the space between the object and the mirror. If the distance of the near surface of the slab from the mirror is 1 cm, then what will be position of the final image?

Answer 1

Let the object be placed at the point `O` on the principal axis of the concave mirror at a distance 21cm from it. The paraxial rays emerging from mirror will be refracted through the 3cm thick glass slab before being incident on the mirror and so the object will appear at the shifted position `O'` on the principal axis due to refraction through glass slab. This shift `OO'` is given by

`OO'=t(1-1/mu)`,where `t=3cm` is the thickness and `mu=1.5` is the refractive index of the slab.

So `OO'=3(1-1/1.5)=1cm`

Hence the apparent object distance will be `u=-(21-1)=-20cm`

Focal length `f=r/2=-10/2=-5cm`

Image distance `v=?` to be found out.

Now applying relation

`1/v+1/u=1/f`

`=>1/v-1/20=-1/5`

`=>1/v=-1/5+1/20=-3/20`

`=>v=-20/3` cm.

Hence the mirror will try to form image at a distance `20/3` cm in front of it. But due to refraction through the glass slab again the image will be shifted by 1cm again towards the direction of reflected light.

Finally image will be formed at a distance `20/3+1~~7.67` cm from the mirror.