Question

A 20-cm-thick slab of glass of refractive index 1.5 is placed in front of a plane mirror and a pin is placed in front of it in the air at a distance of 40 cm from the mirror. Find the position of the image.

Answer 1

The slab of thickness t forms the image of the object O at the point O’.

The slab shifts the image of object by (t t − µ) . O’ serves as an object for the plane mirror. 

Object distance for the mirror is MO.

Since in a plane mirror, object distance = image distance.

MI' = MO' = 100/3 cm.

Now, I' serves as an object for the slab again.

I' is shifted to I" by 20 - 20/1.5 = 20/3 cm.

Distance of the final image (I") from the mirror

= 100/3 - 20/3

= 80/3 cm.