An object is placed 21 cm in front of a conacve mirror of radius of curvature 10 cm. A glass slab of thickness 3cm and refractive index 1.5 is the placed close to the mirror in the space between the object and the mirror.Find the position of the final image. take the distance between the glass slab and the mirror is 1cm.

the position of the object as well as the image will change due to presence of the glass slab as it causes refraction of the light rays.

Now, the shift in the position of the image will be shown sing the following image

The image is positioned at O but due to the refraction from glass slab it will be shifted to O' as seem by the mirror. So, the new image distance will be

u = O'P = OP - OO'

now OO' = t.(1 - 1/u)

here t = 3cm and u = 1.5

so,

OO' = 3.[1 - (1/1.5)]

or

OO' = 1 cm

so, the image distance will be

u = 21cm - 1cm = 20 cm