Question

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 refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror . Find the position of the final image formed.

Answer 1

Given: 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 refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror .

To find the position of the final image formed.

Solution:

There will be change in the position of the object and the image due to the presence of glass slab, as glass slab causes refraction.

The glass slab makes the optical path bigger by $(\mu -1)\times d$, where $d$ is the thickness of the glass slab.

So the optical distance of the object to mirror, $u$, is not 21cm but it is $21+\left(\right(\mu -1)\times d)=21+\left(\right(1.5-1\left)3\right)=21+1.5=22.5cm$

The radius of curvature of the mirror is given as, $R=10cm$

Therefore the focal length becomes, $f=\frac{R}{2}=5cm$

Applying the mirror formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}⟹\frac{1}{5}=\frac{1}{v}+\frac{1}{22.5}⟹\frac{1}{v}=\frac{1}{5}-\frac{1}{22.5}⟹\frac{1}{v}=\frac{4.5-1}{22.5}⟹v=6.43cm$

This is also optical distance of the image.

6.43-1.5=4.93 cm is the actual distance of the image from the mirror.