Question

An object is placed at a distance of 40 cm on the principal axis of a concave mirror of radius of curvature 30 cm . By how much does the image move if the object is shifted towards the mirror through 15 cm?

Answer 1

Given: An object is placed at a distance of 40 cm on the principal axis of a concave mirror of radius of curvature 30 cm .

To find the distance by which the image move if the object is shifted towards the mirror through 15 cm

Solution:

As per the given criteria,

Radius of curvature, $R=-30cm$

So, the focal length, $f=\frac{R}{2}=\frac{-30}{2}=-15cm$

object distance, $u=-40cm$

Applying the lens formula, we get

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}⟹\frac{1}{(-15)}=\frac{1}{v}+\frac{1}{(-40)}⟹\frac{1}{v}=\frac{1}{(-15)}+\frac{1}{\left(40\right)}⟹\frac{1}{v}=\frac{-8+3}{120}⟹v=-24cm$

The distance between the image and the mirror is 24cm when the object distance is 40cm.

Now, the object is moved 15cm towards the mirror, so the new object distance is $u^{\prime }=-(40-15)=-25cm$

Now applying the lens formula we get,

$\frac{1}{f}=\frac{1}{v^{\prime }}+\frac{1}{u^{\prime }}⟹\frac{1}{(-15)}=\frac{1}{v^{\prime }}+\frac{1}{(-25)}⟹\frac{1}{v^{\prime }}=\frac{1}{(-15)}+\frac{1}{\left(25\right)}⟹\frac{1}{v^{\prime }}=\frac{-5+3}{75}⟹v=-37.5cm$

is the new position of the image

So the shift in the position of the image is $=40-37.5=2.5cm$