Question

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

Answer 1

given that

object distance from the mirror is 40 cm

focal length of the mirror is half of the radius of curvature = 15 cm

now we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{-40} = \frac{1}{-15}$

$d_i = - 24 cm$

now when object is shifted by 15 cm towards the mirror

now the distance of object is 40 - 15 = 25 cm

again by the mirror formula

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{-25} = \frac{1}{-15}$

$d_i = -37.5 cm$

$\delta X = d_i' - d_i = 37.5 - 24 = 13.5 cm$

so the image will shift by 13.5 cm

Answer 2

it's best answer and correct also.
I'm sure please mark it as brainlist.