Question

an object of height 12cm is kept in front of a concave mirror. if the magnification of the image is 3 times the height of the object then what is the height of the image​

Answer 1

mirror formula 1/u + 1/v = 1/f

object height h = 3cm

focal length f= 15 cm

magnification m= 3

image position v = ?

size of the image h’ = ?

m =-v/u =h’/h = 3

h’=3*3=9

v=3u

Substituting the values in the mirror formula

1/u +1/3u =1/15

(3+1)/3u =1/15

3u = 60

u =20

Image is formed at a distance of 60 cm from the mirror.

When the object is between the principal focus and center of curvature, a real, inverted and magnified image is formed beyond the center of curvature.

Answer 2

Explanation:

Object distance =12cm

hi=3h0

hi= height of image

ho= height of object

Magnification, m=h0hi

=h0−3h0

m=−2

We know,

m=−uv

−3=(−12)−v

v=−33cm

Image position is 33cm from the mirror on the same side of the object