Question

an object is placed at 4cm distance in front of a concave mirror of radius of curvature 24 cm. find the position of image. is the image magnified?

Answer 1

hello friend
position of image is 6cm and magnified by 1.5

Answer 2

Answer:

Image distance, v = 6 cm. It is 1.5 times magnified.

Explanation:

It is given that,

Object distance, u = -4 cm

Radius of curvature, R = 24 cm

Focal length, f = -12 cm

Using mirror's equation :

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-4}$

v = 6 cm

The image is formed at a distance of 3 cm in front of mirror.

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-6}{-4}$

m = 1.5

The image is 1.5 times magnified. Hence, this is the required solution.