Question

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

Answer 1

Answer:

Explanation:

acc to mirror formula

1/f = 1/v +1/u   here f = r/2  so f = 12

now allpying formula

-1/12=1/v+(-1/4)  ( sign convention)

v = 6 cm ( image is formed 6 cm behind mirror)

image is magnified  - v/u =    -(6/-4) = 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 :

$\frac{1}{f} = \frac{1}{v} + \frac{1}{v}$

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

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

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 :

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

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

m = 1.5

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