Radius of curvature, r = 2f
24 cm = 2f
2f = 24
f = 24/2 = 12
Hence the focal length is = 12 cm.
Object distance, u = 4 cm.
Since the object is placed between the focus and the pole of the mirror, at this position for a concave mirror, the resulting image would be virtual and would be behind the mirror.
1/u - 1/v = 1/f
1/4 - 1/v = 1/12
1/4 - 1/12 = 1/v
1/v = 1/4 - 1/12 = (3 - 1)/12 = 2/12 = 1/6
1/v = 1/6
v = 6
Therefore the image is 6cm behind the mirror.
Magnification = v/u = 6/4 = 3/2 = 1.5
Yes, the image is magnified by 1.5 times.
Simply copy the diagram for an object placed between the focus and the pole of a mirror from your textbook.
