An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Object distance, u = −20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = 2 × Focal length
R = 2f f = 15 cm
According to the mirror formula,
1/v + 1/u = 1/f
1/v = 1/f – 1/u = 1/15 + 1/20 = 4+3/60 = 7/60
V =8.57
The positive value of v indicates that the image is formed behind the mirror.
Magnification, m = – Image Distance/Object Distance = -8.57/-20 = 0.428
The positive value of Magnification indicates that the image is formed is virtual.
Magnification, m = Height of the image / Height of the object = h’/h
h’ m x h = 0.428 x 5 = 2.14 cm
The positive value of image height indicates that the image formed is erect.
Therefore, the image formed is virtual, erect, and smaller in size.

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