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.

Answers (1)

Radius of curvature, R = 30 cm

Focal length, f = R/2 = 15 cm

Object distance, u = -20 cm

Let the image distance be v 

\frac1f=frac1v+frac1u\ frac115=frac1v+frac1-20\ v=8.57 cm

Since v is positive image is formed behind the mirror.

\Magnification=-fracvu\ =-frac8.57-20\ =0.428

Object size = 5 cm

\Image  size= Object size 	imes Magnification\ =5	imes 0.428\ =2.14 cm

A virtual, erect and diminished image of size 2.14 cm would be formed 8.57 cm behind the mirror.

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