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Q 14.     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.

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Radius of curvature, R = 30 cm

Focal length, f = R/2 = 15 cm

Object distance, u = -20 cm

Let the image distance be v 

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{-20}\\ v=8.57\ cm

Since v is positive image is formed behind the mirror.

\\Magnification=-\frac{v}{u}\\ =-\frac{8.57}{-20}\\ =0.428

Object size = 5 cm

\\Image \ size= Object\ size \times Magnification\\ =5\times 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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