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in Physics by (51.7k points)

An object 5 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 of convex mirror, R = 30 cm.

Focal length of convex mirror, f=R/2=30cm/2=15cm

Now h=5cm,u=-20cm,v=?,h'=?

Using the mirror formula 1/f=1/u+1/v, we have

1/v=1/f=1/u=1/15=1/-20=1/15+1/20=(4+3)/60

v=60/7=8.6cm.

Thus, image is formed at a distance of 8.6cm behind the convex mirror. The image is virtual and erect.

m=h'/h=-v/u

h'/5=-8.6/(-20)

h'=8.6/20x5=2.15cm.

Thus, the size of the image is 2.15 cm u,which is positive. It indicates that the image formed is erect, virtual and diminished.

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