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A `4.5 cm` needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

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Here, object size, `h_(1) = 4.5 cm`
object distance, `u = -12 cm`
focal length, `f = +15 cm`
image distance, `v = ?` (to be calculated)
magnification,`m = ?`
As `1/u + 1/v = 1/f` , `1/v = 1/f -1/u`
Putting `u = -12 cm` and `f = + 15 cm` , we get
` 1/v = 1/15 - 1/(-12) = (4+5)/60` or `v = 60/9 = 6.7 cm`
i.e., image is formed `6.7 cm` behind the convex mirror. It must be virtual and erect.
If `h_(2)` is size of the image, then `m = h_(2)/h_(1) = (-v)/u` or `m = h_(2)/h_(1) =(-(6.7))/(-12) = 0.558`
i.e., `h_(2) = 0.558 h_(1) = 0.558 xx 4.5 = 2.5 cm`
As the needle is moved farther from the mirror, image moves away from the the mirror till it is at focus F of the mirror. The size of the image goes on decreasing.

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