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A 4.5 cm needle is placed 12 cm away fro...

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, `h_1 = 4.5 cm, u = -12 cm, f = 15 cm, v = ?, m = ?`
As `(1)/(u)+(1)/(v)=(1)/(f) :. (1)/(v)=(1)/(f)-(1)/(u)=(1)/(15)+(1)/(12)=(4 + 5)/(60)=(9)/(60)`
`v = 60//9 = 6.7 cm`
`:.` Image is virtual, formed at `6.7 cm` at the back of the mirror.
As `m = (h_2)/(h_1) = -(v)/(u) :. (h_2)/(4.5) = -(6.7)/(12)` or `h_2 = (6.7 xx 4.5)/(12) = 2.5 cm`
`:.` Image is erect, and of course virtual.
As needle is moved farther from the mirror, image moves away from the mirror (upto F) and goes on decreasing in size.
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