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.
Text Solution
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Here, `h_(1) = 4.5cm, u = - 12cm, f = 15cm, v = ?, m = ?` As `(1)/(u) + (1)/(v) = (1)/(f) :. (1)/(v) = (1)/(f) - (1)/(u) = (1)/(15) + (1)/(12) = (4 + 5)/(60) = (9)/(60)` `:.` 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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