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  4. An object is placed in front of a convex mirror at a distance equal to focal length of mirror. The magnification of the object is

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Q. An object is placed in front of a convex mirror at a distance equal to focal length of mirror. The magnification of the object is

AMUAMU 1998

Solution:

: For a mirror, $ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} $ For convex mirror $ f $ is positive For virtual image $ v $ is positive $ u $ is always negative as per convention $ \therefore $ $ \frac{1}{v}-\frac{1}{f}=\frac{1}{f} $ or $ \frac{1}{v}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f} $ or $ v=\frac{f}{2} $ $ \therefore $ $ \frac{v}{u}=\frac{f}{f\times 2}=\frac{1}{2} $ $ \therefore $ Magnification $ =\frac{1}{2} $