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Use the mirror equation to deduce that

The virtual image produced by a convex mirror is always diminished in size.

Asked by prakriti12oct | 27 Jun, 2019, 02:20: AM
In Convex mirror , image is always formed behind the mirror.

When object is at infinite distance , a point image is formed behind the mirror  at a distance equals to focal point of mirror .

When object is moved towards pole of mirror from infinite distance , image will move from focal point to pole behind the mirror.

we have mirror equation :-  (1/v) + (1/u ) = 1/f

where v is distance from pole of mirror to image, u is distance from pole of mirror to object and f is focal length.

By Cartesian sign convention, focal length f is positive , image distance v is positive and object distance u is negative.

Hence we get,  (1/v) = (1/f ) + (1/u)

( all distances in above expression is magnitude only )

Hence we get v = ( f u ) / ( f + u )  .........................(1)

when u   , we get ,
Otherwise , eqn.(1) is written as

From above expression we see that , v < u

magnification m = - v / u

Since u is negative by cartesian sign convention  and v < u , then magnification m < 1 and m is positive

Hence , in convex mirror , we always get , erect and diminished virtual image behind mirror

Answered by Thiyagarajan K | 21 Sep, 2021, 01:02: PM

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