Use the mirror equation to deduce that

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


Asked by prakriti12oct | 27th Jun, 2019, 02:20: AM

Expert Answer:

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 begin mathsize 14px style rightwards arrow infinity end style  , we get , begin mathsize 14px style v space equals space fraction numerator f space u over denominator f space plus space u end fraction space equals space fraction numerator f over denominator begin display style f over u end style plus 1 end fraction rightwards arrow f end style
Otherwise , eqn.(1) is written as
begin mathsize 14px style v space equals space fraction numerator u over denominator 1 plus begin display style u over f end style end fraction end style
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 | 21st Sep, 2021, 01:02: PM