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:

mirror equation :-  begin mathsize 14px style 1 over v space plus space 1 over u space equals space 1 over f end style  ................................(1)
where v is distance from pole of mirror to image, u is distance from pole of mirror to object and f is focal length.
 
sign convention :-  v is positive because image always formed right side behind the mirror. 
                              u is -ve because object is at left side in front of mirror
                              f is +ve for convex mirror
 
After applying sign convention ,  we get   begin mathsize 14px style 1 over v space equals space 1 over f plus 1 over u end style ................................(2)
when object is at infinity, (1/u) → 0 , hence v = f ( highly diminished image formed at focal point)
 
when object moves from infinity to pole , (1/u) > 0 and keep on increasing, hence v will keep on decrease.
 
Hence v which is the distance between pole and image is constrained between focal point and pole.
 
Hence v is always less than u. Hence magnification , m = - v/u is always <1 or virtual image size is always diminished

Answered by Thiyagarajan K | 27th Jun, 2019, 10:12: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.