Question

an object 4.0 cm in size is placed at a distance of 25.0 cm in front of a convex mirror of radius of curvature 40 cm. find the position ,size and nature of the image

Answer 1

f = -15 cm; u = -25 cm ; ${ H }{ 0 }$ = 4 cm; v = ?
The mirror formula is 1/f = 1/u + 1/v = 1/-15 = 1/-25 + 1/v
1/v = 1/25 - 1/15
1/v = 3-5/ 75 =-2/ 75
v =-37.5 cm
The screen should be placed 37.5 cm from the pole of mirror and the image is real.
Magnification, m = ${ H }{ i }$ / ho = -v/u = hi / 4 = 37.5 / 25
hi = 37.5 / 25 x 4 = -6 cm
So the image is enlarged and inverted.

Answer 2

Answer:

Here ,h= +4.0 cm

U=–25.0 cm

R= +40 cm

F=R/2 =40/2=20.0cm

(i) DETERMINATION OF THE POSITION OF THE IMAGE

USING ,1/u+1/v=1/f , we get

1/v=1/f–1/u=1/20–1/(–25)=1/20–1/25=9/100

Or v=100/9=11.11cm

THUS THE IMAGE AT 11.11 CM BEHIND THE VONVEX MIRROR

(ii) DETERMINATION OF SIZE

USING ,m=h'/h=–v/u or h'=–v/u×h

Or h'=(100/9)(4)=–16/9 =–1.78

Explanation: