An object `5.0 cm` in length is placed at a distance of `20 cm` in front of a convex mirror of radius of curvature `30 cm`. Find the position of image, its nature and size.

Explanation

Object distance, u = -20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = `2 xx" Focal length"`
`ddotu=ddotu implies = R/2 = ddotu`
According to the mirror formula
`1/f=1/v+1/u`
`1/v=1/f-1/u=(1)/(15)-(1)/(-20)`
`implies 1/v=(1)/(15)+(1)/(20)=(4+3)/(60)=(7)/(60)`
v=8.57 cm
The positive value of v indicates that the image is formed behind the mirror.
Magnification, `m= -("Images distance")/("Object distance")`
`=(-8.57)/(-20) = 0.428`
Positive value of magnification shows that image formed is virtual.
u = -20 cm`" "`m= -3
`implies h.=m xx h=0.428xx 5 = 2.14 cm`
The height of image is positive which shows that image formed is erect.
So, a virtual, erect and smaller than object image is formed.