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 × Focal length
R = 2f
f = 15 cm
According to the mirror formula,
`1/v-1/u=1/f`
`1/v=1/f-1/u`
`=1/15+1/20=(4+3)/60=7/60`
`v=8.57cm`
The positive value of v indicates that the image is formed behind the mirror.
`"Magnification," m= -("Image Distance")/("Object Distance")=(-8.57)/-20=0.428`
The positive value maf=gnification indicates that the image formed is virtual.
`"Magnification," m=("Height of the Image")/("Height of the Object")=(h')/h`
`h'=mxxh=0.428xx5=2.14cm`
The positive value of image height indicates that the image formed is erect.
Therefore, the image formed is virtual, erect, and smaller in size.