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

Here, object size, `h_(1) = 5.0 cm`, object distance, `u = - 20cm`
radius of curvature, `R = 30 cm`, image distance, `v = ?`, image size, `h_(2) = ?`
As `(1)/(v) + 1/u = (1)/(f) = 2/R`, `(1)/(v) = 2/R - 1/u = 2/30 + 1/20 = (4+3)/60 = 7/60` or `v = 60/7 = 8.57 cm`
Positive sign of v indicates that image is at the back of the mirror. It must be virtual and erect.
As `m = (h_(2))/(h_(1)) = - v/u, (h_(2))/(5.0) = (-60//7)/(-20) = (3)/(7)`
`h_(2) = 3/7 xx 5.0 = (15.0)/(7) = 2.1 cm`
This is the size of the erect image.