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 distance, u = -30 cm (To the left of mirror)
Image distance, v = ? (To be calculated)
Radius of curvature, R = 30 cm (It is a convex mirror)
`"So, Focal length, "f=R/2=30/2=+15 cm`
Now, putting these values of u and f in the mirror formula :
`1/v+1/u=1/f`
`"We get: "1/v+1/(-20)=1/15`
` 1/v=1/15+1/15`
`1/v=1/15+1/20`
`1/v=(4+3)/60`
`1/v=7/60`
`v=60/7`
`"or "v = + 8,57 cm`
Thus, the position of image is 8.57 cm behind the mirror (on its right side). Since the image is formed behind the convex mirror, therefore, the nature of imgae is virtual and erect.
Now, For a mirror, Magnification, `m = - v/u`
`"So, "m = -(+8.57)/(-20)`
m = + 0.42
`"Also, Magnification "=("height of image")/("height of object")`
`"or "m=h_(2)/h_(1)`
`"So, "+0.42 = h_(2)/(5.0)`
`h_(2)=0.42xx5.0`
`"Height of image, "h_(2) = 2.1 cm`
Thus, the size of image 2.1 cm.