An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

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,

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\) \(\frac{1}{v}=\frac{1}{15}+\frac{1}{20}\) \(=\frac{4+3}{60}=\frac{7}{60}=8.57 cm\)

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 of mgnification indicates that the image is formed is virtual

Magnification = Height of the image / Height of the object = h’/h

h’= m x h = 0.428 X 5 = 21.4 cm

The positive value of image height indicates that the image formed is erect.

Hence, the image formed is erect, virtual, and smaller in size.

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