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,

\( \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\( \frac{1}{u}=\frac{1}{f}+\frac{1}{u}\)

\( =\frac{1}{15}+\frac{1}{20}=\frac{4+3}{60}=\frac{7}{60}\)

v = 8.57 cm

The positive value of v indicates that the image is formed behind the mirror

\( Magnification,m=-\frac{Image \,Distance}{Object \,Distance}=\frac{-8.57}{-20}=0.428\)

The positive value of magnification indicates that the image formed is virtual

\( Magnification,m=\frac{Height \,of\,the\,image}{Height\,of\,the\,object}=\frac{h^1}{h}\)

\(h^1=m\times h=0.428\times5=2.14\,cm\)

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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