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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