Object distance, \(u = −20\ cm \)
Object height, \(h = 5\ cm\)
Radius of curvature, \(R = 30\ cm \)
Radius of curvature is twice the focal length, \(R = 2f \)
\(f = 15\ cm \)
Using the mirror formula,
\(\frac 1v+\frac 1u=\frac 1f\)
\(\frac 1v=\frac 1f-\frac 1u\)
\(\frac 1v=\frac {1}{15}+\frac {1}{20}\)
\(\frac 1v=\frac {4+3}{60}\)
\(\frac 1v=\frac {7}{60}\)
\(v= 8.57\)
The positive value of v indicates the image is formed behind the mirror.
Magnification, \(m=-\frac {\text {Image\ distance}}{\text{Object\ distance}}\)
\(m =-\frac {8.57}{-20}\)
\(m=0.428\)
The positive value of magnification indicates the image formed is virtual.
Magnification, \(m=-\frac {\text {Height \ of the \ Image}}{\text{Height \ of the \ Object}}\)
\(m = \frac {h'}{h}\)
\(h'=m \times h\)
\(h' =0.428 \times 5\)
\(h'=2.14\ cm\)
The positive value of image height indicates the image formed is erect.
Therefore, the image formed is virtual, erect, and smaller than the object.