Question

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 the image, its nature and size.

Answer 1

Object distance, \(u = −20\ cm \)
Object height, \(h = 5\ cm\)
Radius of curvature, \(R = 30\ cm \)
Radius of curvature \(= 2 × \text {Focal length}\)
\(R = 2f \)
\(f = 15\ cm \)
According to 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 that the image is formed behind the mirror.
Magnfication, \(m=-\frac {\text {Image\ distance}}{\text{Object\ distance}}\)
\(m =-\frac {8.57}{-20}\)
\(m=0.428\)
The positive value of magnification indicates that the image is formed is virtual.
Magnfication, \(m=-\frac {\text {Height \ of the \ Image}}{\text{Height \ of the \ Object}}\) 
\(m = \frac {h'}{h}\)
\(h'=m \times h\)
\(h' =0.428 \times 5\)
\(h'=2.14\ cm\)
The positive value of image height indicates that the image formed is erect. 

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