Question

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.

Answer 1

Object distance, \(u = −27\ cm\)
Object height, \(h = 7\ cm\)
Focal length, \(f = −18\ cm\)
According to the mirror formula,
\(\frac 1v+\frac 1u=\frac1f\)

\(\frac1v=\frac 1f-\frac 1u\)

\(\frac 1v=-\frac {1}{18}+\frac {1}{27}\)

\(\frac 1v=-\frac {1}{54}\)
\(v=-54\ cm\)
The screen should be placed at a distance of \(54\ cm\) in front of the given mirror.
Magnfication, \(m=-\frac {\text {Image\ distance}}{\text {Object\ distance}}\)
\(m =-\frac {54}{27}\)
\(m=-2\)
The negative value of magnification indicates that the image formed is real. 
Magnfication, \(m=\frac {\text{Height\ of\ the\ image}}{\text {Height\ of\ the\ Object}}\)
\(m=\frac {h'}{h}\)
\(h'=m\times h\)
\(h'=7 \times (-2)\)
\(h'=-14\ cm\)
The negative value of image height indicates that the image formed is inverted.