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 focussed image can be obtained? Find the size and nature of image.
[Hint. Find the value of image distance (v) first. The screen should be placed from the mirror at a distance equal to image distance].
Given: \(h_1\) = 7 cm, u = 27 cm, f = 18 cm
We know that
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
⇒ \(\frac{1}{v}\) = \(\frac{1}{f}\) - \(\frac{1}{u}\)
= \(\frac{1}{-18}\) - \(\frac{1}{-27}\)
= \(-\frac{1}{18}\) + \(\frac{1}{27}\)
= \(\frac{-3 + 2}{54}\) = \(-\frac{1}{54}\)
v = 54 cm
The screen should be placed at a distance of 54 cm in front of the concave mirror.
m = \(-\frac{v}{u}\) = \(\frac{h_2}{h_1}\)
⇒ \(-\frac{-54}{-27}\) = \(\frac{h_2}{7}\)
⇒ \(h_2\) = -14 cm
Image is 14 cm in size, real and inverted.