When an object is placed 20 cm from a concave mirror, a real image magnified three times is formed. Find :
(a) the focal length of the mirror.
(b) Where must the object be placed to give a virtual image three times the height of the object?
Given: u = -20 cm, m = -3 cm, for the real image
(a) We know that
m = \(\frac{v}{u}\)
m = -3 = -\(\frac{v}{-20}\)
⇒ v = -60 cm
We have
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
⇒ \(\frac{1}{-60}\) + \(\frac{1}{-20}\) = \(\frac{1}{f}\)
⇒ \(\frac{1}{f}\) = -\(\frac{1}{60}\) - \(\frac{1}{20}\)
= \(-\frac{-1 - 3}{60}\) = - \(\frac{1}{15}\)
∴ f = -15 cm
(b) For virtual image m = 3, and f = -15 cm
We know that
m = \(-\frac{v}{u}\)
m = 3 = \(-\frac{v}{u}\)
⇒ v = -3u
We have,
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
⇒ \(\frac{1}{-3u}\) + \(\frac{1}{u}\) = \(\frac{1}{-15}\)
⇒ \(\frac{-1 + 3}{3u}\)
= -\(\frac{1}{15}\)
⇒ u = -\(\frac{2 \times 15}{3}\)
= -10 cm
So object should be placed 10 cm from the concave mirror.