Here, u = -15 cm, f = +10 cm
\(\frac{1}{v}\) − \(\frac{1}{u}\) = \(\frac{1}{f}\) ⇒ \(\frac{1}{v}\) = \(\frac{1}{f}\) + \(\frac{1}{u}\) = \(\frac{1}{10}\) + \(\frac{1}{-15}\)
= \(\frac{1}{30}\)
∴ ν = + 30 cm
Magnification, m = \(\frac{v}{u}\) = \(\frac{30}{-15}\) = −2
Nature of image: Real, inverted, magnified as |m| = 2>0
Size of image(hi): m = \(\frac{h_i}{h_o}\)
∴ hi = m x ho = -2 x 2 = -4 cm