Question

An object $2\, cm$ high is placed at a distance of $16 \,cm$ from a concave mirror, which produces a real image $3 \,cm$ high. What is the focal length of the mirror?

• A. $-9.6\,cm$
• B. $-3.6\,cm$
• C. $-6.3\,cm$
• D. $-8.3\,cm$

Answer 1

Answer: (A)

Here, $h_1 = 2\, cm, u = -16\, cm$
$h_2 = -3\,cm$ (since image is real and inverted)
$\because m =\frac{h_{2}}{h_{1}} =-\frac{v}{u} $
$\therefore v=\frac{-h_{2}}{h_{1}} u = \frac{3}{2}\times\left(-16\right) $
$= -24\, cm$
$ \frac{1}{f} = \frac{1}{v}+\frac{1}{u} $
$= -\frac{1}{24}-\frac{1}{16} $
$= \frac{-2-3}{48} = \frac{-5}{48};$
$ f = \frac{-48}{5} = -9.6 \,cm $