Q. A cube of side $2\ m$ is placed in front of a concave mirror focal length $1m$ with its face $P$ at a distance of $3\ m$ and face $Q$ at a distance of $5\ m$ from the mirror. The distance between the images of face $P$ and $Q$ and height of images of $P$ and $Q$ are
A) $1\ m,\ 0.5\ m,\ 0.25\ m$
B) $0.5\ m,\ 1\ m,\ 0.25\ m$
C) $0.5\ m,\ 0.25\ m,1\ m$
D) $0.25\ m,\ 1\ m,\ 0.5\ m$
Explanation
(d) For surface $P,\frac{1}{v_{1}} = \frac{1}{- f} - \frac{1}{( - u)} = - 1 + \frac{1}{3} = - \frac{2}{3}$
$\Rightarrow v_{1} = - \frac{3}{2}m$
For surface $Q,\frac{1}{v_{2}} = \frac{1}{- f} + \frac{1}{u} = - 1 + \frac{1}{5} = - \frac{4}{5}$
$\Rightarrow v_{2} = - \frac{5}{4}m\ \therefore v_{1} - v_{2} = 0.25m$
Magnification of $P = \frac{v_{1}}{u} = \frac{3/2}{3} = \frac{1}{2}$
$\therefore\ $Height of $P = \frac{1}{2} \times 2 = 1m$
Magnification of $Q = \frac{v_{2}}{u} = \frac{5/4}{5} = \frac{1}{4}$
$\therefore$ Height of $Q = \frac{1}{4} \times 2 = 0.5m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.