Question

Consider an optical system consisting of a concave mirror $M_1$ and convex mirror $M_2$ of radii of curvature $60 \,cm$ and $20\, cm$ , respectively. Two mirrors are separated by a distance of $40\, cm$ . An object $O$ is placed at a distance $80\, cm$ from $P$ . The final image is formed at a distance

• A. $40 \,cm$ on the right of $M_2$
• B. $40 \,cm$ on the left of $M_2$
• C. $48 \,cm$ on the right of $M_1$
• D. $48 \,cm$ on the left of $M_2$

Answer 1

Answer: (D)

$48 \,cm$ on the left of $M_2$

Answer 2

Concave mirror $M_1$ with radius $r_1 = 60\, cm$
Convex mirror $M_2$ with radius $I_2 = 20 \,cm$
Focus, $f_1 = \frac{60}{2} = 30\,cm$
Focus, $f_2 = \frac{20}{2} = 10\,cm$

$OP = 80 = u$
From mirror formula,
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
From concave minor, the image formed by minor $M_1$ will be
$f_1 = -30\,cm$
$u = - 80\,cm$
$v = ?$
$\frac{1}{-0} = \frac{1}{v} + \frac{1}{-80}$
$\frac{1}{v} = \frac{1}{80} - \frac{1}{30}$
$= \frac{30-80}{80\times 30}$
$\frac{1}{v} = - \frac{50}{80\times 30}$
$v = - 48 \,cm$
The image will be out of reflecting surface of mirror $M_2$. So, this will be the final image and it will form on the left of $M_1$.