Question

For the system shown in the figure, the position of the final image of object $O$ is
[Assume origin of co-ordinate system to be at optical centre of half convex lens]

• A. $(x=60\text{cm},y=-6\text{mm})$
• B. $(x=60\text{cm},y=-10\text{mm})$
• C. $(x=30\text{cm},y=-6\text{mm})$
• D. $(x=30\text{cm},y=-10\text{mm})$

Answer 1

The correct option is D $(x=30\text{cm},y=-10\text{mm})$

For refraction at lens,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{(-60)}=\frac{1}{20}\Rightarrow v=30\text{cm}$
Magnification, $m=\frac{v}{u}=\frac{30}{-60}=\frac{-1}{2}$
So, height of image formed,
$h_i=|-\frac{1}{2}|\times h_0=\frac{1}{2}\times 4=2\text{mm}$ (inverted)
Point $B_1$ is $2\text{mm}$ below the principal axis of the lenses and plane mirror is $6\text{mm}$ below it. Hence, mirror will form virtual image of $A_1{B}_1$ as shown below.
Therefore, final image is at a distance of $x=30\text{cm}$ and $y=-10\text{mm}$