Question

In fig., $L$ is a converging lens of focal length $10cm$ and $M$ is a concave mirror of radius of curvature $20cm$. A point object $O$ is placed in front of the lens at a distance of $15cm$. $AB$ and $CD$ are optical axes of the lens and mirror, respectively. Find the distance of the final image formed by this system from the optical center of the lens. The distance between $CD$ and $AB$ is $1cm$.

Answer 1

$I_1$ is the image of object $O$ formed by the lens.
$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f}{u}_1=-15$ and $f_1=10$
Solving, we get $v_1=30cm$
$I_1$ acts as source for the mirror
$\therefore u_2=-(45-v_1)=-15cm$
$I_2$ is the image formed by the mirror
$\therefore \frac{1}{v_2}=\frac{1}{f_m}-\frac{1}{u_2}=-\frac{1}{10}+\frac{1}{15}$
The height of $I_2$ above principal axis of lens $=\frac{v_2}{u_2}\times 1+1=3cm$
$I_2$ acts as a source for the lens
$\therefore u_3=-(45-v_2)=-15cm$
Hence, the lens forms an image $I_3$ at a distance $v_3=30cm$ to the left of the lens.
The height of $I_2$ above principal axis of lens $=\frac{v_3}{u_3}\times 1+1=3cm$
$\therefore$ Required distance $=\sqrt{{30}^2+6^2}=6\sqrt{26}cm$.