Question

In the figure shown 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 & AB is 1 cm. Take $\sqrt{26}=5.1$.

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 $f_1=10$
$\frac{1}{v_1}-\frac{1}{-15}=\frac{1}{10}$
$v_1=30\text{cm}$

Now, $I_1$ acts as source for mirror
$\therefore u_2=-(45-v_1)=-15\text{cm}$
$I_2$ is the image formed by the mirror
$\therefore \frac{1}{v_2}=\frac{1}{f_m}-\frac{1}{u_2}=-\frac{1}{10}-\frac{1}{-15}\therefore v_2=-30\text{cm}$


The height of $I_2$ above the principle axis of lens is $=\frac{v_2}{u_2}\times 1+1=\frac{-30}{-15}\times 1+1=3\text{cm}$

Now, $I_2$ acts a source for the lens $\therefore u_3=-(45-v_2)=-15\text{cm}$, here we took the $-ve$ sign because the ray is coming from left to right. So the sign convention for such rays reverses.

Hence the lens forms an image $I_3$ at a distance $v_3=30$ cm to the left of the lens.

The height of $I_3$ below the principal axis lens is $=\frac{v_3}{u_3}\times 3=-6\text{cm}$

$\therefore \text{Required distance}=\sqrt{{30}^2+6^2}=6\sqrt{26}\text{cm}$
$\text{Required distance}30.6\text{cm}$.