Physics - Geometrical Optics Question with Solution | TestHub

For the first lens $L_1$ (convex lens):
Object distance, $u_1=-30$ cm
Focal length, $f_1=+10$ cm
Using the lens formula, $\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$
$\Rightarrow \frac{1}{v_1}-\frac{1}{-30}=\frac{1}{10}$
$\Rightarrow \frac{1}{v_1}+\frac{1}{30}=\frac{1}{10}$
$\Rightarrow \frac{1}{v_1}=\frac{1}{10}-\frac{1}{30}=\frac{3-1}{30}=\frac{2}{30}=\frac{1}{15}$
$\Rightarrow v_1=+15$ cm
The image $I_1$ formed by $L_1$ acts as a virtual object for the second lens $L_2$.

For the second lens $L_2$ (concave lens):
Distance between the lenses, $d=3$ cm
Object distance for $L_2$, $u_2=v_1-d=15-3=+12$ cm
Focal length, $f_2=-10$ cm
Using the lens formula, $\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$
$\Rightarrow \frac{1}{v_2}-\frac{1}{12}=\frac{1}{-10}$
$\Rightarrow \frac{1}{v_2}=\frac{1}{12}-\frac{1}{10}=\frac{5-6}{60}=-\frac{1}{60}$
$\Rightarrow v_2=-60$ cm

The negative sign indicates that the final image is formed $60$ cm to the left of the concave lens.

Answer: $60$ cm to the left of the concave lens