A concave mirror of focal length $f_m=-10\text{ cm}$ is placed at a distance of $d=30\text{ cm}$ from a convex lens of focal length $f_l=15\text{ cm}$. An object is placed $u=-20\text{ cm}$ in front of the lens. Find the position of the final image.

Solution

1. Given: lens focal length $f_l=15\text{ cm}$, object distance $u=-20\text{ cm}$, and mirror focal length $f_m=-10\text{ cm}$ at a distance $d=30\text{ cm}$.
2. Using the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_l}$, we find $$\begin{array}{rl}\frac{1}{v}& =\frac{1}{15}-\frac{1}{20}\\ & =\frac{1}{60}\end{array}$$, so the first image forms at $v=60\text{ cm}$ behind the lens.
3. The mirror is $30\text{ cm}$ from the lens, so the object distance for the mirror is $$\begin{array}{rl}{u}_m& =-(60-30)\\ & =-30\text{ cm}\end{array}$$.
4. Using the mirror formula $\frac{1}{v_m}+\frac{1}{u_m}=\frac{1}{f_m}$, we get $$\begin{array}{rl}\frac{1}{v_m}& =\frac{1}{-10}-\frac{1}{-30}\\ & =-\frac{2}{30}\end{array}$$, resulting in $v_m=-15\text{ cm}$ (a real image $15\text{ cm}$ in front of the mirror).
5. This image acts as an object for the lens at a distance of $30-15=15\text{ cm}$ from the lens, and applying the lens formula again yields a final image at $u=-20\text{ cm}$.
6. Hence the answer is (A).