Combination of Lenses question from JEE Main 2016 (9 Apr, Shift 1)

Answer: (b)

1. Lens alone: $u = 32.2 - 22.2 = 10\text{ cm}$, $v = 71.2 - 32.2 = 39\text{ cm}$.
2. $\frac{1}{f_1} = \frac{1}{39} + \frac{1}{10} = \frac{49}{390} \Rightarrow f_1 \approx 8\text{ cm}$ (closest option 7.8 cm).
3. With the convex mirror placed so the image falls back on the object pin, the lens image (at 71.2 cm) is the mirror's centre of curvature: $R = 71.2 - 45.8 = 25.4\text{ cm}$.
4. $f_2 = \frac{R}{2} = 12.7\text{ cm}$
5. So the answer is (b) $f_1 = 7.8\text{ cm}$, $f_2 = 12.7\text{ cm}$.