COMEDK 2025 Evening Solved Paper

Concept:Use the lens formula for the two-lens system. The first lens forms an image that acts as the object for the second lens.Explanation:The diverging lens has focal length $f_1 = -20 \; \text{cm}$ (negative for diverging).Incident parallel light means object distance $u_1 = \infty$ for the first lens.Using lens formula $\frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f_1}$ gives $\frac{1}{v_1} + 0 = \frac{1}{-20}$ ⇒ $v_1 = -20 \; \text{cm}$.So the first lens forms a virtual image $20 \; \text{cm}$ to its left.The second lens is a converging lens of focal length $f_2 = +15 \; \text{cm}$ placed $10 \; \text{cm}$ to the right of the diverging lens.The virtual image from lens 1 is located $20 \; \text{cm}$ left of lens 1, i.e., at position $x = -20$ (taking lens 1 at $x=0$).Distance from that point to lens 2 (at $x = +10$) is $20 + 10 = 30 \; \text{cm}$.Since the object lies on the side from which light approaches lens 2, the object distance $u_2 = +30 \; \text{cm}$.Apply lens formula for lens 2: $\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2}$ ⇒ $\frac{1}{v_2} = \frac{1}{15} - \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}$ ⇒ $v_2 = +30 \; \text{cm}$.A positive $v_2$ indicates a real image formed on the opposite side of the lens from the incoming light, i.e., $30 \; \text{cm}$ to the right of the converging lens.Answer:The final image is real and at a distance of $30 \; \text{cm}$ to the right of the converging lens. Hence the correct option is C.