COMEDK 2025 Afternoon Solved Paper

Concept:The combined effect of a concave lens followed by a convex lens on a parallel incident beam depends on the individual focal lengths and the separation between them.Explanation:For the concave lens $(f_1 = -0.5\ \text{m})$: a parallel beam (object at infinity) gives a virtual image at $v_1 = f_1 = -0.5\ \text{m}$ (left side). Using lens formula: $\frac{1}{v_1} = \frac{1}{f_1} - \frac{1}{u_1} = \frac{1}{-0.5} - \frac{1}{\infty} = -2$, so $v_1 = -0.5\ \text{m}$.The lenses are $1.25\ \text{m}$ apart, so this virtual image is $1.25 + 0.5 = 1.75\ \text{m}$ to the left of the convex lens. Thus, for the convex lens $(f_2 = +1.75\ \text{m})$, the object distance is $u_2 = +1.75\ \text{m}$ (positive since object is on the left). Apply lens formula: $\frac{1}{v_2} = \frac{1}{f_2} - \frac{1}{u_2} = \frac{1}{1.75} - \frac{1}{1.75} = 0$. Hence, $v_2 = \infty$, meaning the emergent rays are parallel to the principal axis.Answer:Option C. The emergent beam will pass parallel to the principal axis.