COMEDK 2025 Evening Solved Paper

Concept:Immersion in water reduces the relative refractive index of the lens, increasing its focal length.Explanation:In air, the object is placed at $u = 1.5f$ (midway between $f$ and $2f$).Using the lens formula $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$, we get $v = 3f$ – a real, inverted image beyond $2f$.When the setup is immersed in water of refractive index $n_w > 1$, the lens maker’s equation becomes$\frac{1}{f_{\text{water}}} = \left( \frac{n_{\text{lens}}}{n_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Because $\frac{n_{\text{lens}}}{n_w} - 1 < n_{\text{lens}} - 1$, the new focal length is larger: $f_{\text{water}} > f$.The object distance $u = 1.5f$ remains unchanged, so now $u < f_{\text{water}}$.For a convex lens, when the object lies within the focal length, the image is virtual, erect, and magnified, located on the same side as the object.Therefore the original real inverted image is replaced by a virtual erect magnified image on the object’s side.Answer:Option A: Magnified virtual erect image on the same side of the lens where the object is placed.