NEET 2026 Re Exam Physics and Chemistry Paper

Concept:Use the lens formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ with proper sign convention for each lens separately. The image formed by the first lens acts as the object for the second lens.Explanation:For the convex lens $L_1$: object distance $u = -30 \text{ cm}$ (real object, left side), focal length $f = +10 \text{ cm}$.Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$:$$\frac{1}{v}+\frac{1}{30}=\frac{1}{10}$$→ $\frac{1}{v}=\frac{1}{10}-\frac{1}{30}=\frac{2}{30}$ → $v=+15 \text{ cm}$ (right of $L_1$).This image is 15 cm to the right of $L_1$. Distance between lenses is 3 cm (from figure). So image formed by $L_1$ is $15-3 = 12 \text{ cm}$ to the left of $L_2$ (object for $L_2$). For concave lens $L_2$: object distance $u' = +12 \text{ cm}$ (virtual object, since image lies to its left), focal length $f' = -10 \text{ cm}$.Using $\frac{1}{v'}-\frac{1}{u'}=\frac{1}{f'}$:$$\frac{1}{v'}-\frac{1}{12}=\frac{-1}{10}$$→ $\frac{1}{v'}=\frac{1}{12}-\frac{1}{10}=\frac{-1}{60}$ → $v'=-60 \text{ cm}$ (negative sign means image is on the same side as the object, i.e., to the left of $L_2$).Hence final image is 60 cm to the left of the concave lens.Answer:60 cm to the left of the concave lens (Option C).