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(D) The system acts as a combination of a plano-convex water lens and the concave mirror.Let $R$ be the radius of curvature of the mirror.The focal le

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real and between $O$ and $C$

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(D) The system acts as a combination of a plano-convex water lens and the concave mirror.Let $R$ be the radius of curvature of the mirror.The focal length of the mirror is $f_m = -R/2$.The focal length of the plano-convex water lens (refractive index $\mu = 4/3$) is given by $1/f_l = (\mu - 1)(1/R - 1/\infty) = (1/3)(1/R) = 1/(3R)$. So,$f_l = 3R$.When the object is at $C$ (distance $R$ from the mirror),the light rays pass through the water lens,reflect from the mirror,and pass through the water lens again.$1$. Refraction at the water lens: $1/v_1 - 1/u_1 = 1/f_l \implies 1/v_1 - 1/(-R) = 1/(3R) \implies 1/v_1 = 1/(3R) - 1/R = -2/(3R) \implies v_1 = -1.5R$.$2$. Reflection at the mirror: $1/v_2 + 1/u_2 = 1/f_m \implies 1/v_2 + 1/(-1.5R) = 1/(-0.5R) \implies 1/v_2 = -2/R + 2/(3R) = -4/(3R) \implies v_2 = -0.75R$.$3$. Final refraction at the water lens: $1/v_3 - 1/u_3 = 1/f_l \implies 1/v_3 - 1/(-0.75R) = 1/(3R) \implies 1/v_3 = 1/(3R) - 4/(3R) = -1/R \implies v_3 = -R$.Since the final image distance is negative,it is a real image formed between $O$ and $C$.

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