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(C) Let the thicknesses of the upper and lower liquid layers be $t_1$ and $t_2$ respectively.When light from a point source at height $d$ enters the l

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$\mu_1 d$

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(C) Let the thicknesses of the upper and lower liquid layers be $t_1$ and $t_2$ respectively.When light from a point source at height $d$ enters the liquid layers,it undergoes refraction.The apparent shift produced by a slab of thickness $t$ and refractive index $\mu$ is $\Delta x = t(1 - \frac{1}{\mu})$.The total apparent shift is $\Delta x_{total} = t_1(1 - \frac{1}{\mu_1}) + t_2(1 - \frac{1}{\mu_2})$.Since the source coincides with its image,the rays must strike the mirror normally.For the rays to strike the mirror normally,the apparent position of the source as seen from the mirror must be at the center of curvature $R$ of the mirror.However,in this specific configuration where the source is at a large distance $d$,the effective distance of the source from the mirror after refraction is $d' = \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + (d - t_1 - t_2)$.For the image to coincide with the source,the rays must return along the same path,which happens if the effective position of the source is at the center of curvature $R$.Given $d$ is very large,the thickness of the layers is negligible compared to $d$,so the effective distance is $d_{eff} = \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + (d - t_1 - t_2) \approx d$.For the rays to be normal to the mirror,the radius of curvature $R$ must be equal to the effective distance of the source from the mirror,which simplifies to $R = d$ if $\mu_1 = \mu_2 = 1$. Considering the refractive indices,the effective depth is $d_{eff} = \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + (d - t_1 - t_2)$. As $d \gg t_1, t_2$,the result is $R = d$ is not correct; rather,the condition for normal incidence implies $R = d$ is only true if $\mu=1$. Given the options,the correct relation is $R = d$ is not listed,but evaluating the effective path,the answer is $d$.

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