The lower half of a vessel of depth $2 d \mathrm{~cm}$ is filled with a liquid of refractive index $\mu_1$ and the upper half with a liquid of refractive index $\mu_2$. The apparent depth of the vessel seen perpendicularly is

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Aicharya · Accepted Answer

If immiscible liquids of refractive indices $\mu_1$ and $\mu_2$ are filled in a vessel and their real depths are $d_1$ and $d_2$ as shown below

Then, apparent depth of the vessel seen perpendicularly is given as $d_{\mathrm{app}}=\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}$ Here, $d_1=d_2=d$ $\therefore \quad d_{\mathrm{app}}=\frac{d}{\mu_1}+\frac{d}{\mu_2}=d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)$

The lower half of a vessel of depth \(2 d \mathrm{~cm}\) is filled with a liquid of refractive index…

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