The lower half of a vessel of depth 2d text{ | vedclass

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(B) When a vessel is filled with multiple immiscible liquids of refractive indices $\mu_1, \mu_2, \dots, \mu_n$ and real depths $d_1, d_2, \dots, d_n$

Vedclass · Accepted Answer

$d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)$

Vedclass · Answer

(B) When a vessel is filled with multiple immiscible liquids of refractive indices $\mu_1, \mu_2, \dots, \mu_n$ and real depths $d_1, d_2, \dots, d_n$,the total apparent depth $d_{app}$ when viewed perpendicularly is given by the formula:$d_{app} = \sum_{i=1}^{n} \frac{d_i}{\mu_i} = \frac{d_1}{\mu_1} + \frac{d_2}{\mu_2} + \dots + \frac{d_n}{\mu_n}$In this problem,the total depth is $2d 	ext{ cm}$,which is divided into two equal halves. Thus,the real depth of each liquid is $d_1 = d$ and $d_2 = d$.The refractive index of the lower half is $\mu_1$ and the upper half is $\mu_2$.Substituting these values into the formula:$d_{app} = \frac{d}{\mu_1} + \frac{d}{\mu_2}$$d_{app} = d\left(\frac{1}{\mu_1} + \frac{1}{\mu_2}ight)$

The lower half of a vessel of depth $2d \text{ cm}$ 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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