Two immiscible liquids of refractive indices | vedclass

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(D) The apparent depth of an object in a medium of refractive index $\mu$ and real depth $h$ is given by $h_{app} = \frac{h}{\mu}$.For multiple layers

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(D) The apparent depth of an object in a medium of refractive index $\mu$ and real depth $h$ is given by $h_{app} = \frac{h}{\mu}$.For multiple layers of immiscible liquids,the total apparent depth is the sum of the apparent depths of each layer:$h_{app} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$Given:$h_1 = 6 \,cm, \mu_1 = \frac{8}{5}$$h_2 = 6 \,cm, \mu_2 = \frac{3}{2}$Substituting the values:$h_{app} = \frac{6}{8/5} + \frac{6}{3/2}$$h_{app} = \frac{30}{8} + \frac{12}{3}$$h_{app} = \frac{15}{4} + 4 = \frac{15 + 16}{4} = \frac{31}{4} \,cm$Comparing this with the given apparent depth $\frac{\alpha}{4} \,cm$,we get $\alpha = 31$.

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