Three immiscible transparent liquids with ref | vedclass

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(D) The apparent depth of a combination of immiscible liquids is given by the formula: $d_{app} = \sum \frac{d_i}{\mu_i}$,where $d_i$ is the real dept

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$10$

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(D) The apparent depth of a combination of immiscible liquids is given by the formula: $d_{app} = \sum \frac{d_i}{\mu_i}$,where $d_i$ is the real depth and $\mu_i$ is the refractive index of the $i$-th liquid.Given:$d_1 = 3 	ext{ cm}, \mu_1 = 3/2$$d_2 = 4 	ext{ cm}, \mu_2 = 4/3$$d_3 = 6 	ext{ cm}, \mu_3 = 6/5$Calculating the apparent depth:$d_{app} = \frac{3}{3/2} + \frac{4}{4/3} + \frac{6}{6/5}$$d_{app} = (3 	imes 2/3) + (4 	imes 3/4) + (6 	imes 5/6)$$d_{app} = 2 + 3 + 5 = 10 	ext{ cm}$.Therefore,the apparent depth of the vessel is $10 	ext{ cm}$.

Three immiscible transparent liquids with refractive indices $3/2, 4/3$ and $6/5$ are arranged one above the other in a container. The depths of the liquids are $3 \text{ cm}, 4 \text{ cm}$ and $6 \text{ cm}$ respectively. The apparent depth of the vessel is (in $\text{ cm}$)

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