Question:
Three immiscible liquids of densities d1 > d2 > d3 and refractive indices µ1 > µ2 > µ3 are put in a beaker. The height of each liquid column is h/3. A dot is made at the bottom of the beaker. For near-normal vision, find the apparent depth of the dot.
Solution:
h1 = -μ2/ μ1 h/3
h2 = – μ3/ μ2(μ2/ μ1 h/3 + h/3) = -h/3 (μ3/ μ2 + μ2/ μ1)
h3(1/ μ1 + 1/ μ2 + 1/ μ3) is the required depth.
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