Two immiscible liquids of refractive indices 8/5 and 3/2 respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is α/4 cm. The value of α is _________.

Solution

Physics sir · 4 steps

Each layer lifts the coin on its own: add the reduced depths t/μ of the two liquids.

1. Layer by layer. Near normal viewing, a layer of thickness t and refractive index μ looks only t/μ thick from air. With stacked layers, the apparent depths of the layers add up.

   dₐₚₚ = t₁/μ₁ + t₂/μ₂

2. Put in numbers. Lower liquid (μ₁ = 8/5): 6/(8/5) = 30/8 = 3.75 cm. Upper liquid (μ₂ = 3/2): 6/(3/2) = 4 cm.
3. Add them. dₐₚₚ = 3.75 + 4 = 7.75 cm = 31/4 cm.
4. So. Comparing with α/4 cm gives α = 31.

The common slip: 7.75. Stopping at the apparent depth gives 7.75, but the question asks for α, which is 4 times that, so 31.

The idea to keep. For several slabs viewed from air at near normal incidence, apparent depth is the sum of t/μ for every slab. The order of the layers does not matter.

Answer from the NTA final answer key. Solution by PrepWiser.