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NCERT Exemplar · Q20

Q.Three immiscible liquids of densities d1>d2>d3d_1 > d_2 > d_3 and refractive indices μ1>μ2>μ3\mu_1 > \mu_2 > \mu_3 are put in a beaker. The height of each liquid column is h3\dfrac{h}{3}. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

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The apparent depth is found by summing the apparent shifts from each layer using the formula apparent depth=∑actual thicknessμ\text{apparent depth} = \sum \frac{\text{actual thickness}}{\mu}. For three layers each of height h/3h/3, the result is h3(1μ1+1μ2+1μ3)\frac{h}{3}\left(\frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3}\right).

The key insight here is Index Matching — when light travels from a denser to a rarer medium (or vice versa), the apparent shift depends on the refractive index of the medium the light is leaving. For near-normal viewing from air above, each liquid layer acts like a "compressor" of the apparent depth beneath it.

Think of it this way: if you have a single liquid of refractive index μ\mu and actual depth hh, the apparent depth when viewed from air is h/μh/\mu. Why? Because light bends away from the normal when exiting into air, making the bottom look shallower. Now, when you stack multiple liquids, the light passes through each interface sequentially. The total apparent depth is just the sum of the apparent thicknesses of each layer — because each layer's effect is independent when the viewing angle is near normal.

Let's work it out step by step.

  1. Set up the problem.

    The beaker has three immiscible liquids. Densities decrease upward (d1>d2>d3d_1 > d_2 > d_3), so the densest liquid (d1d_1) sits at the bottom, then d2d_2, then d3d_3 on top. The refractive indices also decrease upward (μ1>μ2>μ3\mu_1 > \mu_2 > \mu_3). Each liquid column has actual height h/3h/3. A dot is at the very bottom of the beaker. We view from above, near normal incidence.

  2. Understand the apparent depth for a single layer.

    For a single medium of refractive index μ\mu and actual depth HH, the apparent depth when viewed from air (refractive index 1) is H/μH/\mu. This comes from Snell's law for small angles: n1sin⁡θ1=n2sin⁡θ2n_1 \sin\theta_1 = n_2 \sin\theta_2 reduces to n1θ1≈n2θ2n_1 \theta_1 \approx n_2 \theta_2, and geometry gives apparent depth=actual depthμ\text{apparent depth} = \frac{\text{actual depth}}{\mu}.

  3. Extend to multiple layers — the principle of additivity.

    When light from the dot passes through the bottom liquid (index μ1\mu_1), it first encounters the interface with the middle liquid (μ2\mu_2). For an observer in the middle liquid, the bottom of the beaker appears at depth (h/3)/μ1(h/3)/\mu_1 as measured from that interface. But we're not stopping there — the light then enters the middle liquid, then the top liquid, then air.

    The neat trick: treat each layer as if it were the only one, and add the apparent thicknesses. Why does this work? Because at each interface, the apparent depth of everything below gets "compressed" by the ratio of the refractive indices. For near-normal rays, the total apparent depth from the top is:

apparent depth=h1μ1+h2μ2+h3μ3\text{apparent depth} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2} + \frac{h_3}{\mu_3}

where h1,h2,h3h_1, h_2, h_3 are the actual heights of the layers from bottom to top. …

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