Physics, asked by anyasingh8120, 1 year ago

The 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 h3. 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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Answered by SteffiPaul
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When immiscible liquids of densities d1>d2>d3 and refractive indices μ1>μ2>μ3 are put in a beaker, the height of each liquid column being h3, for near normal vision, the apparent depth of the dot is given as follows:

1. The apparent depth of a liquid medium of depth h, having refractive index μ is h/μ.

2. It is independent of the density of the liquid.

3. If n liquids having different refractive indices are present then the net refractive index becomes μ= μ1μ2*...*μn.

4. Apparent depth of the dot at the bottom of the beaker 'ha' is h3/μ1μ2μ3.

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