Question

By how much distance, a coin at depth h below the surface in a liquid ( refractive index n) appear to be shifted when observed from top, outside the liquid?​

Answer 1

Answer:

real depth of the coin =h

apparent depth of coin= h1. (let

refractive index of liquid (n)=c/v

=h/t×t/h1

=h/h1

h1=h/n

shift=real depth-apparent depth

=h-h1

=h(1-1/n)

So, the coin appeared to be raised up by h(1-1/n)

Answer 2

Answer:

Distance, a coin at depth h below the surface in a liquid-

Real depth $=n\times Apparent Depth$

Explanation:

Figure depicts an observer's perspective of a coin submerged in water. The refraction at the water's surface is what causes this effect.

As they emerge from the water, light rays from the coin are deviated from their typical path. They appear to come from a virtual coin that is above the real coin when they reach the eye. The apparent depth is the separation between the virtual image and the water's surface. The distance between the actual object, O, and the water's surface is the true depth.

Refractive index, n, and real depth and apparent depth are related by:

$n=\frac{Real Depth}{ Apparent depth}$

Real depth $=n\times Apparent Depth$

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