20 Physics -- Reflection at Plane Surfaces

A diamond ring is placed inside a liquid. If the critical angle is found to be 37.40 and the refractive index of liquid is given to be 1.47, find the refractive index of diamond.

A diamond ring is placed inside a liquid. If the critical angle is found to be 37.40 and the refractive index of liquid is given to be 1.47, find the refractive index of diamond.

Refractive index of diamond is 2.42.

Solution:

We have,

i_c = 37.40

mu_liquid = 1.47

mu_diamond = ?

The critical angle is given for diamond-liquid interface. It is applicable for a ray of light passing from diamond through the liquid and out of this interface. When we apply these conditions and related formula, we get to our answer.

Let mu be the refractive index of diamond-liquid interface then,

mu = mu_diamond/mu_liquid

We know,

mu = 1 / sin(i_c)

= 1 / sin (37.4)

= 1.65

Now,

Using the fact: mu = mu_diamond/mu_liquid

or, 1.65 = mu_diamond / 1.47

or, mu_diamond = 1.65 x 1.47

So, mu_diamond = 2.42

Hence, the required refractive index of diamond is 2.42.

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