Question

A diamond ring is placed in a container full of glycerin. If the critical angle is found to be 37.4º and the refractive index of glycerin is
given to be 1.47, find the refractive index of diamond.​

Answer 1

Explanation:

the refractive index is 2.42

Answer 2

Answer:

The refractive index of diamond will be equal to $2.42$ .

Explanation:

We know that the sine of critical angle is the ratio of refractive index of low dense medium to the refractive index of high dense medium.

$n_{2}$ is refractive index of low dense medium

$n_{1}$ is the refractive index of high dense medium

$\theta_{c}$ is critical angle.

Therefore,    $Sin\theta_{c}=\frac{n_{2}}{n_{1}}$                                         ................(1)

Given: $\theta_{c}=37.4^{o}$

refractive index of glycerin (low dense) $n_{2}=1.47$

$n_{1}$ is refractive index of diamond (high dense)

Put the values in eq.(1);

$Sin37.4^{o}=\frac{1.47}{n_{1}}$

$n_{1}=(1.47)(0.607)$

$n_{1}=2.42$

Therefore refractive index of diamond is $2.42$.