Question

The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the nearest hundredth. nair = 1.00 Answer

Answer 1

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, $\theta_1=47.7^{\circ}$

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

$n_1\ sin\theta_1=n_2\ sin(90)$

$n_1\ sin(47.7)=1$

$n_1=\dfrac{1}{sin(47.7)}$

$n_1=1.35$

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.