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

The index of refraction of the liquid is 1.35.

Step-by-step 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=(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.