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

Respuesta :

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, [tex]\theta_1=47.7^{\circ}[/tex]

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 :

[tex]n_1\ sin\theta_1=n_2\ sin(90)[/tex]

[tex]n_1\ sin(47.7)=1[/tex]

[tex]n_1=\dfrac{1}{sin(47.7)}[/tex]

[tex]n_1=1.35[/tex]

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