Question

A light ray in air strikes an oil
surface at a 61.0 deg angle, and
moves inside the oil at 38.3 deg.
What is the index of refraction of
the oil?
(Water n = 1.33, Air n = 1.00)
(No Unit)

Answer 1

Answer:

The index of refraction of the oil is 1.41.

Explanation:

We can find the index of refraction of the oil using Snell's law:

$n_{1}sin(\theta) = n_{2}sin(\alpha)$

Where:

n₁: is the index of refraction of medium 1 (air) = 1.00

n₂: is the index of refraction of medium 2 (oil) =?

θ: is the angle between the light ray and the medium 1 = 61.0°

α: is the angle between the light ray and the medium 2 = 38.3°

Hence, the index of refraction of the oil is:

$n_{2} = \frac{n_{1}sin(\theta)}{sin(\alpha)} = \frac{1.00*sin(61.0)}{sin(38.3)} = 1.41$

   

Therefore, the index of refraction of the oil is 1.41.

I hope it helps you!                                      

Answer 2

Answer:

1.41

Explanation: