Snells law and EM wave properties

Question

1.What is the index of refraction of a refractive medium if the angle of incidence in air (n=1.0003) is 87 degrees and the angle of refraction is 5 degrees?

2.What is the critical angle for the boundary between an unknown medium (n=3.97) and water (n=1.33)?

3.The critical angle for a certain liquid-airsurface is 27°. What is the index of refraction of this liquid? 

*answer and explanation please*

Answer 1

Hi!

For all of these questions (of which there are 3), you're going to need to use Snell's Law:

n1 sin (theta1) = n2 sin (theta2)

The angles are measured from the normal line. (You don't list the angle measurement reference point, but I think that assuming that it's the normal line is the best way to continue.)

1. The first material (air) has an angle of 87 degrees, and the angle of refraction is 5 degrees

(1.0003) (sin 87 degrees) = n2 (sin 5 degrees)

(1.0003) (sin 87 degrees) / (sin 5 degrees) = n2 = 11.46

(That's a really huge value for index of refraction, but this is a large amount of refraction, so it's consistent)

For #2 and 3, critical angle occurs when the angle of refraction is 90 degrees. It only occurs when electromagnetic waves travel from high index to low index of refraction, cases where the angle increases as the wave goes to the second material. At angles of incidence greater than the critical angle, the wave will not leave the material and will reflect back into the original material (a process called total internal reflection).

2. The light is traveling from a material of n=3.97 into water, where n=1.33

Using Snell's Law: (3.97 sin (theta1)) = (1.33)(sin 90 degrees)

theta1 = sin^-1(1.33 / 3.97) = 19.58 degrees

1. The wave is travelling from an unknown material into air (n=1.0003)

Using Snell's Law: n1 (sin 27 degrees) = (1.0003)(sin 90 degrees)

n1 = 1.0003/(sin 27 degrees) = 2.20

Hope this helps!