00:01
So in this problem, we are told that light is moving from water, from water to air.
00:13
No, sorry, from air to water.
00:16
So we are being asked to find out what is the angle of refraction, as well as the wavelength that the light will be refracted at.
00:33
So the important thing that we need to keep in mind is that the refractive index is going to help us solve this problem.
00:43
So the n of air is basically 1 .0 .0 and the n of water is 1 .333.
00:52
So now we're going to be able to use the fact that the initial wavelength is 589 nanometers and that the initial angle is 35 degrees to solve this problem.
01:07
So our first equation that we're going to use is n1, sine of theta 1 is equal to n2, sine of theta 2.
01:15
So that means that if we just take, we know that this is simply equal to 1.
01:20
So we're going to take sine of theta of 1 and divide that by n of 2, and find the inverse sign by the inverse sign of this value, and that will give us our theta 2.
01:38
So we will have the inverse sign of the sign of 35 divided by 1 .333.
01:55
That will give us our theta 2...