A horizontal, parallelsided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of 35.0° with the normal to the top surface of the glass.(a) What angle does the ray refracted into the water make with the normal to the surface?

Video transcript

Welcome back, everyone. We are making observations about a light beam here. Now, we are told that the light beam in the air strikes a flat layer of pure quartz that is in contact with a thick layer of turpentine oil. Now, the beam is incident on the air quartz interface at an angle of 28 degrees with respect to the normal to the surface. And we want to find at what angle does our beam emerge from the turpentine oil. Now, we are told that the refractive indices of quartz and turpentine are respectively 1. and 1.47. Before getting started here, I do wish to acknowledge our multiple choice answers on the left hand side here, those are the values that we are going to strive for. So without further ado let us go ahead and begin. Now, according to Snell's law of refraction here, if we're looking at any interface, we have the following equality that the first refractive index times the sign of the first angle is equal to the second refractive index. The sign of the second angle. Let's just take a look here first at the air courts interface, we have that the index of air times the sign of the incident angle is equal to the index of quartz time. The sign of the angle uh that we have that the beam makes with the court surface, we also have. However, if we're looking at the court's turpentine interface, we have the NQ sign of theta Q is equal to and T sign of beta R which is the angle at which our beam emerges the turpentine oil. Now you can see that we have a common term in both of these equations. So let's combine our equations here. What we get is that the index of air times the sign of the initial incident angle equal to NT times the sign of our angle at which the beam leaves the turpentine. I'm gonna divide both sides by N oh Sorry, that should be T we have NT I'm gonna divide both sides by NT here. You will see that the sign of our desired angle is going to be equal to the index of air times the sign of our original incident angle divided by the index, the refractive index of our turpentine. So let me go ahead and change colors here so that we can go ahead and calculate that, that the sign of the R is equal to while the index of air is just one times the sign of our original angle 28 divided by the index of our turpentine, which is 1.47 which gives us 0.319. Now, to figure out what our theta R is, I'm going to take the inverse sign of both sides of our equation. You see on the left hand side here, the sine functions cancel out and we get that theta R is equal to 18.6 degrees corresponding to our final answer. Choice of a. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.