A ray of light is incident on a glass pane with an angle of 60°. The light partially reflects and partially refracts. What is the angle θ between the reflected and refracted rays?

Welcome back, everyone. Hopefully, you had a chance to try this practice problem. So we have light that's gonna bounce off of a glass pane. We know that light is always gonna reflect and refract. So this light is gonna partially reflect and go off in this direction, and then part of it is also gonna refract and go through the glass pane. Now what we're looking for in this problem is not the angle of reflection or the angle of refraction, but actually the angle that is between the reflected and refracted rays. So in other words, this is my reflected and this is my refracted. It's the angle between those. Alright? So let's go ahead and see if we can solve this. Now the first thing we know about this problem is we know that light is gonna partially reflect. And we know from the law of reflection that theta 1 is equal to theta 1 prime. So in other words, if this is my incident ray here, I'm gonna call this theta 1, then this is my reflected ray, and it has to be the exact same angle of 60 degrees. That's the that's theta 1 prime. Alright? Now we also know that this this angle here of the refracted ray is not going to be 60 degrees because in general, what happens is that lend light has to bend at the boundaries. If this is air, if this is glass, then what happens is we have a higher end here. And so the words, the, angle, the refraction has to sort of bend more towards the normal. Alright. So this is not gonna be theta 2 or sorry. This is not gonna be 60 degrees. I'm going to call this my theta 2. So how do we actually go ahead and solve for this this theta here, this angle between the reflected and refracted rays? Well, if you look at what's going on here, what happens is we can look at this normal line that makes a straight line. Right? This normal line that goes, perfectly vertically up and down. We know that a perfect straight line is a 180 degree angle. Right? So this is a 180 degrees. And, by the way, this a 180 degrees means that this is also a 180 degrees. So if you look at sort of what's going on here, this 180 degrees is gonna go through the 60 degrees. It's gonna pass through our our angle that we're interested in, and it's also going to encompass the theta 2. It's actually going to be all 3 of those angles put together. So I can actually come up with a, a sort of relationship here, and I'm going to say that my theta 1 prime plus my theta 2 plus the angle that's my target variable that my my target theta, is all actually equal to 180 degrees. So if I want to figure out this theta here, then I just have to do a 180 minus my theta 1 prime minus my theta 2. And we've actually already figured out what theta one prime is, so we've already knocked out one of the variables. Right? We said that this, this because of the law of reflection, that if this is 60, this also has to be 60. So in other words, we've already got that's this is a 180 minus 60. Now we just have to figure out what theta 2 is, and that's gonna be my target variable. Alright? So that's what's going on here. We kinda have to solve for this theta by actually solving for the other angles that are around it. So it's kind of a weird way to think about this problem, but that's what's going on. Alright? So then how do we solve for this theta 2 now? Well, what happens is with the refraction, we could always just use Snell's law. So actually, let's just go over here and solve for Snell's law. So n1 sine theta 1 equals n2 sine theta 2. So in other words, my n1 is actually just air, so this is just n one equals 1, and my n two is gonna be for glass, so this is gonna be 1.46. So I've got this is just 1. This is my angle of incidence, which I have. This is my index, so I can just solve for the theta 2, and we've done this before. So this is actually just gonna be, this is just gonna be, I've got n one equals one times the sine of this is my angle of 60 degrees. This equals the 1.46 times the sine of theta 2. Now when I work this out, what I'm going to get here, is I'm going to get this is going to be 0.87 divided by, and then I just divide the 1.46 over, and this equals the sine of theta2. So my theta2 is equal to the inverse sine of 0.87 divided by 1.46. Now when you work this out, what you should get is 36.6 degrees. Now this makes sense because we're gonna get a smaller angle. We came in at 60 degrees. It bends toward the normal. We get a smaller number of 36.6. So now we're actually ready to get, we're actually ready to plug this into our final equation over here. So this theta is equal is equal to 180 minus 60 and then minus 36.6. When you work this out, what you should get is an angle of theta, my target variable, which is 83.4 degrees. So that is your final answer for this angle here. This is 83.4 degrees. Alright? So anyway, folks, that's it for this one. Hopefully, that made sense. Thanks for watching.

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A ray of light is incident on a glass pane with an angle of 60°. The light partially reflects and partially refracts. What is the angle θ between the reflected and refracted rays?

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