Refraction and Lenses Legacy Problem #10 Guided Solution

This problem is often described as a layer problem, which involves several layers layered on top of one another. Each layer has a boundary which is parallel to the previous layer. And as such, as light refracts through, it's going to refract at one boundary and then approach the next boundary. And what's going to be true, geometrically, is that the angle of refraction at the first boundary is going to be equal to the angle of incidence at the very next boundary. Because these angles of refraction and angle of incidence at consecutive boundaries are alternating interior angles formed by the normal lines intersecting the boundaries. And so as I approach this problem, I'm going to have to do four different Snell's Law calculations. One for each boundary. You can count the boundaries and you should notice that there are going to be four different boundaries. As a ray of light comes in from the top, approaching the layer one boundary with water, we can do a Snell's Law calculation there to find the angle of refraction layer one. Then continue the light ray to the next boundary and do a Snell's Law calculation for layer one, layer two. And then continue for layer two, layer three, and then layer three in the water as it exits. Now in each case, we know the index of refraction of the incident and the refractive material. And so all we have to do is apply Snell's Law in one sine theta one equal to two sine theta two in order to solve for the unknown angle. And then we just take that same angle we calculated as an angle of refraction and use it as the angle of incidence in the next n1 sine theta one equal to n2 sine theta two calculation. Now it involves a lot of organization and work, so I wouldn't even think of doing this problem without actually having a sheet of paper out and a pencil or pen to write with and a calculator. So here we go. Starting with the first boundary, the water layer one boundary. I'm going to use the equation 1.33 times the sine of 62.8 degrees equal 1.91 times the sine of theta in the layer one. Then I'm going to get sine of theta layer one by itself. So I divide each side of the equation by 1.91 and that leaves me 0.61933 is equal to the sine of theta in layer one. I can use the inverse sine function or the second sin button on the calculator to find the angle of refraction at the layer one water boundary. It comes out to be 38.2674 and I can round that to three significant digits. Now in my next calculation I'm going to use that same 38.2674 as the angle of incidence. So setting up Snell's Law for the layer one, layer two boundary, I have 1.91 times the sine of 38.2674 is equal to 1.52 times the sine of the angle in layer two. When I get the sine of the angle in layer two by itself, I end up with 0.77824 equal the sine of the angle in layer two. Then I can use the inverse sine function or the second sin button on my calculator and I find that the angle of refraction in layer two equal 51.0997. I can round that to three significant digits, 51.1 degrees. Now I go to the layer two, layer three boundary and the angle of incidence there is the same 51.0997 degrees. I use Snell's Law so I get 1.52 times the sine of 51.0997 degrees equal 1.36 times the sine of the angle in layer three. Then I divide each side by 1.36 and I get 0.86980 equal the sine of the angle in layer three. I use the inverse sine function or the second sine button and I get the angle of refraction entering layer three is 60.4350 degrees. Now I go to the layer three, water boundary and I say 1.36 times the sine of 60.4350 is equal to 1.33 times the sine of the angle in the water. I divide each side by 1.33 and I get 0.88942 equal the sine of the angle in water. I push the inverse, the second sine button and I get 62.8 degrees. And then I say, wow, that's the same 62.8 degrees that we had entering the water, entering layer one from the water. And that's no coincidence as long as you have the same material on the top of layer one as on the bottom of layer three in parallel layers, you're going to end up having the light emerge from the last layer at the same angle as it entered the first layer.