Eye can see you. Just kidding, that is the best I can for an intro. We don't do about eyes anyway, I guess that is saved for biology (ew). Although we don't learn about penis in Physics or Chem, so that is an advantage to doing bi. Though studying bi would give you more insight into cock. Oops went on a tangent there, onto optics. This topic is mainly about light. Let's shine a light on this knowledge.
Just as a note, there is a bit about reflection, refraction and diffraction in my waves review, which I probably won't repeat. This topic goes much more in depth into the concepts, so that was like a little taster of what was to come. Except reflection, I guess that was all there was to know.
As we know, refraction is when light is bent when it goes through a different substance, meaning the light is angled closer to the normal. Now we have equations and calculations to do with it! First, we have a way to indicate how refractive a substance is. Refractive Index is a measure of the ratio of the speed of light in comparison with a vacuum. This is absolute RI, whereas relative RI compares 2 substances. We usually use absolute RI. Refractive index can be found with the equation n=c/cₛ, where n is the RI, c is the speed of light (3*10⁸), and cₛ is the speed of light in the substance for which the RI is being found. RI doesn't have any units, because you are dividing speed by speed, so ms⁻²/ms⁻²=0. An example of this is RI of air, which is useful to know anyway (you might need to, not sure. You'll soon realise it is very easy to remember). We say the speed of light in air is 3*10⁸, as it is very close to in a vacuum. That means that the RI is 3*10⁸/(3*10⁸), which is 1. RI of air is 1. This is the smallest possible RI, as if the RI is smaller than 1, you have done something wrong. This is because that would mean that light travels faster through the substance than in a vacuum, which isn't possible. So any RI bigger than 1 travels slower in a substance than air.
But measuring speed of light through a substance would be very hard to do. Luckily, Snell came along to make everything easier. Snell's law is n₁sinθ₁=n₂sinθ₂. n₁ is the RI of the first substance, with n₂ being the RI of the second substance. θ₁ is the angle of incidence, and θ₂ is the refracted angle. With this, as long as you know 3 of the values, you can work out the last one. The most common use will be to work out an n, usually the second one. However, you could always work out an angle if you have 2 RIs and the other angle.
We know about normal refraction, where light is bent. However, if the angle of incidence≥ an angle called the critical angle (θ꜀(that is supposed to be a subscript c, but subscript characters are limited)), TIR occurs. Total internal reflection is when light is reflected back inside a medium. When demonstrating this, a semi-circle glass block is used, because the curve reduces the initial refraction of air to glass (as the angle of incidence in the air is 0 all along the curve, so there is no refraction). If θᵢ<θ꜀, there isn't TIR. If θᵢ=θ꜀, the ray is reflected along the boundary. If θᵢ>θ꜀, then the ray is refracted back into the glass (TIR).
We can work out the critical angle using snell's law. For this, n₁ will be the glass/substance TIR happens in, and n₂ is air/substance on the other side. At the critical angle, the ray is refracted along the boundary, meaning θ₂=90. sin(90)=1, so the equation to find the critical angle will be θ꜀=sin⁻¹(n₂/n₁).
There are 2 requirements for TIR. The first is that the angle of incidence must exceed the critical angle (θᵢ>θ꜀), which we covered above. The other is that the RI of the first medium must be bigger than the RI of the second (n₁>n₂). This is just a fact, but can also be shown with the critical angle equation above, as if n₁ is smaller, the you be trying to inverse sine a number bigger than 1, which isn't possible.
