Snell's Law Example 1. Created by Sal Khan.
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- At6:28maybe a more useful way or remembering is that when you go from
MORE dense -> LESS dense, rays bend away from normal
LESS dense -> MORE dense, rays bend towards the normal.
Or am i completely wrong(11 votes)
- Indeed your completely correct as when light ray travels from a less dense medium into a denser medium, velocity of the light way changes (velocity is a vector quantity so both direction and speed will change). When you enter a denser medium, speed of light ray slows down and hence the refracted ray bends towards the normal. When you enter a less dense medium, speed of light ray increases and hence the refracted ray will bend away from the normal. Keep in mind, that the angle of incidence is measured from the incidental light ray to the normal and the angle of refraction is measured from the refracted light ray to the normal.(5 votes)
- How is Snell's Law derived?(6 votes)
- Hi Agarwal,I'll answer your question now.Snell's Law is a formula used to discribe the relationship between the angles of incidence and refraction,when referring to light or other waves passing through a boundary between to different isotropic media,such as water,glass and air.
In optics,the law is used in ray tracing to compute the angles of incidence or refraction,and in experimental optics and gemology to find the refractive index of a material.The law is also satisfied in metamaterials,which allow light to be bent "backward" at a negative angle of refraction with a negative refractive index.(13 votes)
- Can reflection and refraction happen at the same time?(2 votes)
- Absolutely. For example:
If light travels between water and glass then refraction as well as reflection occurs. When a light ray strikes the border that separates the to mediums, a part of that light ray is reflected back into the same medium it came from (in this case the medium is water). The rest of that light ray travels from water to glass and undergoes refraction. Therefore both reflection and refraction can occur at the same time.(4 votes)
- what happened? when angle of incidence is goes from denser medium to lesser medium(2 votes)
- Snell's law works whether n1>n2 or n2>n1. Working the numbers, you'll see that in your case (the incident side is usually denoted the 'n1' side) that the angle between the normal and the ray will be greater than the angle between the incident ray and the normal. [Later you will note that going from higher index to lower allows a beam to be refracted as far as 90 degrees. This is called total internal reflection. This doesn't happen going from a lower to higher index.](4 votes)
- what does the 1.000000 ( vacuum ) mean(3 votes)
- It's the refractive index (n). it's defined as the speed of light in a vacuum divided by the speed of light of the medium in question.(1 vote)
- Is this the only way to solve such a question?(3 votes)
- Being a mass-less particle, photons (light) should move with the unchanged velocity(as in vacuum) in any other medium. Then why the velocity of light changes ?(2 votes)
- The velocity of light in a vacuum doesn't change, it's always c
In materials other than a vacuum, it's just not the case that it "should move with unchanged velocity". I don't know why you have asserted that.(2 votes)
- why does speed of light becomes slow when it enters from rarer to denser medium? But speed of light is constant! Does this violate the law of einstien that speed of light is constant?(2 votes)
- At2:00, is there not also a formula to get that which is just n = sin i / sin r . What is the difference between the formula he gives and this one. Am just a bit confused as this is the formula we are meant to learn/understand for class but this is also similar.(2 votes)
- In the second example of the vacuum that Mr. Sal had solved its refractive index using the Snell's law that states : n of the first medium*sin of the angle of the incidence = n of the second medium*sin of the refraction angle ,he got the refractive index = 1.29 , while when I used that same Snell's law but of the form that states : sin of incidence angle over sin of the refraction angle = the speed of light in the first medium over the speed of light in the second medium , I got the refractive index = 0.77786 approximately and the same speed as he got 233 million m/s , my question is why did I get a different refractive index ? And as I got a different refractive index , this symbolizes that there is a difference in my medium than his . Could someone explain the wrong step ?
Second Question , what is the relationship between the optical density of the medium and its refractive index ?(2 votes)
As promised, let's do a couple of simple Snell's law examples. So let's say, that I have two media-- I guess the plural of mediums. So let's say I have air right here. And then right here is the surface. Let me do that in a more appropriate color. That is the surface of the water. And I know that I have a light ray, coming in with an incident angle of-- so relative to the perpendicular-- 35 degrees. And what I want to know is what the angle of refraction will be. So it will refract a little bit. It will bend inwards a little bit, since this outside's going to be in the air a little longer, if you buy into my car travelling into the mud analogy. So it will then bend a little bit. And I want to figure out what this new angle will be. I want to figure out the angle of refraction. I'll call that theta 2. What is this? This is just straight up applying Snell's law. And I'm going to use the version using the refraction indices, since we have a table here from the ck12.org FlexBook on the refraction indices-- and you can go get it for free if you like. And that just tells us that the refraction index for the first medium-- that is air-- times the sine of the incident angle, in this case is 35 degrees, is going to be equal to the refraction index for water, times the sine of this angle right over here-- times the sine of theta 2. And we know what the refraction index for air and for water is, and then we just have to solve for theta 2. So let's just do that. The refraction index for air is this number right over here, 1.00029. So it's going to be, there's three 0's, 1.00029 times the sine of 35 degrees, is going to be equal to the refraction index for water, which is 1.33. So it's 1.33 times sine of theta 2. Now we can divide both sides of this equation by 1.33. On this side, we're just left with the sine of theta 2. On the left-hand side, let's get our calculator out for this. So let me get the handy calculator. And so we want to calculate-- and I made sure my calculator is in degree mode-- 1.00029 times the sine of 35 degrees, so that's the numerator of this expression right here-- the green part-- that's 0.5737 divided by 1.33. I'm just dividing by the numerator here. When you just divide this answer, it means your last answer. That's the numerator up here divided by that denominator. And so I get 0.4314. I'll just round a little bit. So I'll get-- I'll switch colors-- 0.4314 is equal to sine of theta 2. And now to solve for theta, you just have to take the inverse sine of both sides of this. This doesn't mean sine to the negative 1. You could also use the arcsine. The sine inverse of 0.4314 is going to be equal to-- the inverse sine of sine is just the angle itself or, I guess when we're dealing with angles in a normal range, it's going to be the angle itself. And that's going to be the case with this right over here. And if any of that is confusing you might want to review the videos on the inverse sine and the inverse cosine, and they're in the Trigonometry playlist. But we can very easily figure out the inverse sine for this right over here. You literally, you have sine here, when you press Second you get the inverse sine. So it's the inverse sine, or the arcsine of that number right over there. And instead of retyping it, I can just put Second, and then Answer. So I'm taking the inverse sine of that number. And that will give me an angle. And I get 25.55, or I'll round it, 25.6 degrees. So this theta 2, is equal to 25.6, or I'll say approximately equal to some 25.6 degrees. So Snell's law goes with our little car driving in to the mud analogy. It's going to be a narrow degree. It's going to come inwards a little bit closer to vertical. And theta 2 is equal to 25.6 degrees. And you could do the other way. Let's do another example. Let's say that we have, just to make things simpler, that I have some surface right over here. So this is some unknown material. And we're traveling in space, we're on the space shuttle, and so this is a vacuum. Or pretty darn close to a vacuum. And I have light coming in at some angle, just like that. Let me drop a vertical. So it's coming in at some angle. Actually, let me make it interesting. Let me make the light go from the slower medium to the faster medium, just because the last time we went from the faster to the slower. So it's in a vacuum. So let's say I have some light traveling like this. And once again, just to get the "get" of whether it's going to bend inward or bend outward, the left side is going to get out first, so is going to travel faster first. So it will bend inwards when it goes into the faster material. So this is some unknown. This is some unknown material, where light travels slower. And let's say we were able to measure, the angles. So let me drop a vertical right here. And so let's say that this right here, is 30 degrees. And let's say we're able to measure the angle of refraction. And the angle of refraction over here is, let's say that this is 40 degrees. So given that we're able to measure the incident angle, and the angle of refraction, can we figure out the refraction index for this material? Or even better, can we figure out the speed of light in that material? So let's figure out the refraction index first. So we know the refraction index for this questionable material times the sine of 30 degrees is going to be equal to the refraction index for a vacuum. Well, that's just the ratio of the speed of light in the vacuum to the speed of light in the vacuum. So that's just going to be 1. This is the same thing as n for a vacuum-- and I'll just write a 1 there-- times the sine of 40 degrees. Or If we wanted to solve for this unknown refraction index, we just divide both sides of the equation by sine of 30. So our unknown refraction index is going to be-- this is just the sine of 40 degrees-- over this-- over the sine of 30 degrees. So we can get our handy calculator out. And so we have the sine of 40 divided by the sine of 30 degrees. Make sure you're in degree mode, if you try this. And you get, let's just round it, 1.29. So this is approximately equal to-- so our unknown refraction index for our material is equal to 1.29. So we were able to figure out the unknown refraction index. And we can actually use this to figure out the velocity of light in this material. Because remember, this unknown refraction index is equal to the velocity of light in a vacuum, which is 300 million meters per second, divided by the velocity in this material, the unknown material. So we know that 1.29 is equal to the velocity of light in a vacuum. So we could write 300 million meters per second, divided by the unknown velocity in this material. I'll put a question mark. And so we can multiply both sides times our unknown velocity. I'm running out of space over here, I have other stuff written over here. So I could multiply both sides by this v and I'll get 1.29 times this v with a question mark, is going to be equal to 300 million meters per second. And then I could divide both sides by 1.29. v question mark is going to be this whole thing, 300 million divided by 1.29. Or another way to think of it is, light travels 1.29 times faster in a vacuum than it does in this material right over here. But let's figure out it's velocity. So in this material, light will travel a slow-- so the 300 million divided by 1.29. Light will travel a super slow 232 million meters per second. So this is approximately, just to round off, 232 million meters per second. And if we had to guess what this material is, let's see-- I just made up these numbers-- but let's see if there's something that has a refraction index close to 1.29. So that's pretty close to 1.29 here. So maybe this is some type of interface with water in a vacuum, where the water somehow isn't actually evaporating because of the lack of pressure. Or maybe it's some other material. Let's keep it that way, maybe it's some type of solid material. But anyway, those were, hopefully, two fairly straightforward Snell's law problems. In the next video, I'll do a slightly more involved one.