reflection and refraction going to be the topics of this lesson in my new General Physics playlist which when complete will cover a full year of University algebra based physics now when light is traveling from one medium to the next couple different things might happen at that medium and that's what we're going to concern ourselves with this lesson now the first might be reflection and we'll spend a very little bit of time talking about the law of reflection but the light that is actually transmitted into the next medium provided it's transparent so what happens is it bends a little bit or at least it changes Direction and we call this refraction we're going to talk about the index of refraction we're going to talk about Snell's law of refraction we'll talk about the special case of total internal reflection and spend just a little bit of time talking about dispersion my name is Chad and welcome to Chad's prep where my goal is to take the stress out of learning science now if you're new to the channel we've got comprehensive playlists for General chemistry organic chemistry General Physics and high school chemistry and on Chads prep.com you'll find premium Master courses for the same that include study guides and a ton of practice you'll also find comprehensive prep courses for the DAT the MCAT and the oat so let's first describe reflection and refraction and if we take a look at say maybe some uh light rays entering water and we use these lovely Ray diagrams where you straight line Rays to kind of identify light rays if you will so and if we're going from say air to water and this doesn't have to be air and water but if we're going from one medium to the next and assuming that they are both transparent mediums so we got a couple different things what might happen at this interface now one is you might get some reflection here so and it turns out we draw the normal here the perpendicular to the surface and the angle that the light Ray makes with respect to that the one that's entering we're going to call this Theta so and it turns out we have one simple Law of Reflection the angle of incidence we call this is equal to the angle of reflection and so couple different ways we might def find this we might call this Theta 1 we might call this Theta 1 Prime so and big thing here with angle of incidence equals angle of reflection angle or Theta 1 equals Theta 1 Prime That's the Law of Reflection so and it doesn't matter what this angle is that angle of incidence will equal the angle of reflection assuming this is a fairly flat smooth surface you know if it's choppy water and stuff like that well you're you're going to get a little uh more complicated effect that's beyond the scope of this course but for a smooth interface between the two mediums angle of incidence equals the angle of reflection and we don't really want to go too much more into it than that the rest of this lesson is going to be spent talking about refraction now in most cases not all of that light is going to be reflected some portion of it is actually going to get passed along into the next medium which might be water as in this case or something like this so and it turns out though when you go from one medium to the next typically the light rays are going to we say Bend so it might be better just to say that they change direction so it's usually fairly slight but it can be more profound in in certain cases and stuff like this so in the case of going from uh air to say water we might find that they would actually change directions so towards the normal and again this angle right here we'd Now call our angle of refraction so in this case we often call it Theta 2 and so turns out that it's typically not the case that your angle of incidence equals your angle of refraction so the only way that would be true is if the two substances you're dealing with happen to have the exact same index of refraction which we'll talk about next so now let's focus just on the refraction in fact the rest of this lesson we will only be focusing on the refraction so is there a potential for a reflected ray yes and we're just not considering it we want to focus just on the refraction now before we do that we've got to talk about uh some properties of the electromagnetic waves as they're being transmitted from one medium to the next and to do that we have to identify what we call the index of refraction so n which is equal to the ratio of the speed of light in a vacuum relative to the speed of light in that medium and the reason we identify this is because the speed of light in a medium is going to change relative to the speed of light in a vacuum it turns out no matter what medium you shine light through it's going to be slower than the speed of light in a vacuum and so if you look then if the speed of light in a vacuum relative to the speed of light in that Medium if this is always going to be slower than C the speed of light in a vacuum then this ratio is always going to be larger than one and the bigger it gets gets the slower the light is in that particular medium now if you take a look at air speed of light in air is almost exactly 3.0 * 10 8 m/s almost it's like 2.99 and some change so and as a result turns out the index of refraction for air is like 1.003 whereas this you know index of refraction for a vacuum would be I guess one by definition so but notice that what that means then is that the speed of light in air is almost exactly the same as the speed of light in a vacuum in fact we treat them as being the same value 3.0 * 10 8 because just to like two sig figs it's exactly true you've got to carry out much more sig figs to see where they start to differ more appreciably now in the case of say water it turns out the index of refraction for water is actually 1.33 for certain types of glass it's 1.52 depending on the substance it changes and so the speed of light is changing now you might be asking the question be like I thought the speed of light was a constant well the speed of light in a vacuum is a constant it's a constant 3.0 * 10 e m/s but it turns out once you take it out of a vacuum even going through air what you're going to find out is that the electric field of the electromagnetic waves is going to start interacting with matter while in air matter is pretty diffuse and spread out so there's not a lot of matter for it to interact with and so it doesn't change things appreciably but once you get into liquids and solids that are transparent as the light is transmitted into to those uh you get stronger and stronger interactions between the electric field of the electromagnetic waves and any kind of uh protons electrons dipoles uh that are present in that matter and the stronger they are often times the greater the interaction and they ultimately tend to decrease the electric field and overall slow down the overall speed of light in that Medium so this index of refraction is kind of our our big go-to characteristic for defining a medium and the larger it gets the slower the speed of uh of light in that particular medium okay so if the speed of light is changing what else is changing because if you recall we said that the frequency times the wavelength was equal to the speed of light and that is true but it's true in the medium as well and so if you're in some other medium the frequency times the wavelength is going to equal the velocity as well now here's the deal if you were going from one medium to the next the frequency doesn't change if the frequency either sped up or slowed down you would have a problem with what's going on at the interface it was slowing down you get this buildup of wavelengths piling up at the interface waiting to get transmitted because the frequency was slower in the new medium so it turns out it's just impossible the frequency can't speed up or slow down the frequency from one medium to the next is exactly the same and so any change in the speed of light is not due to a change in the frequency it's due to a change in the wavelength so if the speed of the light in the medium goes down it's because the wavelength is going going down proportionally now one thing to keep in mind like when I talk about violet light and I say violet light has wavelength of right around 400 nomers well I'm referring to the wavelength of violet light in a vacuum which is almost exactly the same as the wavelength of violet light in air so however if you shine that violet light in you know underwater or something like this the color of that light is not going to change I want to make that clear it's still violet light but it will have a different wavelength in that new Medium say water or whatever so so keep in mind that color doesn't change frequency doesn't change but the wavelength and the speed are what are changing so just keep that in mind all right now the reason we're kind of introducing this index of refraction is that we can predict the angle of refraction using what's called Snell's law of refraction which is what we're going to take a look at next so here we have Snell's law of refraction N1 sin Theta 1 equal N2 sin Theta 2 where N1 is the index of refraction of the medium the light is coming from N2 is the index of refraction of the medium where the light is ending up at going to being transmitted into Theta 1 is the angle of incidence Theta 2 the angle of refraction so angle of incidence angle of refraction N1 N2 so and if you notice like if we you know you envision this as being air and water then air would be N1 correspond to N1 anyways and water would correspond to N2 but nothing says we couldn't shine light from underneath the water out out and in that case then the water would have been N1 and the air would be N2 again wherever the light is beginning the medium of origin that's going to correspond to N1 and wherever it is transmitted into the next medium that's going to correspond to N2 now big consequence here I want you to realize something so if we remind yourself what a sign function looks like so we've got here 0 degrees s function looks like this so all the way to 360° well the big thing I want to make a cut off here is at 90° and so if you look what is the smallest angle we could possibly have relative to the normal for light being shined from one medium to the next well the smallest it could be is zero if it's going straight in so and it turns out in such cases then it just comes straight out the other side right along the normal as well it's only when you come off the normal that you're going to see this bending of the light we call refraction and so in this case it turns out then what's the biggest angle of incidence it could be well to still be being trans ited from this medium the biggest it's going to be away from that normal it's going to approach 90° but it's never going to be larger than 90° so the reason I highlighted 90 here is I wanted you to see what the sign function does just in this little narrow window the sign function only grows so it only gets larger as long as we're restricting ourselves to talking only about angles from 0 to 90° as we are doing right here and so when we're talking about reflection I'm sorry when we're talking about uh incident angles and refracted angles 90° is the maximum you're going to be away from the normal no never anything larger all right so if we take a look at Snell's law one more time then so we can make some conclusions about when N1 is larger or N2 is larger and so let's just start off talking about when N1 is larger than N2 and then we'll talk about when N2 is larger than N1 now if N1 is larger than N2 so what that would mean here is that N1 is larger N2 is smaller well then to maintain that equality if n1's larger then sin Theta 1's going to have to be smaller than sin Theta 2 now signs again if we hadn't kind of made this determination well again as long as we're restricting ourselves just talking about from 0 to 90° as Theta goes up sin Theta goes up and so instead of saying with N1 being larger sin Theta 1 has to be smaller we can actually make it even simpler and just say Theta itself is going to have to be smaller whereas over here Theta 2 is going to have to be larger because Theta and sin Theta correlate at least from 0 to 90 and so whichever side has the larger value of the index refraction is going to have the smaller angle so it's always going to work that way with refraction so whichever side has the smaller index of a fraction is going to have the larger angle and so going over here if you look we can see that our angle of incidence was bigger than our angle of refraction well why would that be if this has got the bigger angle so then N1 versus N2 here if this has got the bigger angle it must have had the smaller index of refraction so this would be a case where N1 was less than N2 or if you want to say N2 is greater than N1 sameed diff and so if the refracted Ray bends toward the normal that's indication that you've gone from a lower index of refraction to a higher index of refraction or another way of phrasing that would be to say that the speed of light has slowed down from one medium to the next so let's take a look at the opposite possibility here we'll draw in the normal so but in this case we're going to have this Bend away from the normal and so now again our incident angle or angle of incidence Theta 1 as compared to our angle of refraction ction now Theta 2 is bigger and the reason Theta 2 is bigger must be because instead of the speed of light slowing down the speed of light must have sped up so N2 down here versus N1 up here if I get the bigger angle it must be because we have the smaller index of refraction and again I'm not I haven't provided any kind of like explanation for why this is or how this works I'm just interpreting the math here in light of the relationship between Theta and sin Theta and this equality here we call Snell's law so but it's really convenient you want to be able to predict whether or not uh your refracted angle is going to bend toward the normal or away from the normal and it's all did the index or fraction you enter is it higher than the one you started with or lower than the one that you started with that's kind of the key so before we move on we want to get some practice in I've got a few sample calculations want to work out and so the first one uh just says what is the speed of light in a medium having an index of refraction of 2 so here n is equal to 2.0 well again if you remind yourself what an index or fraction is it's the ratio of the speed of light in a vacuum to the speed of light in the medium and if that ratio is equal to two that just simply means that the speed of light in the vacuum is twice as big as the speed of light in that medium well we know the speed of light in a vacuum is 3.0 * 10 8 m/ Second and if that is twice as big of the speed of light in that Medium you can you know do the math pretty well in your head which is why I chose a nice round number for that index of refraction here so but if we take nals C over V rearrange it V is going to equal C Over N which in our case is 3.0 * 10 8 m/s all over 2.0 and that velocity is therefore going to come out to 1.5 * 10 8 m/s so ultimately you can kind of look at it if your index refraction is two it's because the speed of light in the medium is half that of what it is in in a vacuum if your index of refraction was 3.0 it would mean that the speed of light in that Medium is onethird of what it is in a vacuum so on and so forth now obviously if you don't have a nice round number you can use the equation but I wanted you to at least in principle kind of see what this index or refraction implies about the speed of light in that Medium relative to a vacuum now the next question we're going to look at the wavelength and frequency of orange light in a vacuum are 600.0 nanometers and 5.0 time 10 14th Hertz respectively so let's start that off in a vacuum okay so the wavelength is 6 100.0 nanom frequency is 5.0 * 10 14 Herz and the question is what are the frequency and wavelength in a medium having an index of refraction of 2.0 so now we've got some new medium with an index of refraction of 2.0 what's the new frequency what's the new wavelength and again the big thing you're supposed to remember is that as light is transmitted from one medium to the next the frequency remains constant so that hasn't changed frequency is still 5.0 * 10 14th Hertz but the question is what is the new wavelength and again the light is still orange light but it's no longer going to have a wavelength of 600.0 nanm now a couple different ways to look at this so if you look at the relationship between frequency wavelength and the speed of the wave in that Medium the frequency hasn't changed but the Velocity in this case has gone down relative to a vacuum by factor of two well velocity and wavelength are going to be proportional if the frequencies remaining constant so if the velocity is gone down by factor of two the wavelengths gone down by a factor of two and if you know that you're like well going down by a factor of two would mean that our wavelength is going to drop down to 300.0 nanom and that is true now there's a couple different ways you could go about figuring this out so one is you could figure out what that velocity of the wave is based on your index reaction which we already did so and then you could say again that your frequency times your wavelength equals that velocity and so your wavelength must equal the velocity over the frequency which in this case would be that 1.5 * 10 8 m/s all over 5.0 * 10 14th Hertz which is a inverse second so these will cancel it'll come out meters then You' convert it to nanometers and stuff like this so but lo and behold if you did this with 3.0 * 10 8 all over the frequency you'd get the 600 nanometers I mean you you'd get like 6.0 * 107 met something like that and convert to nanometers so if we're doing it with half as big a number in the numerator that's why it's going to give us half as big a wavelength in this case but by all means let your calculator do the heavy lifting for you but it will indeed be 300.0 nanometers here okay now the other way way you could approach this so keep in mind that the index of refraction is equal to C over V where this is the frequency times the wavelength and I'll call them fnot and Lambda KN for being in a vacuum all over F in that medium and Lambda in that Medium so but in this case the frequencies don't change and so another way to look at this is that the index refraction is not only equal to the speed of light in a vacuum relative to the speed of light in that Medium it's equal to the wavelength of the light in the vacuum relative to the wavelength of light in the medium as well and we could have just gone about it this way as well and just said okay well the index of refraction was 2.0 that's equal to wavelength vacuum of 600.0 nanom all over the wavelength in that Medium move this over divide by two and you get your 300.0 nanometers that way as well so the next three questions are going to involve mostly Snell's law of refraction and you're going to do a lot more calculations with Snell's law of refraction then you are with the law of reflection and it's not really a calculation right the angle of incidence equals the angle of reflection but we're so used to getting bombarded with questions regarding refraction that be careful they don't just slip one about reflection in like I'm about to here the question says if the angle of incidence of a flashlights beam into a lake is 30.0 deges what is the angle of reflection and notice in this question I'm nice enough to then provide you with indices of refraction and index of refraction of a is 1.0 and the index of refraction of water is 1.33 but you don't need the indexes of refraction because I'm asking about the angle of reflection and so in this case uh in fact I'm not even going to justify this by writing it on the board but if the angle of incidence is 30° then the angle of reflection is also 30° law of incidence equals Law of Reflection done or I'm sorry angle of incidence equals angle of reflection done that's the law of reflection all right but the next question might have been the one you thought you were being asked and it says if the angle of incidence of a flashlight beam into a lake is 30° and so we'll shine this in here angle of 30° so let's get that normal on there so in this case this angle of incidence relative to the normal again is 30° the question is what is the angle of refraction and then the index refraction of air is given as 1.0 and for water so N1 N2 1.33 now notice I divided you know identified them as N1 and N2 but again wherever the light starts that's N1 according to s's law wherever the light ends up going to next is N2 all right so those IND indices of refraction were provided and notice if we're going from a smaller index of refraction to a larger index of refraction well again if the index of refraction goes up then the angle must go down and so this is going to bend towards the normal that way we get a smaller angle Theta 2 right there and so we know it's going to be less than 30° question is what is that angle that's what we're going to use Snell's law for and we are definitely going to rely on our calculator to accomplish this and so we've got N1 sin Theta 1 equals N2 sin Theta 2 and N1 is just the 1.0 so s of 30.0 De equal N2 1.33 * the S of theta 2 and so we're going to divide over here by 1.33 and then take the inverse sign and we'll definitely let our calculator do the heavy lifting for us here uh obviously anything times one is itself and S of 30 is 1/2 I made the the math nice here so we got 1 time a half and then divided by 1.33 so 0.5 divided 1.33 equals like 3759 something something change and then we'll just take the inverse sign of that answer and we'll get 22 08 and I'm going to round it to three sig figs here so 22.1 de okay so there's your first application of Snell's law this next question is going to look a little bit familiar uh it says an underwater diver shines a flashlight toward the surface if the angle of incidence is 22.1 De what is the angle of refraction so this is kind of the exact reverse of what we just did here so now instead of the lights originating from the side with the air so the light is going to originate now from the side of the water from this underwater diver we'll get the normal on there and we're starting with an angle of 22.1 De that should look pretty familiar here that was this angle right here the angle of refraction that's going to be our angle to start with and we're starting with N1 being 1.33 and then N2 now being 1.0 and so let's go back and set up Snell's law one more time so we're going to have N1 sin Theta 1 = N2 sin Theta 2 N1 is now 1.33 time the S of 22.1 De equals N2 which is the 1.0 * s of theta 2 and if you look at the math on this compare it to this equation right here you had 1 * sin 30 1 * s of theta 2 but on the other side notice it's exactly the same here it's 1.33 * s of theta 2 where that Theta 2 was 22.1 de so the 1.33 was ultimately being multiplied by S of 22.1 so if this side is the same as this side well then the other side which has a 1.0 here is also going to have a 30° Theta 2 so and we'll confirm that with our calculator so but Snell's law here is completely reversible if you kind of reverse the the ray diagram here and so in this case if we take one point 1.33 times the S of 22.1 and then divid it by 1 and then we're going to take the inverse sign of all of that and we're going to get 30.02 5 which I'm going to round to 30 0° at the very least if we were drawing the ray diagram in we're going from a larger index of refraction to a smaller and so if the index of refraction is getting smaller then the angle must be getting larger so we're bending away from the normal so Theta 2 is going to be bigger than Theta 1 over here and in fact we were just doing this exact diagram in reverse so the next topic involves refraction or really the absence of refraction and it's called total internal reflection now it turns out when you go from an area where you got a higher index of refraction to an area of lower index of refraction we saw that the angle is going to kind of Bend away from the normal so we got this angle of incidence and because you're going to a lower index of refraction you've got to get a larger angle of refraction it turns out this idea of total internal reflection is only possible when you get a larger angle of refraction than the angle of incidence when that uh transmitted Ray bends away from the normal and if you make this angle of incidence bigger the angle of refraction will get bigger and at some point you've made this angle big enough that the angle of refraction is going to be 90° and it turns out we refer to this angle right here as the critical angle and ultimately What's Happening Here is if if the angle of refraction is 90° and the light's being refracted we'll call it right along the interface between the two mediums but none of the light is actually being transmitted into that second medium whatsoever and so at this point no refraction into that next medium is ultimately taking place you're only getting reflection no refraction hence the name total internal reflection now if we take a look at what this is means in terms of Snell's law so a couple things here so one we're going to input the critical angle here and when you hit that critical angle the angle of refraction comes out to 90° well the S of 90 is one and anything times one is itself so that ultimately just kind of falls out of the equation and so if you go ahead and solve for the sign of that critical angle you get N2 and will divide through by N1 so if you take the inverse sign you solve for that critical angle and ultimately any light being shined towards the surface at any angle equal to or larger than that angle will lead to no refraction whatsoever and only reflection so let's give an example of this so next question says what is the critical angle for total internal reflection for light traveling from from glass to air so and the index of refraction of air is given as 1.0 and for glass as 1.50 so and again you always you know plug and chug this with Snell's law but typically this is an equation that is presented and definitely kind of uh already reduced down for us for quick calculation it's what we'll use we'll take the inverse sign N2 over N1 so 1.0 over 1.50 let our calculator do the heavy lifting for us so the inverse sign parentheses of 1 / 1.5 close the parenthesis we're going to get 41.8 de with three sigfigs cool so we kind of showed why this only works when you're going from medium of higher index refraction to lower index refraction because we need this uh refracted rate to bend away from the normal to the point where it no longer actually is being refracted now you can look at it mathematically as well so here we're taking the inverse sign of this ratio well you can only take the inverse sign of things that are actually in the range of what the sign function is and the sign goes from negative 1 to a maximum positive one it can't be larger than positive one uh if you're going to take uh uh if you're going to if it's part of the sign function so if you're going to take the inverse sign you can only take the inverse sign of a number that's between negative 1 and positive one and so here N2 has to be smaller than N1 if it was the other way around you actually mathematically couldn't take an inverse sign and so whether you want to look at it conceptually or mathematically so total internal reflection is only possible when you're going for from a region with a higher index of refraction to a region with a lower index of refraction now one common example and really useful use of total internal reflection is fiber optic cables and so fiber optic cables are either a glass or plastic cable that they coat with a special coating and this special coating has a much lower index of refraction than the glass or the plastic that the cable is made out of so in such a way so that we can shine light in one end of this cable and it's going to reflect so but not refract whatsoever because you're always going to be greater than that uh critical angle and so this is going to keep reflecting all the way down and get transmitted through this cable and it's nice and flexible and we have a light signal if you will well these can replace electric signals which involve copper wires which are much larger and more expensive to manufacture and things of the sort uh and and also these light signals travel faster they're traveling at the speed of light now they're not going in a straight line or anything like that but still if you're traveling at the speed of light you're going to Trump any kind of electric signal which does not travel at the speed of light uh hands down it's not even close so really useful in fiber optic cables which are commonly used for the internet and the phone systems uh that still have landlines anyways and things of this sort now one last topic we want to talk about it's called dispersion so and we're going to look at it purely conceptually no math involved in this part whatsoever but dispers results from the fact that uh turns out the index of refraction is wavelength dependent now it turns out it also has a temperature dependence that we're totally not going to look at but it is wavelength dependent the different wavelengths of light actually have slightly different indices of refraction in a particular medium so like when we gave you an index of refraction all throughout this lesson we're usually giving you some sort of average across all the wavelengths so and that average is pretty effective like let's say you're using like white light which is really kind of all the wavelengths and stuff like that like is good so but when you start doing refraction with white light so the longer that light travels it might actually separate into colors because of these slightly different indices of refraction so one thing you should know is that for most mediums there's an inverse relationship the index refraction is going to be higher for the shorter wavelengths so for example like we gave you the index of refraction of water is 1.33 well it turns out it's not quite so easy so we can talk about the index of refraction for red light in water so versus blue light in water and so if you look Blue's definitely got the shorter wavelength uh around 400 NM Red's closer to 700 NM so for the shorter wavelength we should expect a higher index of refraction so and that's what we see if I round it to actually give it an extra decimal place here it's 1343 and for red light at the same temperature it's like 1.33 0 and so very subtle difference in the index of refraction well what this do let's take an a look here so let's get a interface between air and water now let's shine some light through so let's start with the red light let's get our normal on the diagram and as we're going from a lower index of refraction to a higher index of refraction therefore if you're going from lower to higher then your angle has to get smaller so it's going to bend towards the normal so here Theta one versus here Theta 2 it got smaller okay so but instead of white light let's say this really was red W uh red red wavelengths of you know right around 700 nmet or so where the index of ref fraction was 1.33 but let's say we also shine blue light right along that same path so and now it's got the same incident angle but with an a little bit larger index of refraction it's going to want to get the angle small smaller that angle of refraction even smaller it's going to want to bend towards the normal even more if you will and follow a slightly different path here for a smaller Theta 2 if you will cool and I I kind of took the edges of the visible spectrum with red light on the longer end uh for wavelength and red light uh I'm sorry blue light on the shorter end for wavelength so but you'd have all the different colors in between if it was actually white light being shined through and so this is what happens when you shine white light through a prism because of those different uh slightly different uh uh values for the index of refraction for the different colors they get split onto slightly different paths and if you shine them on a screen you can see the entire spectrum that white light gets broken up into its colors so it is also a consequence of this that we can see rainbows and things of this sort you're getting dispersion in raindrops as well now a little bit more of a complex phenomenon but it's still a result of dispersion if you found this lesson helpful consider giving it a like happy studying