Transcript for:
Overview of Waves for A-Level Physics

let's go over the Hall of AQA waves for a level physics we're going to start off with the oscillations of the particles in a media we need to define a few terms let's start off with the amplitude this is defined as the maximum displacement from the equilibrium position and it's Illustrated with these blue lines across here in this displacement against distance sketch that I have across here onto our next definition which is the frequency remember the equation for the frequency should we just write this over here as well is that frequency is defined as one over the time period And as such it defines the number of completed cycles per unit time please know that sometimes you ask to Define units the SI unit for frequency is the Herz how would we Define the Hertz itself well that is the number of cycles per second remember if we're defining a unit we can only Define it in terms of other units we can also Define the wavelength which is the minimum distance between two points in a wave that are in phase for instance this point over here let's say 0. one and point 2 they are in phase and the distance between them is known as the wavelength the speed of a wave on the other hand is the distance traveled by the wave per unit time and the famous equation um the famous wave equation is that wave speed is equal to frequency multiplied by wavelength and now let's talk about the phase of a wave the phase describes the fraction of a cycle covered up to that point and it can be measured in either degrees or radians here's a little example I just have a displacement against distance please know that this could be for instance displacement against time as well the concept is exactly the same each point is going to be assigned a value which is a fraction of a completed cycle please note that one complete cycle is equivalent to 360° which is also equivalent to 2 pi radians why is that we're borrowing some mathematics from a circle essentially because after we complete one cycle we come back back to where we started kind of like if we're moving in a circle and we come back to the point where we started each point will be assigned a phase for instance this point number one over here has a phase of 0° 0 radians the next one here is a quarter of a cycle so 2 which is 90° which is Pi / 2 this one here 3 is going to have a phase of 180° which is pi radians Point number four will be 270 which is 3 piun / 2 radians and the last point here5 is going to have a phase of 360° and 2 pi radians now a phase difference describes how far out of sync two points on a wave or two points on separate waves are in degrees or radians for instance the phase difference between let's take points three and four so to go from here to here I just need to go up with 90° which is also Pi / 2 radians let's do one more I don't know shall we just say1 and 3 well the phase difference between them is simply 180° which can also be expressed as Pi radians here's an example question on phase difference so the distance between point 1 and point 2 is shown in the diagram this here is just .5 CM if the wavelength of the wave is 1.2 cm find the phase difference between points 1 and two we need to express our answer in radians up to the appropriate number of significant figures there's a formula that we're going to find very very useful and that is that phase difference is going to be equal to the distance between the two points on the actual waves basically this distance here D whatever that may be divided by Lambda this here is going to give us the fraction of the cycle and then multiply this by 2 pi radians please note that if this was in degrees rather than 2 pi we would write 360° but we're going to stick to radians because that's what the question is actually asking us our phase difference is the distance between the two waves which is just .50 divide that by the wavelength which is given in the question 1.2 multiply this by 2 pi because the question is asking me to do this in Radiance up to the appropriate number of significant figures I'm not going to leave this as a fraction of Pi putting this into the calculator we get 2 61 7199 but all of our answers are up to two six figs so we're just going to write 2.6 radians just a quick little note that I thought would be very useful for you well here are all of the phase difference equations if we have a displacement against distance graph they can be given different symbols but if the distance between two points is D then the phase difference is going to be that distance divided by the weight wavelength time 360 if we're in degrees if we're in radians it's going to be D over the wavelength time 2 pi now we can also find the phase difference from a displacement against time graph now the separation between the two points will now be in units of time so if that's T then the phase difference is going to be a fraction of a full cycle so it's going to be T / capital T where capital t is actually just the time period if we're in degrees the phase difference is going to be T over capital T * 360 if we're in radians it's going to be T over the time period multiply this by 2 pi let's cover transverse and longitudinal waves so a transverse wave is a wave where the oscillations are perpendicular to the direction of energy transfer whereas in a longitudinal wave the oscillations are parallel to the direction of energy transfer this also leads to compressions and rare factions an example of transverse waves are em or electromagnetic waves it's important to knowe that all electromagnetic radiation travels at the same speed in a vacuum this is given in the formula booklet but the speed of line typically given the symbol C is right around 3.0 multipli by 10^ 8 m/s you can also have have waves on a piece of string for instance as long as the oscillations are perpendicular to the direction of energy transfer it's a transverse wave you can also have longitudinal waves an example of those for instance are sound waves and now let's talk about polarization what is the definition of a plain polarized wave so the oscillations of the particles or the fields are restricted onto a sing single plane because of that only transverse waves can be polarized because in a longitudinal wave the oscillations are always parallel to the direction of motion anyways now if you have a transverse wave you can have a whole bunch of different planes through which you can oscillate you can have a vertical plane polarization which essentially all of the individual particles are kind of going up or down like this and that's going to correspond to this case sometimes you will see in textbooks and exam questions this symbol for a vertical plane polarization you can also have a horizontally plain polarized wave in this case the particles are going to be moving horizontally like so along this line and so on and so forth you can also have a mixture of these in which the wave is oscillating along all planes or has a component of polarization in the vertical plane a component in the horizontal Al plane and if that's the case you have unpolarized wave the symbol that you sometimes see is this one here signifying that it has a vertical component and a horizontal component just as a side note but very often uh sunlight is an example of an unpolarized waves in particular non-reflected sunlight we're about to go into this in a little bit more detail and now let's have a look at an application of polarization namely Polaroid materials but first let's take a moment to appreciate my drawing so this drawing is definitely not up to scale by the way we have some sunlight which is being emitted from the Sun in all directions initially this light is unpolarized and then that light let's say it travels to the surface it strikes the surface but then after it is reflected most most of the light is horizontally polarized so only this component is left maybe we have like a tiny vertical component and this reflection over here which is mostly horizontal I know I'm exaggerating it this is what reaches our Polaroid however our Polaroid only transmits the vertical component of the light the horizontal component is absorbed that means that that most of the reflected light will actually be absorbed so most of the reflections will be absorbed by the Polaroid and that's going to reduce glare and Reflections if there are other things around here I don't know let's say that there's a piece of rock or something like that I know this looks like a square but if there's something over here then the light from it will not necessarily be horizontally Pol polarized mostly and we'll be able to see this with approximately only 50% drop in intensity because only the component that is horizontal is going to be absorbed and we still have quite a large vertical component so to sum up the Polaroids absorb the horizontal component of the polarization of light please note that in some questions you may be given a Polaroid that's been that has been rotated so the horizontal and vertical Direction may change on the question so please read the question really carefully but in this case it absorbs the horizontal component the Polaroid transmits the vertical component if light is unpolarized 50% of the light will pass through the Polaroid the reflected light is mostly horizontally polarized so most of the reflected oh this here should say light will be absorbed reducing the glare and the reflections a different application of polarization is to be found in AOS for transmission and reception let's have a look at an example problem we've got a transmitter of microwaves there it is and a receiver with an aerial which is shown and we need to plan an experiment to determine if the transmitter emits waves which are polarized either in the vertical or the horizontal plane so the first thing to note in this question is that we need some form of rotation typically what we tend to be able to rotate is to rotate the area we can start rotating it in one of the planes for instance in the vertical plane if the question specifically mentions the vertical plane in the question itself we need to mention the vertical plane if the area was vertical when the signal shown on the receiver is at a maximum the transmitter admits waves which have been polarized in the vertical plane on the other hand if the area is horizontal basically this thing tilted on its side when the signal shown on the receiver is a maximum then the transmitter the waves of the transmitter are polarized in the horizontal plane and now let's talk about how a stationary wave is formed an incident wave is reflected of a boundry or a fixed point the incident and the reflected wave will then superpose and places where the waves interfere constructively are known as antinodes these are points of Maximum amplitude places where the waves interfere destructively are known as nodes and a node is simply a point of zero amplitude what are the differences between Progressive waves and stationary waves well first of all Progressive waves waves actually transfer energy in the direction of travel whereas in stationary wave the energy is actually stored between the two points where the wave is additionally in a progressive wave the amplitude is constant across all points now what do I mean by that imagine I have a rope between those two points and I send a progressive wave to travel then let's say if the wave is going this way each of those points is going to go up and down by the same amount which is known as the amplitude the amplitude itself varies on a stationary wave so here is a stationary wave uh what let's just draw another one over here so here is my station wave let's say we have something like this just differentiate it further from a progressive wave let's draw it like this so if we're at this point we will constantly be going up and down up and down like so with the maximum amplitude whereas if we are on a point right next to it we are going to be going up and down with a slightly lower amplitude and then slightly lower amplitude still until we reach a node where we're going to have zero amplitudes so this here is zero this here is a Max standing for maximum amplitude and let's also talk about phase difference in station waves which is a little bit tricky to understand the two rules that I like to remember are that between adjacent nodes all the particles are in phase and particles on different sides of a node are in antias now what does that mean imagine that we have a stationary wave so let's say we draw a wave that looks something like this this here is going to be a node this here is going to be a node and this here will be an antinode antinode node so on and so forth our first rule says that between adjacent nodes all the particles are in Phase well those are two nodes and the particles between them are going to be in phase meaning that all of those particles between the nodes are going to be going either up or down together like so on the other hand between those two adjacent nodes all the particles are going to be going down and this is where our second rule comes in place and that is that particles on different sides of a node are in anti base so those two regions on the different sides of the same node are in antiphase just while we talking about useful rules the distance between adjacent nodes or adjacent antinodes is actually half of the wavelength so for instance let's say this distance between those two nodes well we can clearly see that this will will be half a wavelength and this one here will also be half a wavelength but also between antinodes as well so that will be this distance here this will also be Lambda over 2 by the wavelength we mean the wavelength of the progressive wave which formed the stationary wave the harmonics can be demonstrated by this lovely experiment in which we have a vibration generator the vibration generator is typically by the way connected to something known as a signal generator as well which can produce an electrical signal at a given frequency and that moves the vibration generator that particular frequency we have a piece of string of a certain length between the pulley and the vibration generator and then a mass underneath we're going to notice that at a certain frequencies specific patterns appear so the first harmonic occurs whenever the length is exactly equal to half of the wavelength well how do we remember this in this particular experiment our wave is right between two fixed points and those have to be nodes well the first possible solution is actually just fitting in about well exactly half the wav length so L is equal to Lambda / 2 one wavelength is the next possible solution which means that between this um node and node uh over a distance L we can actually fit two half wavelengths and this case L is equal to Lambda notice that this is not a progressive wave but a stationary wave and then of course three halfes Lambda would be the third harmonic each time adding in more half wave lengths there's also something really important that will allow us to calculate any subsequent harmonics so all the harmonics are multiples of the first harmonic in this case here I've just made up some numbers just for the sake of revision our first harmonic is let's say 30 htz so if the first is 30 htz then the second will be found at 2 * 30 which is equal to 60 hertz the third harmonic will be found at * 30 which will be at 90 htz and so on and so forth and this is very often a very nice easy shortcut to solve a question Let's do an example problem a stationing wave is set up on a string it's shown below and it's oscillated at the second harmonic what we need to do is compare the motion of the following points we have initially point1 and then also 3 the best way to tackle these questions is by identifying which are the nodes and which are the antinodes so in the second harmonic we have two half wavelengths so this here will be a node and this here will be a node uh this here will be an anti node and this here 2 will be an antinote now notice that1 and 3 are on the opposite sides of a node and this means that there will be an an T phase please note that we can also say that they have a phase difference of Pi radians or 180° now what if this question was let's say for two marks each another aspect that we can write is that all the points will have the same frequency and if they have the same frequency they will also have the same time period if there was a grid behind this this is just merely a sketch we could probably compare the amplitudes as well now tell you what let's compare the amplitudes for Part B in which we need to compare the motion of points one and two now Point 2 is the antinode therefore 2 will have the higher amplitude alternatively we can say that uh 0.1 will have a lower amplitude compared to two and also those two points will also uh have exactly the same frequency but also they will have the same phase so we can say that points one and two are in Phase if we ever asked to talk about the nodes remember the nodes are actually stationary they have zero displacement zero amplitude in the AA formula booklet you will find this equation that the frequency of the first harmonic only is given by 1 2 L root of the tension in the string divided by the mass per unit length now where does this equation actually come from even though the derivation is not actually on the spec I find that it's pretty crucial to understanding and being able to apply this to new situations which is what the exam is all about so where does this come from well if we have the first harmonic notice how in the length L we can only have half a wavelength you can think of this as two nodes so this here from here to here is half a wavelength well this here would mean that the wavelength will just be equal to 2 L but now remember that V the wave speed is equal to the frequency C multiply by the wavelength this means that the wavelength will be equal to V over F so we can plug this into here to get that V over f is actually just equal to 2 L let's be consistent with our L's we didn't have a capital l so I'll just change that to a lowercase L we're almost there now should we just flip this and rearrange for the frequency so I'm going to do f/ V is equal to 1/ 2 L now let's carry on across here F will be equal to 1 / 2 L multiplied by the speed V and now another equation that is sometimes given in the question itself is that V is equal to the square root of the tension divided by the mass per unit length okay well we can just sub this back into here and we are going to get that F isal to 1 2 L and then the speed is equal to T / um the mass per unit length now what do all of those symbols actually mean so the tension is just equal to essentially the force which is is acting on this piece of string so let's say if we had a mass here M then the tension will just simply be equal to mg mu here which is the mass per unit length and mu which is the mass per unit length the equation for that will just be equal to the mass of the string which is divide that by the total length of the actual string now something which I've noticed quite a lot is that sometimes you given the density in the cross-sectional area of the cable or the actual string so a really useful equation to keep in mind for Mu because this was equal to the mass of the string divided by the length of the string so but mass in general is equal to density time volume so one thing that we can say is that this is equal to the density of the string or the C multiply by the volume divided by the length but because our volume for a circular cable is just going to be equal to the cross-sectional area multiplied by the length then this here will be equal to the density of the string multiplied by the cross-sectional area multiply by the length divide that by the length and notice how the two lengths are going to cancel so this means that our mass per unit length will just be equal to the density of the string multipli by its cross-sectional area it's typically circular but obviously do watch out uh what the question actually says so this here will be given by pi d/ 2 s leading to the density of the string multip by pi d^2 divide that by four and this is something that I've seen quite a lot in AQA questions we also need to go over a really important definition and that is the definition of superposition so when two waves interact the result in displacement is the vector sum of the individual displacements the word displacement is typically underlined in marks games meaning that we need to mention it to score full marks but what does this definition actually mean so let's say I have a point x and a wave arrives there and as it's arriving it's in this part of the cycle let's say that another wave arrives at that point at exactly the same time and in this case their overall displacement is going to be equal to zero when one is going up the other one is going down meaning that the overall motion is cancelled in a way on the other hand if we had two waves that were arriving in Phase so let's draw the other one let's say a little bit small like smaller like this then the resultant wave will have a displacement which will be the sum of this plus that which is going to be something like something over here so it's going to look something like this uh roughly speaking of course moving on to interference we first need to consider a couple of really important terms we're going to start off with path difference path difference is defined as the difference in path lengths between the waves that has been traveled that have been emitted by the sources so what do I mean by that imagine that I have a couple of sources of w these could be speakers with sound waves they could be two separate lasers they could be two slits in the double slit experiment Etc now if I pick a point p and let's say that the point p is over here then the distance traveled by Source One to P is going to be this much however the distance traveled by Source 2 to P will clearly be bigger and that difference in this will be the path length we can even illustrate this by plugging in some numbers let's say that this distance from Source 2 to P was let's say 7 m we say that this here was 5 m then in this case the path difference would actually just be equal to 7 - 5 which is equal to 2 m please not that if we pick a point which is is exactly between the two sources the paths will be exactly equal because they're going to form this equilateral triangle like so so we can say that if the point is between the sources the path difference would simply be equal to zero and this is simply because the waves from Source One travel the same distance as they do from source too very often by the way in AQA we end up having to use Pythagoras to figure out some distances so sometimes we may be given let's say this distance here so just keep in mind that you may have to use the Pythagoras Theorem at some point okay now what is coherence mean so two waves are defined to be coherent and this is typically a two marker the first and really important point is that they must have a constant phase difference and the terminology is really important the word constant is crucial the other important uh aspect is that they need to have the same frequency which is equivalent to saying that they have the same time period now in practice what does that mean it means that the two waves do not overtake one another so let's see if I can visualize this without an animation so let's say that I have one wave which is like this and then another wave which is in antiphase now those two waves will have a phase difference of 180° or Pi radians now imagine we see this wave that is uh traveled to the right at some point into the future then I have to draw this quite carefully then the two waves will still be in antiphase we didn't have a wave overtaking uh another wave they still have a phase difference of Pi radians this can only happen of course if the two waves have the same time period SL frequency because otherwise they'll keep overtaking one another so these uh this here is an example of two coherent waves if for instance we looked at it and then suddenly they were in Phase like so then this here would be an example of non-coherence because their phase difference has changed from PI to let's say 2 pi radians Etc and now let's talk about Jung's double slit experiment so this here was one of the original experiments to prove the wave nature of light so what we need to know for the exam is the use of two coherent sources or the use of a single source with double slits to produce an interference pattern what I've drawn over here is the use of a single source with double slits to produce this interference pattern over here what we actually have is a laser which is essentially a monochromatic Source if you remember monochromatic simply means a source with that consists only of a single wavelength that's why lasers appear to be the same color as well because the fluctuation wavelength is very very small so I've kind of drawn this pretty badly over here but what actually happens is that you get defraction you get the wave spreading as it leaves each slit and then those two uh sources end up overlapping I.E superposing and places in which the waves meet Peak to Peak so you can sort of Imagine like a wave here and then a wave here so at a maximum the waves meet Peak to Peak which means that they are in phase on the other hand if we have a minimum for instance here that's the interference pattern really it's this continuation of these bright spots which are Maxima and these dark spots which are known as Minima and kind of intermediate intensity in between but at a Minima the waves actually meet peaked trough so if we were able to kind of zoom in we would see something like this the two waves essentially um superposing and causing the structive interference we can summarize this into the conditions for interference the waves from the two sources must superpose Maxima are caused by constructive interference and for that the waves must be in phase we can also say that the path difference is a multiple of the wavelength Minima on the other hand are caused by by destructive interference and the condition for this is that the waves are in antias and what path differences would actually work well it actually has to be equal to a multiple of this expression which is n plus a half multiplied by the wavelength so for constructive interference the path difference would simply be need to be equal to a multiple of the wavelength which is n Lambda but for destructive interference we are going to need a multiple of half a wavelength or one full wavelength plus half a wavelength so we're going to need n plus a half Lambda where n is an integer in both of these Expressions calculations in these types of questions typically involve the famous yung's double slit equation now in this equation W is The Fringe separation which is the distance between the center of a maximum to the center of another maximum please note that if you're doing a paper three question what you should really do is measure multiple of these for instance 10 of them or so and divide that number by 10 to reduce the percentage uncertainty Lambda is the wavelength of light the distance D is the distance from the slits to the screen and S is the slit spacing this here is usually measured with a traveling microscope if that ever appears on a question and now let's let's go through a proof of yung's double slit equation so let's pick a point on the screen that will end up being a maximum so let's just pick this point the path difference or the path from one of these slits let's go from the middle of the slit is going to be approximately this one over here on the other hand the path to the other slit will be just a little bit bigger let's just draw that over here we can find the PATH difference which is simply going to be just this distance over here which will end up having a 90° triangle like so so the key bit about this proof is that our distance to the screen D has to be significant ific anly bigger than the slit separation s I've only drawn this relatively big simply so that we can visualize this a little bit better let's just say that this here is the central maximum and this here is let's say the first Maxima that we can see after the central Maxima okay so the path difference in this case should be exactly equal to the wavelength and the two rays in a way are going to differ by this distance over here now let's pick one of the angles so I'm going to call this angle Theta I can say that the sign of the angle s of theta which will be equal to the opposite divided by the hypotenuse now my opposite in this case is simply the wavelength length which is the path difference for the first Fringe divide that by the hypotenuse which in this case is just the slit spacing because this triangle is actually really stretched out in the limit that D is significantly bigger than the slit separation then we can say that sine of theta is approximately equal to Theta so therefore we can say that Theta is approximately equal to Lambda / s now what we want to do as well is have a look at the two points across here so this here was the central maximum let's call that .1 and. 2 and the distance between them is actually W let's so let's just be a little bit more precise the we can actually invoke them into another triangle because this this distance D here will have a 90° to um to the screen and meaning that those two triangles will actually be just similar so if I was to make this triangle like this I mean this one right over here well because of that similarity this angle here is also Theta so by looking at this second triangle we can say that tan of theta which is approximately equal to Theta is then going to be equal to the opposite ided by the adjacent which is W / D so this means that Lambda / s is equal to w/ D in the limit that the distance D is significantly larger than the the slit separation just rearranging for w we get the W is going to be equal to Lambda D / by S and here's a really important example question we have a microwave transmitter and those are picked up by a microwave receiver some of the microwaves will go in this path which is the direct path others are going to go up and there will be reflect Ed by this aluminum reflective plate and then they will also end up back the receiver imagine that we take this aluminum plate and we simply move it upwards what will happen to the signal picked up at the receiver the first thing to note is that there will be interference because the reflected and the direct microwaves will end up superposing then a really useful tip would be to comment on the 2 PDS path difference and also potential difference sometimes in these questions we may even need to do a calculation for the path difference but in this case we just need to explain it so what we're going to say is that as the plate goes up the path difference increases which also means that the phase difference will be continuously changing the signal itself will fluctuate between Maxima and Minima this signal typically is actually the reading of a sensitive am meter that's connected to the actual receiver I'm just going to carry on up here because I'm running out of space Whenever there is a maximum the path difference will be a multiple of the wavelength and whenever there is a minimum reading the path difference will be equal to m + a half multiplied by the wavelength let's do another one of these because these questions are just so incredibly important we have two speakers that produce sound waves and then we have a student that finds a maximum intensity using a microphone connected to an oscilloscope at a certain point P what we need to do is explain the conditions for a maximum the first thing that we need to say is that the waves at Point P the waves need to superpose or interfere now in order for there to be a maximum we need to comment on the two PDS let's start off with path difference we could even write down an equation for it and we can say say that PA difference I'm going to write PD for now is equal to n Lambda the other PD that we need to comment on is the phase difference so the waves from both speakers need to arrive at p in Phase so what does in Phase means a phase difference of zero well in this case it's not going to be uh zero because it's not exactly right at the center but it could be zero if P was right at Center but this is just an explanation in Brackets so that would be 0 360 720 Etc so that's n ulti by 360 where n is just an integer and now we're going to do exactly the same question however I've changed the word maximum to a minimum so imagine the reading at P was a minimum what are the conditions for minimum intensity well the waves also need to superose to produce a minimum but now we have a different equation the path difference needs to be half a wavelength plus another wavelength or plus two wavelengths Etc so multiples of these so the PA difference will be equal to n + a half multiplied by the wavelength and finally the waves from from both speakers need to be in antiphase so for instance this could be 180° out of phase or Pi uh radians and let's do one more example so we have the same diagram however now it is definitely not drawn to scale and we can assume that the distance to the microphone D is significantly larger than the separation of the speakers so a student moves the microphone along the vertical line PQ and finds the separation between maximum readings to be 0.45 explain what will happen to the distance between adjacent Maxima if the separation between the centers of the speakers is half okay so we can definitely use our Fringe spacing equation our fringes in this case are going to be the Maxima so we know that W will be equal to Lambda D / by S so assuming that the wave length is constant we can say that the um spacing between the fringes in other words the spacing between the Maxima will double because if we were to have this we are decreasing it by a factor of two and if Lambda and constant W will increase increase so we can say that this is because W um is inversely proportional to the uh separation of the speakers and we have also assumed that the wavelength is constant the final bit that I'm just going to probably write over here is to actually give a value because we are given a value in the actual question we can say that the uh new W or the new uh separation between two readings will be 2 * .45 which is going to be .90 M there's also a spec point that sometimes makes an appearance in exams and that is the production of interference patterns using uh white light so if we end up using white light in the double slit experiment what we are actually going to get is all the color on the actual screen why is that well because the different colors are going to have different wav lengths they're also going to have different Fringe separations so The Fringe separation is the um distance let's say from a Max to a Max or from a Max in a Min now everything else will be the same however because the FR separation is equal to Lambda d/ s so if D and S are constant the longer the wavelength well the higher The Fringe separation because they are directly proportional let's try and draw it maybe so blue is going to have a very short wavelength so we're going to get some sort of a blue spot let's say the next blue spot is over here let's say that green is something in between so the central one will be here but then the next one will just be a little bit longer somewhere over here red will also will be sort of here and then the next one will be sort of over there now the combination of all of those arriving at the central uh Maxima will make a central Max which appears white but then all of the other colors are going to spread out with blue/ Violet the closest that has the lowest W and this is simply because blue has the lowest wavelength whereas red has the highest wavelength so it will have the greatest separation hopefully that makes sense and now let's talk about defraction first of all what is defraction well it's the spreading of light as it passes through a gap or around an obstacle the condition for defraction is relatively simple the gap size has to be approximately equal to the wavelength in other words this distance here I don't know should we just say that the gap size I'm just going to call it a has to be approximately equal to Lambda at least of the same order for maximum defraction our next point from the specification is the defraction pattern from a single slit with monochromatic light remember monochromatic light simply means light source that produces just individual single wavelengths which tend to translate to just one color for instance red is quite commonly used in the lab due to safety as the source p passes through the Gap it spreads out and then it's going to create a wide Central maximum with narrower side fringes and we need to remember the explanation of this pattern remember there's only two points really that we need to remember for the exam it has a white Central maximum which is kind of like I've tried to show that over here and then narrower side fringes on the side but what if we used just one light to produce a defraction pattern so if you have some incident white light and here is the gap the wave will still spread out however not at the central maximum so all the wavelengths are going to arrive together at the central maximum so the central spot will appear white at all the other orders though I've only drawn the first order here we're essentially going to get a little Rainbow the colors are going to separate and each wavelength is going to spread with a different angle from The Gap red is going to have the largest angle it's going to be right at the top blue/ Violet is going to have the smallest angle um and hence it will be down here important to note that this is just the first order of refraction and this whole rainbow pattern would repeat sort of let's say over here and then over here as well for however many orders we can actually see there's also a really specific specification point which is the variation of the width of the central defraction maximum essentially how wide this thing here is should we just call that W compared to the wavelength and also the width of the slit should we just call that s so what we need to remember for the exam is that increasing the wavelength increases the width of the central maximum increasing the slit width on the other hand decreases the width of the central maximum this is for qualitative treatment only we don't really need to remember any equations I personally find it quite useful just to remember that the width of the central maximum is proportional to Lambda / s assuming everything else is constant for instance the distance to the screen typically given the capital d letter is constant what you will typically use in class though is a defraction grating now what is a defraction grating so it looks like this and there's multiple different openings this is not drawn up to scale you can't actually see the openings with your eyes because there's many of them over a very small distance a very typical defraction grating might have something of the order of 300 l L per millim what you will then see are distinct dots producing what is known as a defraction pattern the equation that governs this is the famous defraction grading equation D sin Theta is equal to n Lambda first of all we're going to explain the symbols then we're going to do a small example problem and finally we're going to derive the equation as well D in this case is the line separation I often like to remember that D is equal to 1 / n where n is the number of lines per meter we need to be really careful with the units uh in this why is that because n being the number of lines per meter not per millimeter typically in the question and in the fraction grading with the lines per millimeter so we need to convert that 300 lines per millimeter is actually going to equate to 300,000 lines per meter now what are the other symbols in the equation actually mean the sign of the angle is the angle corresponding to the order of the fraction in other words this integer n so for n is equal to Z the beam just goes straight through or part of the beam just goes straight through with an angle of zero for n is equal to 1 is going to have its corresponding angle Theta 1 this angle is going to be the angle between the defraction grating and the first Fringe because the pattern is fully symmetric this will be 100% equal to the angle here which is also going to be Theta 1 Theta 2 the angle will be the corresponding angle between the defraction grating and N is equal to 2 so that's the angle let's use a different color like so this one here is Theta 2 correspondingly this angle here will also be equal to Theta 2 now let's just showcase how this equation Works shall we say that Theta 1 is equal to 10° so D sin Theta is 1 / n s of the angle and this will be equal to just Lambda for n is just equal to 1 so that's going to give us 1 over 300,000 so I'm going to add a factor of 10^ 3 multiply by the S of 10° is going to be equal to the wavelength so if we were to find the wavelength in this particular case Lambda will have to be equal to this expression pluging in some numbers into the calculator we're going to get about 5.8 * 10 ^ -7 M and here's a really important question on the maximum number of fringes so we've got a laser light of a certain wavelength is used to produce a defraction pattern we're given the slit separation of the grating directly what is the total number of fringes that can be observed so in all cases if we have some light heading into a defraction ating the angle between the fringes can at the very maximum be equal to 90° so we can say that D sin Theta is going to be equal to n Lambda then we can set Theta to be equal to 90° for a maximum which means that D sin of 90 which is just one is going to be equal to n Lambda now we can find the maximum order of defraction which is going to be given by D over Lambda now in this case D is directly given rather than the number of lines per millimeter so that makes our job a little bit easier D is going to be directly 3.3 * 10^ of -6 and then we need to divide that by 640 multiply this by 10^ of - 9 as a decimal we're going to get around 5.16 or so but we need to round this up to the closest whole number of completed integers so in this case n is just going to be equal to five but there is a trick the trick here is that the total number of fringes okay so what is what will the picture of this actually look like we're going to have a central maximum here that's going to be n is equal to0 then we're going to have five more spots above it but that's just one side of the symmetric pattern we're also going to have five more spots that's going to give us a grand total of maximum number of fringes to be 11 and this is really really important as a RAR thumb unless there's something that's um obstructing it we're always going to have this to be um an odd number which uh because we're going to have the same number on both sides plus the central maximum so it could be a useful technique for multiple choice questions in particular and let's go over the derivation of D sin Theta is equal to n Lambda so imagine that I have a defraction grating and this here are just two slits and the distance between them is just D if p is the first order defraction uh Maxima in other words N is equal to 1 and right over here we're going to have the central maximum and is equal to zero so the first step in this derivation is to consider the path difference so those two paths are going to differ by this distance here if the angle to the first Fringe is Theta the yellow triangle right over here is typically assumed to be similar to this one over here which I believe is actually a tiny approximation but this is typically true because the slit separation is significantly smaller than the distance to the actual screen so we can say that the blue triangle essentially is similar to this yellow uh triangle and if those triangles are similar then this angle here will just be Theta and now we have this triangle in which we're going to have 90 de here D And then Theta well because s of theta is equal to the opposite divided by the hypotenuse and our opposite in this case is actually just the path difference so in this case sine of theta will just be equal to the PA difference over the hypotenuse which is just meaning that our path difference will just be equal to D multiplied by sin Theta but because the waves arrive in phase at P for constructive interference the PA difference is just going to be equal to n lamb this is just a condition for constructive interference and there we have it that n Lambda is equal to D sin of theta we also need to talk about safety issues associated with using lasers some common sense supplies here do not Shine the laser towards people do not shine it on reflective surfaces and also it's quite useful to have a warning sign display to warn people that you're actually using a laser sometimes in questions you also get one or two markers on the application of defraction gradings so there's multiple different ones a couple of obvious ones is that they're used in atomic spectrum to identify elements in stars in astronomy also on Earth as well but x-ray defraction is used to study the structure of crystals in x-ray crystallography this is very very interesting it's because the X-Ray's wavelength is of similar order to the spacing between the atoms so the atomic lattice itself is going to act like a defraction grating and now let's talk about refraction the refractive index is defined as the ratio of the speed of light of vacuum to the speed of light in a substance we have this classic formula that n which is the refractive index is equal to C which is the speed of light in a vacuum 3.0 * 10^ 8 m/s divided by the speed of light in that particular substance it is important to know that the refractive index is actually unitless why is that well if we consider the units of c those are just met per second / m/s they're just going to cancel out leading us to the fact that n is unitless please also note that because nothing can exceed the speed of light n has to be greater or equal to 1 so for air we actually have a very tiny refractive index like 1.006 or something like that but it's approximately equal to one um this is approximately equal to n in a Val vacuum and we just assume that to be equal to one and here's a really popular style of question so the refractive index of the glass shown is 1.5 what is the speed of light in this glass well n is equal to C over CS meaning that we can just rearrange for CS which is going to be C Over N so that's going to be the speed of light in the vacuum which is 3.0 * 10^ 8 divide that by my n which is 1.5 well this here just going to give me 2.0 * 10 8 m/ second now how can that be remember that V is equal to F multiplied by Lambda so if the speed was to change should we even call that CS just to be absolutely consistent but if the speed is to change then either the frequency of the wavelength must change it's really important to note that the frequency always remains constant during refraction it is the wavelength that changes and also the speed that actually changes let's also revise some rules about refraction if I have a beam which enters at 90° let's say let's draw it over here then it's going to enter at 90° but it also it's going to travel through it's going to leave at 90 90° if I have a beam which arrives at an angle so let's draw this one over here the first thing to do is to draw the normal which is going to be around here somewhere let's say that here is the normal so if we're going to a high refractive index the beam will reflect towards the normal so the beam itself would have change direction this one here is the angle of incidence it's always the angle to the norm let's call it Theta 1 and this one here is the angle of refraction let's call it Theta 2 then on the other hand at the second boundary we're going from a higher refractive index to a lower refractive index well if that's the case the beam will bend away from the normal so it's going to do something like this something to note is that this line here ends up actually being parallel to this line here due to Pure geometry and sometimes this is asked in questions as well but what will this angle of refraction actually be exactly well this can be determined by snail's law of refraction for a boundary this says that the initial refractive index let's call it N1 multiplied by S of theta 1 being the angle of incidence is going to be equal to N2 which is the refractive index of the second substance multiplied by the the S of the angle of refraction Theta 2 and this is often applied to lots of problems let's talk about total internal reflection imagine that situation that we had before we had a glass block and then we're going from glass n is equal to 1.5 to Air and is equal to 1 we can go down a few different paths for instance if we go down this path then we're going to emerge a little bit away from the normal and towards the boundary if we take this path along here we're going to get a little bit further away from the normal and a little bit closer to the boundary but if we take a special path which is going to correspond to this special angle to the normal then we're going to emerge exactly parallel to the boundary and this angle of incidence is known as the critical angle and above this angle only pure reflection occurs rather than any refraction if we're going to go above that critical angle we are going to remain perfectly inside totally internally reflected inside of the original substance and this is known as total internal reflection how do we derive this angle well let's use snail's law so I'm going to write write that N1 sin Theta 1 is equal to N2 sin of theta 2 I'm going to call the angle of incidence at which total internal reflection occurs to be the critical angle and this happens when Theta 2 is equal to 90° and then we get that N1 sin of theta critical is equal to N2 2 sin of 90° but remember s of 90 is just one therefore we get that N1 sin Theta critical is equal to N2 in other words s of theta critical will just be equal to N2 / N1 and this is a Formula given in your formula sheet remember in order for this to mathematically work as well s of the angle has to be smaller than than one it's sandwich between 1 and minus one therefore this only works if N1 has to be bigger than N2 so it's going to work on the boundary going from let's say glass to air but it's not going to work the other way around and let's also talk about fiber optics which are a fantastic way to transfer information here is the step index fiber optic we have the core at the center which is surrounded by cladding the core actually propagates light WS by total internal reflection the individual beam itself passes through the core totally internally reflecting we can sort of uh imagine it doing something like this in order for us to have maximum transmission the core needs to have low absorption in exams you may also come across the word attenuation but that exactly me just means absorption the function of the cladding is twofold it needs to protect the core from damage but it also provides a boundary for total internal reflection because it has a lower refractive index compared to the core remember this is actually the condition for total internal reflection to actually occur there's several different problems we have material dispersion and that is that different wavelengths are going to have different speeds due to the different relative refractive indices in the core for instance red might have a very different speed to blue one way of fixing this is by using a monochromatic beam of just one wavelength you can also have model dispersion where different paths have different lengths so the effective time along the fiber differs it can be used by using a single mode fiber which has a narrower core smaller difference between the relative refractive indices of the core and the actual cladding sometimes this appears as Delta n in the actual questions just a little note about the absorption of the core it has a low absorption but if if it had a higher absorption this would lead to a loss in the amplitude of the signal and let's talk a little bit more in detail about modal dispersion this is really caused by the light rays entering the fiber at different angles so this actually leads to what should say this leads to spreading of the pulses due to the pules themselves taking different times to travel through the fiber a different way of saying the same thing is the pulse is actually broaden due to entering the optical fiber a different angle the greatest transmission rate occurs when there is less model dispersion one way of achieving that is the smaller diameter which means a smaller range of path which means less broadening well done for revising the Hall of waves if you are revising for paper one have a look at this video which covers the whole of AQA particle physics click over here