Wave Optics: Interference and Diffraction

Jai Hind everyone, myself Dr. Vanna Sharma from Ajay Kumar Garg Engineering College. The subject which I am teaching, subject is Engineering Physics, code is KS101T. Today the topic which I am going to start, topic is Wave optics. So when I am saying optics, optics means we are dealing with light and you know in case of light there are three basic phenomenas which are involved in light. One is interference, second one is diffraction and the third one is polarization. So in this wave optics we are going to discuss the two main phenomenas of optics. One is interference that is the first part of this wave optics and the second one is diffraction. So I am going to start with the first part that is interference. So the topic or the topic which today I am going to start that is interference. And these are the reference books which we are going to follow for this interference. The first one is a textbook of Engineering Physics by Namneet Gupta. Then a textbook of Optics by Ajay Ghatak. Engineering Physics by SK Gupta and SL Gupta. and some internet links are also there from where you can have this information about interference and refraction. So, the topics various topics which we are going to cover under this interference those topics are the first one is we are going to introduce what is the meaning of interference of light introduction then interference then type of interference. So, we are going to discuss the type of interference here. Condition for interference, the next one is coherent sources, the next one is methods to obtain coherent sources. So here we will discuss that how we can get coherent sources. Interference in uniform and wedge-shaped thin films. So these are two topics actually, one is thin films of uniform thickness and the other one is thin films of non-uniform thickness which is also known as wedge-shaped thin films. After that we will discuss that what is a need, what is a necessity, of extended sources in order to get a sustained interference pattern in thin films. After that a very important and interesting topic we will discuss in this chapter and that topic is Newton's ring experiment. So this is a very beautiful example of interference in thin films and we will also discuss the applications of interference here, application of basically Newton ring experiment. So, now I am going to start with the introductory part of interference or this wave optics. So, as you have seen some beautiful colors on a soap bubble when it is illuminated with white light. Suppose you have a soap bubble and when the soap bubble is illuminated in white light, then you can see the beautiful colors in that soap bubble. So, basically those colors which are there in that soap bubble, Those colours, they are the outcome of interference of light. You might have seen that if there is a drop of oil on the surface of water, then if that drop of oil is illuminated by white light or sunlight, then again you can see colours in that layer of oil. So again those colours, they are because of interference of light. So all these are the examples of interference in thin films. So, how the light will get, how you are getting the interference in light, what is the meaning of this wave optics that we are going to discuss in today's lecture. So, optics is, this is a branch of physics in which we study about light or phenomena related to light. So, optics is that particular branch of physics where you are studying about light. So, you might have gone through this topic in your 12th class about interference, diffraction and polarization. So, this is... The next level of that optics on the basis of dual nature of light optics can be divided in two parts as you know That light shows dual nature sometime light is showing particle nature and sometime light is showing now wave nature So since light is having dual nature. So on the basis of that Optics is divided into two parts. The first one is Geometrical optics which is also known as ray optics. So this geometrical optics actually it is the evidence for particle nature of light and the second one which is known as physical optics or you can say wave optics it is an evidence or it is a proof that light is having wave nature. So these are the two different branches or two different parts of optics on the basis of this dual nature on the basis of particle nature you have that ray optics and on the basis of wave nature you are having this wave optics. In geometrical optics, we consider that light travel in the form of ray. This also proves the particle nature of light. So you might have gone through this ray optics in your 12th class. The next one is physical optics, which is also known as wave optics. Here we consider the light travels in the form of So, this is an evidence for wave nature. So, these are the two different parts of optics. Interference now we will try to understand what is the meaning of interference of light. Due to superposition of two coherent light waves of same frequency and nearly same nearly equal amplitude the intensity of light is redistributed in space. This phenomena is known as interference so what is the meaning of interference of light what you have you are having two light wave and those light waves are of same frequency and they are having nearly equal amplitude then the intensity of these light under suitable conditions these light waves will superimpose and after the superimposition you are going to get a resultant light vector and the that resultant light vector Either the intensity of that resultant light vector is greater than the sum of individual intensities or lesser than the sum of individual intensities. What I am trying to say here, suppose you are having two light waves. One is having intensity I1 and the second one is having intensity I2, right? Now, these two of intensity I1 and I2 and they are having same frequency and they are having slightly same. amplitude. Now under suitable condition they will get superimposed and after that superimposition you will get a resultant light vector. Now that resultant light vector now the intensity of that resultant light vector either that intensity is greater than the sum of individual intensities means this I either that is greater than this I1 plus I2 or that intensity you is lesser than the sum of individual intensities. You are having these two possibilities. Either the resultant light vector which is having intensity I, either that intensity is greater than I1 plus I2 or it is lesser than I1 plus I2. Now depending upon these two, you are having either you are having get you are getting maxima or you are getting maxima or you are getting constructive interference or destructive interference. which we will discuss in next slide. So, the intensity distribution at any point in space is the resultant of individual intensities at that point. So, at different different points you are going to get different intensities either the intensity is greater than the sum of individual intensities or that is less than the sum of individual intensities. So, depending upon that either at certain point you are going to get a maxima or at certain points you are going to get minima. So, I hope the meaning of interference is clear. Interference means when you are having two current light waves of same frequency and of slightly and of nearly equal amplitude and when these will superimpose then you are going to get the resultant light vector and that resultant light vector either the intensity of that resultant light vector is greater than the sum of individual intensities or it is less than the sum of individual intensities. So, depending upon that at certain points you are going to get maximas and at certain points you are going to get minimas. Now, the types of interference. So, we are having two type of interferences as I had discussed in my last slide that at certain points the intensity of the resultant light vector is greater than the sum of individual intensities. So, those points will be the points of maximas or you can say those points will be the points of constructive interference and there will be certain points where the intensity is less than the sum of individual intensities and those points will be the points of minima or you can say destructive interference so two type of interferences are there one is known as constructive interference and the second one is known as destructive interference so there are certain conditions If those light rays, those light waves which are superimposing, if they will satisfy those conditions at certain points, then at those points you are going to get maximas and there are certain conditions. And if those conditions will be satisfied at certain points, then you are going to get minima there. So now let's try to understand that what are those conditions under which you can get maximas at certain points and minimas at certain points. So for that... Here directly the conditions are written but how you are getting those conditions that I will explain in next slides. When two waves are superimposed in the same phase that is same that is phase difference between them is integral multiple of 2 pi then constructive interference is produced. So, what is the meaning of that? You are having two light waves. So, this is the one light wave and this is another light wave right. So, these are the two light waves and if they are in same phase means if the phase difference between them is 0 or integral multiple of pi Then, sorry that is integral multiple of pi. Then the interference between them that is going to give you constructive interference. So, when two waves are superimposed in the same phase that is phase difference between them is integral multiple of 2pi then constructive interference is obtained. So, this is the condition to get that maxima. This is the condition to get that constructive interference in terms of phase difference. And the second condition is when two waves are superimposed in opposite phase when they are out of phase that is phase difference between them is odd multiple of pi. If it is even multiple of pi then you are going to get constructive interference. If it is odd multiple of pi then you are going to get destructive interference. So if in one line I am writing these two conditions then the conditions are like this. If the phase difference between the interfering light waves is even multiple of pi where this n starts from 0. So, you are getting the conditions are when delta is 0, when delta is 2 pi, when delta is 4 pi, when delta is 6 pi then the interfering light waves when they will interfere they are going to give constructive interference at those points where the phase difference is even multiple of pi and if the phase difference is odd multiple of pi means if it is 2 n plus minus 1 pi. if it is like this where this 2 and if you are taking 2 and plus 1 then n is going to start from 0 and if you are taking the condition 2 and minus 1 then n is going to start from 1 but in both the cases you are you can see here that delta is having values pi 3 pi 5 pi 7 pi 9 pi and so on. So what is the meaning of this? Those points where the interfering light waves they interfere in such a way that the phase difference between them is either zero or even multiple of pi then at those points the condition of maxima will be satisfied and at those points the intensity will be greater than the sum of individual intensities so those points will be the points of constructive interference there will be certain points where where the phase difference between the interfering light wave light waves is odd multiple of pi. Odd multiple of pi means either the phase difference is pi or 3 pi or 5 pi or 7 pi and so on. And if the phase difference is like this then at those points the condition of minima will be satisfied means you are going to get the resultant intensity lesser than the sum of individual intensities or you can say those points will be the points of destructive interference. So there are certain points. where the condition of maxima will be satisfied there will be certain points where the condition of minima will be satisfied so you are going to get maxima and minimas in your interference pattern so two type of interferences are there one is constructive and the other one is destructive constructive in other way you can say maxima condition of maxima destructive means the condition of minima for maxima the phase difference between the interfering rays that should be even multiple of pi And for destructive interference, the phase difference between the interfering light wave, that should be odd multiple of pi. So, these are the conditions to get destructive and constructive interference. So, this is an example of your constructive and destructive interference. So, diagrammatically, you can understand from here. As you can see here, these are the two interfering light waves. Now, you can see that there is crust to crust and trough. to 12 correspondence means these two light waves they are in same phase either the phase difference is zero or that is even multiple of pi. So, if you are getting this type of case then there will be enhancement of the intensity means when these light waves they are in same phase. Then the resultant intensity that will be greater than the sum of individual intensities. So you are going to get here constructive interference. So if there is crust to crust or trough to trough correspondence then you are going to get maxima. Second one is, second is this. This is out of phase means either the phase difference is pi or that is 3 pi or 5 pi or 7 pi or so on. So when you are having two interfering light waves and if they are out of phase you are going to get this. then in that case there will be destructive interference. So you can see here at this point for this interfering ray you are getting maxima but here you are getting minima. Here you are getting minima but in second case you are getting maxima. So there is these two light the interfering light waves they are out of phase. The phase difference between them is odd multiple of pi. Here the phase difference is even multiple of pi. Even multiple of pi means either the phase difference is zero. or that is 2 pi or that is 4 pi or that is 6 pi and so on. So, in this case you are going to get constructive interference but in the second case the phase difference is odd multiple of pi so that is 2n plus minus 1 pi. So, the phase difference that is of this order either pi or 3 pi or 5 pi or 7 pi and so on. So, if you are getting this type of phase difference then you are going to get destructive interference. So in case of destructive interference the intensity that will be lesser than the sum of individual intensities right and if you are having two interfering rays with exactly same amplitude or with same intensity then in case of minima the result in case of destructive interference the resultant intensity will be 0. Why I am saying so? I will explain in next slide. What I am saying here if you are having two interfering rays which are of exactly same amplitude and if the phase difference between those interfering rays at certain points is odd multiple of pi then at those points the resultant intensity will be zero and if you are having two interfering rays and if the phase difference at certain points between those light waves is even multiple of pi then you are going to get constructive interference at those points. So, this is your constructive interference and the second one is your destructive interference. Now, the conditions for interference means now we are going to write the conditions in terms of phase difference or we are also writing the intensity of constructive and destructive interference. So, conditions for interference in this particular slide we can we will see these conditions. We know that intensity due to superposition of 2 wave is given by this. From where we are getting this result? So for this you have to remember, you have to just recall about your Young's double slit experiment. You have gone through that Young's double slit experiment in your class 12 and you know if you are having two light rays, sorry if you are having two light waves, one is of amplitude a1 and the second one is amplitude a2. If a1 is the amplitude then I1 will be equal to a1 square and your I2 will be equal to a2 square. So these are the intensities of those two interfering rays right. So if delta is the phase difference between these two light waves then the resultant that according to the principle of superposition we have solved that and we have calculated the resultant intensity. So that resultant intensity is of this particular form. I hope you have recalled about this expression. That resultant intensity because of the superposition of two light waves of intensity I1 and I2 and if delta is the phase difference between them is given by this equation. I is I1 plus I2 plus 2 root I1 I2 cos delta. So what is the meaning of all these terms? What is I1 here? That I1 signifies that I1 denote the intensity of first light wave. I2 that represent the intensity of second light wave. And this delta that represent the phase difference between the two interfering rays. So between those two interfering light wave right. So where delta is a phase difference between them. So this delta actually the value of this delta is going to decide what you are going to get at a certain point. Whether you will get maxima or minima that is completely decided by this delta. Why I am saying so? Because you can see in this equation I1 that is the intensity of first light wave constant. I2 that is again fixed. So only this particular factor only this cos delta is going to change the whole scenario. So this cos delta will decide whether a point will be a point of maxima or that point will be a point of minima. So you know this cos delta. that will give you maximum value and that maximum value will be plus 1 and this cos delta will give you minimum value and that minimum value will be minus 1. So now this cos delta when it is taking plus 1 value you will get maxima and if this cos delta is taking minus 1 value then you are going to get minima because only these are the possibilities to get the maximum value of i or minimum value of i. So now we are going to discuss the condition of maxima that is constructive interference. and condition of minima. So, the first one condition for constructive interference. So, when phase difference so for maxima what should be there cos delta should be equal to plus 1 and you know when this cos delta as you know cos 0 is 1, cos 2 pi is 1, cos 4 pi is 1, cos 6 pi is 1. It means when this delta will be equal to even multiple of pi. where this n that starts from 0. So, when n will take 0 value, delta will be 0. When n will take 1 value, delta will be equal to 2 pi. When n will take 2 value, delta will be equal to 4 pi. So, when delta is 2 and pi means when this phase difference between the interfering light waves, if that phase difference is even multiple of pi, then delta is 2 and pi and starts from 0. and that is 0, 1, 2, 3 and so on means delta starting from 0 next value which it is taking that is 2 pi then 4 pi and 6 pi and so on. So, for those values you are going to get maxima. Why you are getting maxima? Because cos delta will be equal to plus 1 there. So, if delta is taking 2 and plus sorry 2 and pi value then what will be the value of i? If delta is cos delta is 1 then I is I1 plus I2 plus 2 root I1 I2 right. So, this is the resultant intensity for maxima I is equal to I1 plus I2 this is for maxima I1 plus I2 plus 2 root I1 I2 right. It means now the resultant intensity of that constructive interference of that maxima is greater. Then the sum of individual intensities by this factor 2 root I1 I2. Let me explain it. What is the condition of maxima in terms of phase difference? The condition of maxima is delta should be even multiple of pi. And what is the value of this I? I for maxima is let me write it properly. It is I1 plus I2. plus 2 root I1 I2. So, what is the final conclusion you can draw from here? The conclusion you can draw from here that when two light waves of intensity I1 and I2 when they are interfering then after their interference after the superimposition there are certain points where the intensity is greater than the sum of individual intensity. and that is greater by this factor 2 root i1 i2 right. Now this is a condition in terms of phase difference. You can also write the condition in terms of part difference. So you know what is the relationship between phase difference and part difference? The relationship is the part difference is lambda upon 2 pi times phase difference that is a relationship. So if you know the phase difference you can easily write the part difference. So the part difference will be equal to lambda upon 2 pi and what is the phase difference for maxima? It is 2 n pi. So, this is this 2 pi will get cancelled and what you are getting? You are getting n lambda. So, what is n lambda? This n lambda is the condition for maxima that is the condition for for part difference condition for constructive interference in terms of part difference. So, delta is n lambda and you know n starts from 0, 1, 2, 3. that n starts from 0 and it can take all the integers values like 1, 2, 3, 4 and so on. So, delta is 0, then lambda, 2 lambda. When part difference is of this particular order, when part difference is 0, when part difference is lambda, when part difference is 2 lambda, when part difference is 3 lambda, for all those values, what you are going to get? You are going to get maxima. So, those points where the part difference is 0, lambda, 2 lambda, 3 lambda, 4 lambda and so on. Those points will be the points of maxima. So, I hope now the condition for maxima is condition for constructive interference is clear in each and every sense. What is the condition of maxima in terms of phase difference? In terms of phase difference, you can say when the phase difference is even multiple of pi, then you are going to get a constructive interference. In terms of part difference, you can say when the part difference is n lambda. then you are going to get maxima right. So, and the intensity of maxima is I1 plus I2 plus 2 root I1 I2. So, condition for maxima. Now, we will discuss the condition for minima or you can say condition for destructive interference. So, you know that I what we have written there I is I1 plus I2 plus your 2 root I1 I2 cos delta. Now, from according to this equation this i will be minimum if cos delta is minus 1. So the condition of destructive interference should be cos delta that should be minus 1 and you know when this cos delta will be minus 1. If this delta is taking pi value if this delta is taking 3 pi value if this delta is taking 5 pi value then for all these odd multiple of pi's for cos pi, for cos 3 pi, for cos 5 pi, for cos 7 pi, for cos 9 pi. For all these odd multiple of pi's, cos will take minus, cos odd multiple of pi will take minus 1 value. So, all these values or all these points will be the points of minima. So, when this delta is odd multiple of pi. So, here this 2 and plus 1 is written. You can also write 2 and minus 1 here because you can write this. Whether you are writing 2n plus 1 or you are writing 2n minus 1 in both the cases this particular term is going to give you odd value. But the difference is if you are writing 2n plus 1 then n will start from 0 as mentioned over here. But if you will write 2n minus 1 then n is going to start from 1 right. So the meaning is only this that when the phase difference between the super between the interfering light waves is odd multiple of pi then you are going to get the condition of minima at those points. So the points are when delta is pi when delta is 3 pi 5 pi or you can say odd multiple of pi then you will get minima. So this is the condition of minima in terms of phase difference and you know the relationship between part difference and phase differences. The part difference sorry if you know the phase difference And the part difference is lambda upon 2 pi times of phase difference. That is the relationship. So now if you want to write the part difference, it will be equal to lambda upon 2 pi times the phase difference. And phase difference is odd multiple of pi here. So this pi will get cancelled by this. So what you are getting? You are getting this 2 and plus 1 lambda by 2. Or in other words, you can say the part difference is odd multiple of lambda by 2. So, here that power difference is written delta is 2n plus 1 lambda by 2. So, for minima or you can say for destructive interference phase difference should be odd multiple of pi and power difference should be odd multiple of lambda by 2 where n starts from 0 here. So, what are the values for values in terms of power difference for minima? The values are the first one is lambda by 2 when this 2n plus 1 when it is take one value. So, power difference will be lambda by 2. Next odd is 3. So, you will get 3 lambda by 2. Next one is 5. So, you will get 5 lambda by 2, 7 lambda by 2, 9 lambda by 2 and so on. So, the condition for minima is this, right. So, what we have done? We have drawn the conditions for maxima and minima. So, I am just concluding these conditions here. So, on one side, I am writing the conditions of constructive interference. And on this side, I am writing the conditions of destructive interference. So, first I am writing the intensity of constructive interference. So, that intensity is I1 plus I2 plus 2 root I1 I2. This is the intensity of maxima. And if you want to write the intensity this is for maxima. And if you want to write the intensity of minima then what will be that? Intensity of minima that is equal to I1 plus I2. And you know the last term is plus 2 I1 I2 cos delta. For destructive interference that cos delta is minus 1. So, what you are going to get here? You are going to get here minus 2 root I1 I2. This is for minima and that one is for maxima. So, there is one important point and very interesting point here. Whatever is the gain for maxima? What is the gain for maxima with respect to sum of individual intensities that is I1 plus I2? The gain is in terms of 2 root I1 I2. So whatever is the gain in case of maxima that is a loss for minima because in minima the intensity is less than the sum of individual intensities and the factor which is less that is 2 root I1 I2. So the loss for minima or destructive interference that will appear as gain for maxima means that light energy During the formation of interference that light energy will remain conserved. Energy that will remain that will follow the law of conservation of energy law of conservation means energy will remain conserved. There is only redistribution of energy between maxima and minima. Why I am saying there is redistribution of energy between maxima and minima? Because whatever the laws for destructive interference that is appearing as gain for constructive interference right. Now, I am writing the conditions for maximas and minimas. So, in terms of phase difference, you know the condition for maxima is this phase difference is even multiple of pi where n starts from 0, right? So, this is the condition for maxima here in terms of phase difference. And in terms of phase difference, the condition for minima is odd multiple of pi. Phase difference should be odd multiple of pi. And this n starts from 0, right? So, what are the points? What should be the phase difference between the interfering rays? That should be 0 or 2 pi or 4 pi that is even multiple of pi and so on. And here this delta will be, this will be pi, this will be 3 pi, this will be 5 pi, this will be 7 pi. It means odd multiple of pi, right? So, this is for constructive interference, this is for destructive interference. And in terms of power difference. What we have concluded for part difference in terms of part difference the condition is part differences n lambda right. So, part difference for maxima should be 0 or lambda or 2 lambda or and so on. And in case of destructive interference part differences odd multiple of lambda by 2 that is 2 1 plus 1 lambda by 2. So, what is the meaning of this? The part difference between the interfering light wave for destructive interference that should be lambda by 2 or 3 lambda by 2 or 5 lambda by 2 and so on. So, these are the conditions for constructive interference and these are the conditions for destructive interference. So, you have to remember all these conditions because in order to understand the interference in different different cases you should know for when you can get maxima at a certain point and you should know for when you can get maxima at a certain point and Under which condition you can get a minima at a certain point. So, all this is decided by the phase difference between the interfering light wave and the part difference between them, right. So, these are the conditions. Then the next one is coherent sources. So, the coherent sources, this particular condition that the light sources from where you are getting the light that those light sources should be coherent in nature. So, Now, we will try to understand what is the meaning of coherent sources. Two sources are said to be coherent if they emit light waves of same frequency, nearly the same amplitude and the most important condition is the phase difference that which is the phase difference between the light waves which are emitted from those sources, that phase difference should be either that should be 0 or that should be constant. So, that is the main condition to get coherent sources. So, what are coherent sources? two sources are said to be coherent if they emit light wave of same frequency and nearly the same amplitude and the phase difference between the light waves which are emitted from those sources that phase difference either that should be zero or that should be constant so that is the main condition for the interference process the coherent sources are essential so whenever you want to get the interference of light then The main condition is the first condition is the first requirement is that you should have coherent sources. So, that is the first condition. two sources of light are said to be incoherent if they emit light wave whose phase changes with time if the phase changes with time then those light sources they are said to be incoherent in nature so we always require coherent light sources which can maintain a constant phase difference irrespective of time so those sources are said to be current in nature and there is a very important point that if you are having a single source and if you can divide that single source in two parts then those two sources which are generated from same source only the light waves which are emitted from those sources will be coherent in nature but if you are having two independent sources even the similar type of sources i'm taking an example here suppose you are having two sodium lamps two similar type of sodium lamps but those sodium lamps are independent of each other means to you are having two independent sodium lamps there and those two sodium lamps they are emitting light then the light which is emitted from those two independent sodium lamps that light will not be in co that light will not be coherent in nature you can get the coherent light only if the the light rays or the light waves are coming from a single source so two independent sources they can never give you coherent light because The initial phase of each and every source that is independent of each other. So if there is any change in the phase of one source, you cannot maintain the similar type of change for the other source. So that is the reason whenever you want a coherent source, you have to divide a single source in two parts. So you know the main condition to get interference is you should have coherent sources of light only. Then you can get in sustained interference pattern right. So that is about coherent light. Methods to obtain coherent sources in practice. So as I have mentioned that if you have two independent sources then you can never get coherent light. So now the question arises that how can you get coherent light. So as I mentioned that you can get coherent light only by using a single source. So there are two methods through which you can get coherent light. from a single source. One is division of wavefront. This is the first method through which you can get coherent light from a single source. You can divide a single light source in two parts and you can get coherent light from those two parts. So, division of wavefront is one example and the second one is division of amplitude. So, using these two methods you can always divide a single source in two coherent sources. Now, the first one is division of wave front how you can get light there a coherent light there the in the incident wave front is divided into two parts these two parts travel and superimpose each other and produce interference pattern so the example one example one experiment through which you have study in your 12th class famous young's double set experiment so what you have in young's double set experiment you are having a single slit here which is illuminated by a monochromatic source of light. After that there are two slits which are placed at an equal distance from the single slit and there will be division of wave front at these two slits and these two that is S1 and S2. Since they are eliminated from the single slit they are eliminated from the single source. So these two slits S1 and S2 which are eliminated from the single slit from the single source These two are known as coherent sources. From these two slits you will get the coherent light. Reason is because they are illuminated from a same source. So by the division of wave front you are going to get coherent light. So the examples of division of wave front are the first one is Young's dubestate experiment and the second one is freshness by prism experiment. There are some more experiments of more examples of this division of division of wave front like Lloyd's mirror matrix. method. So, all in all these examples by the division of wave front a single source is divided into two coherent sources. So, always a single source is used to generate two coherent sources. So, this is one method and we have gone through the examples of this division of wave front in our previous classes. Now, in our syllabus here we have to discuss the division of amplitude that is the second type of. method through which you can get coherent light and in this entire chapter we are going to discuss this division of amplitude right so the second one is sorry just in the exam experiment the setup of that young's double slit experiment what you have a source of light which will illuminate a single slit then two slits s1 and s2 which are placed at same distance from the single slit then these two slits this s1 let i'm writing it as let s1 and the second one s2 so s1 and s2 these two will act as current sources so there will be division of wave front and these two will act current sources and you will get the interference pattern over the screen. So, this is an example of division of wave trend. Now, the next one is division of amplitude. The intensity of incident light is divided into two parts, right? These light waves superimpose afterwards and produce interference pattern. So, what will happen here? A single light ray, its amplitude is divided into two parts, its intensity is divided into two parts and two. those two parts when a single ray is divided into two parts those two parts since they are generated from a single ray that is generated from a same source so if there is any change in the phase of that ray parent ray you will get the similar type of changes for the rays which are generated from that single ray it means you can maintain a constant phase difference here you can always maintain that constant phase difference for those um those divide those two rays which are obtained from a single ray. You have gone through this division of amplitude in Michelson model experiment. In Michelson model experiment there was an extended source of light from that extended source of light the light will incident over a glass plate which is inclined at 45 degree. So when light will incident over that glass plate which is inclined at 45 degree and that is also semi-silver some part will be reflected and some will be transmitted. So a single ray. that will divide into two parts reflected and transmitted now that reflected and transmitted ray the amplitude of a single ray is divided into two parts reflected and transmitted now if there is any change in the phase of that parent ray you will get the similar type of changes for reflected and transmitted ray also it means you can maintain a constant phase difference here so those two rays will act as two current sources so that is the method to generate two current sources from a single source So, in division of amplitude the intensity of incident light is divided into two parts. As I have given you example some reflected portion, some transmitted portion. These light waves superimpose and then they produce interference pattern. Now the examples of division of amplitude Newton ring experiment, Michelson interferometer, interference and thin films of uniform thickness, non-uniform thickness, all these are the examples of division of amplitude. So, in this chapter where we are discussing this interference basically what we are going to discuss we are going to discuss Newtonian experiment here we are going to discuss interference and thin films of uniform thickness and non-uniform thickness in detail so that is our objective here right so the what are your current sources and what are the methods to generate those current sources so what I told you current sources are those sources from where we are getting the light waves of same frequency. of a slight of nearly equal amplitude and the phase difference is either zero or constant between them. So, if you are having a two independent sources then those two independent sources can never give you coherent light. So, how you can get coherent light? In order to get coherent light what you have to do you have to divide you have to convert a single source into two parts and what are the methods to generate those. those coherent sources the first method is by the division of wavefront and the second one is division of amplitude examples of division of wavefronts wavefront are young's double state experiment and frational biprism experiment and lloyd's mirror method all these are the examples of division of wavefront which you have gone through in your previous classes and the second one is division of amplitude which is the point of discussion in our chapter here So, division of amplitude means a single ray, the amplitude of that single ray or the intensity of that single ray is divided into two parts and then those two parts will act as two coherent sources right. And there are different examples of this division of amplitude. One is Michelson model experiment where the Michelson interferometer. In that interferometer the amplitude of single rays divided into two parts. Newton ray experiment that is an example of division of amplitude. So, these are the examples. So, in this chapter we will discuss this division of amplitude. So, now a quick this review of Newton ring experiment here. This is the experimental setup. So, in this experimental setup you can see here that this is a source of light, monochromatic source of light and this is a glass plate which is inclined at 45 degree. So, when light will incident over this glass plate the light will divide into two parts. So, there will be division of amplitude there right. So, because of the division of amplitude you will get the interference pattern and the interference pattern in newtonian experiment that will be like this. That will be in the form of concentric circles or rings. So how you are getting this interference pattern in the form of concentric circles or rings, why the pattern is like this that we will try to understand when we will discuss this Newton ring experiment in detail. Right so this is Newton ring experiment an example of division of amplitude. Now the conditions for sustained interference. To obtain a clear and stable interference pattern the following conditions must be fulfilled. The light source must be monochromatic. means of single wavelength the light source must be current in nature you know the phase difference should be zero or constant the amplitude of both waves should must be nearly equal for good contrast between the fringes so as i mentioned that you should have the light waves which are having a slightly which are having a nearly equal amplitude the wave must propagate in the same direction so these are the conditions in to get sustained interference pattern so you That's all about today's lecture. These are the reference books which you can follow. In next lecture we will start our division of amplitude with interference and thin films of uniform thickness. Thank you for today. Thank you very much.

So when I am saying optics, optics means we are dealing with light and you know in case of light there are three basic phenomenas which are involved in light. One is interference, second one is diffraction and the third one is polarization. So in this wave optics we are going to discuss the two main phenomenas of optics. One is interference that is the first part of this wave optics and the second one is diffraction. So I am going to start with the first part that is interference.

So the topic or the topic which today I am going to start that is interference. And these are the reference books which we are going to follow for this interference. The first one is a textbook of Engineering Physics by Namneet Gupta.

Then a textbook of Optics by Ajay Ghatak. Engineering Physics by SK Gupta and SL Gupta. and some internet links are also there from where you can have this information about interference and refraction.

So, the topics various topics which we are going to cover under this interference those topics are the first one is we are going to introduce what is the meaning of interference of light introduction then interference then type of interference. So, we are going to discuss the type of interference here. Condition for interference, the next one is coherent sources, the next one is methods to obtain coherent sources.

So here we will discuss that how we can get coherent sources. Interference in uniform and wedge-shaped thin films. So these are two topics actually, one is thin films of uniform thickness and the other one is thin films of non-uniform thickness which is also known as wedge-shaped thin films.

After that we will discuss that what is a need, what is a necessity, of extended sources in order to get a sustained interference pattern in thin films. After that a very important and interesting topic we will discuss in this chapter and that topic is Newton's ring experiment. So this is a very beautiful example of interference in thin films and we will also discuss the applications of interference here, application of basically Newton ring experiment.

So, now I am going to start with the introductory part of interference or this wave optics. So, as you have seen some beautiful colors on a soap bubble when it is illuminated with white light. Suppose you have a soap bubble and when the soap bubble is illuminated in white light, then you can see the beautiful colors in that soap bubble.

So, basically those colors which are there in that soap bubble, Those colours, they are the outcome of interference of light. You might have seen that if there is a drop of oil on the surface of water, then if that drop of oil is illuminated by white light or sunlight, then again you can see colours in that layer of oil. So again those colours, they are because of interference of light. So all these are the examples of interference in thin films. So, how the light will get, how you are getting the interference in light, what is the meaning of this wave optics that we are going to discuss in today's lecture.

So, optics is, this is a branch of physics in which we study about light or phenomena related to light. So, optics is that particular branch of physics where you are studying about light. So, you might have gone through this topic in your 12th class about interference, diffraction and polarization. So, this is... The next level of that optics on the basis of dual nature of light optics can be divided in two parts as you know That light shows dual nature sometime light is showing particle nature and sometime light is showing now wave nature So since light is having dual nature.

So on the basis of that Optics is divided into two parts. The first one is Geometrical optics which is also known as ray optics. So this geometrical optics actually it is the evidence for particle nature of light and the second one which is known as physical optics or you can say wave optics it is an evidence or it is a proof that light is having wave nature. So these are the two different branches or two different parts of optics on the basis of this dual nature on the basis of particle nature you have that ray optics and on the basis of wave nature you are having this wave optics.

In geometrical optics, we consider that light travel in the form of ray. This also proves the particle nature of light. So you might have gone through this ray optics in your 12th class.

The next one is physical optics, which is also known as wave optics. Here we consider the light travels in the form of So, this is an evidence for wave nature. So, these are the two different parts of optics.

Interference now we will try to understand what is the meaning of interference of light. Due to superposition of two coherent light waves of same frequency and nearly same nearly equal amplitude the intensity of light is redistributed in space. This phenomena is known as interference so what is the meaning of interference of light what you have you are having two light wave and those light waves are of same frequency and they are having nearly equal amplitude then the intensity of these light under suitable conditions these light waves will superimpose and after the superimposition you are going to get a resultant light vector and the that resultant light vector Either the intensity of that resultant light vector is greater than the sum of individual intensities or lesser than the sum of individual intensities. What I am trying to say here, suppose you are having two light waves.

One is having intensity I1 and the second one is having intensity I2, right? Now, these two of intensity I1 and I2 and they are having same frequency and they are having slightly same. amplitude. Now under suitable condition they will get superimposed and after that superimposition you will get a resultant light vector.

Now that resultant light vector now the intensity of that resultant light vector either that intensity is greater than the sum of individual intensities means this I either that is greater than this I1 plus I2 or that intensity you is lesser than the sum of individual intensities. You are having these two possibilities. Either the resultant light vector which is having intensity I, either that intensity is greater than I1 plus I2 or it is lesser than I1 plus I2. Now depending upon these two, you are having either you are having get you are getting maxima or you are getting maxima or you are getting constructive interference or destructive interference.

which we will discuss in next slide. So, the intensity distribution at any point in space is the resultant of individual intensities at that point. So, at different different points you are going to get different intensities either the intensity is greater than the sum of individual intensities or that is less than the sum of individual intensities.

So, depending upon that either at certain point you are going to get a maxima or at certain points you are going to get minima. So, I hope the meaning of interference is clear. Interference means when you are having two current light waves of same frequency and of slightly and of nearly equal amplitude and when these will superimpose then you are going to get the resultant light vector and that resultant light vector either the intensity of that resultant light vector is greater than the sum of individual intensities or it is less than the sum of individual intensities. So, depending upon that at certain points you are going to get maximas and at certain points you are going to get minimas.

Now, the types of interference. So, we are having two type of interferences as I had discussed in my last slide that at certain points the intensity of the resultant light vector is greater than the sum of individual intensities. So, those points will be the points of maximas or you can say those points will be the points of constructive interference and there will be certain points where the intensity is less than the sum of individual intensities and those points will be the points of minima or you can say destructive interference so two type of interferences are there one is known as constructive interference and the second one is known as destructive interference so there are certain conditions If those light rays, those light waves which are superimposing, if they will satisfy those conditions at certain points, then at those points you are going to get maximas and there are certain conditions.

And if those conditions will be satisfied at certain points, then you are going to get minima there. So now let's try to understand that what are those conditions under which you can get maximas at certain points and minimas at certain points. So for that...

Here directly the conditions are written but how you are getting those conditions that I will explain in next slides. When two waves are superimposed in the same phase that is same that is phase difference between them is integral multiple of 2 pi then constructive interference is produced. So, what is the meaning of that? You are having two light waves. So, this is the one light wave and this is another light wave right.

So, these are the two light waves and if they are in same phase means if the phase difference between them is 0 or integral multiple of pi Then, sorry that is integral multiple of pi. Then the interference between them that is going to give you constructive interference. So, when two waves are superimposed in the same phase that is phase difference between them is integral multiple of 2pi then constructive interference is obtained.

So, this is the condition to get that maxima. This is the condition to get that constructive interference in terms of phase difference. And the second condition is when two waves are superimposed in opposite phase when they are out of phase that is phase difference between them is odd multiple of pi.

If it is even multiple of pi then you are going to get constructive interference. If it is odd multiple of pi then you are going to get destructive interference. So if in one line I am writing these two conditions then the conditions are like this.

If the phase difference between the interfering light waves is even multiple of pi where this n starts from 0. So, you are getting the conditions are when delta is 0, when delta is 2 pi, when delta is 4 pi, when delta is 6 pi then the interfering light waves when they will interfere they are going to give constructive interference at those points where the phase difference is even multiple of pi and if the phase difference is odd multiple of pi means if it is 2 n plus minus 1 pi. if it is like this where this 2 and if you are taking 2 and plus 1 then n is going to start from 0 and if you are taking the condition 2 and minus 1 then n is going to start from 1 but in both the cases you are you can see here that delta is having values pi 3 pi 5 pi 7 pi 9 pi and so on. So what is the meaning of this? Those points where the interfering light waves they interfere in such a way that the phase difference between them is either zero or even multiple of pi then at those points the condition of maxima will be satisfied and at those points the intensity will be greater than the sum of individual intensities so those points will be the points of constructive interference there will be certain points where where the phase difference between the interfering light wave light waves is odd multiple of pi.

Odd multiple of pi means either the phase difference is pi or 3 pi or 5 pi or 7 pi and so on. And if the phase difference is like this then at those points the condition of minima will be satisfied means you are going to get the resultant intensity lesser than the sum of individual intensities or you can say those points will be the points of destructive interference. So there are certain points.

where the condition of maxima will be satisfied there will be certain points where the condition of minima will be satisfied so you are going to get maxima and minimas in your interference pattern so two type of interferences are there one is constructive and the other one is destructive constructive in other way you can say maxima condition of maxima destructive means the condition of minima for maxima the phase difference between the interfering rays that should be even multiple of pi And for destructive interference, the phase difference between the interfering light wave, that should be odd multiple of pi. So, these are the conditions to get destructive and constructive interference. So, this is an example of your constructive and destructive interference.

So, diagrammatically, you can understand from here. As you can see here, these are the two interfering light waves. Now, you can see that there is crust to crust and trough.

to 12 correspondence means these two light waves they are in same phase either the phase difference is zero or that is even multiple of pi. So, if you are getting this type of case then there will be enhancement of the intensity means when these light waves they are in same phase. Then the resultant intensity that will be greater than the sum of individual intensities. So you are going to get here constructive interference. So if there is crust to crust or trough to trough correspondence then you are going to get maxima.

Second one is, second is this. This is out of phase means either the phase difference is pi or that is 3 pi or 5 pi or 7 pi or so on. So when you are having two interfering light waves and if they are out of phase you are going to get this.

then in that case there will be destructive interference. So you can see here at this point for this interfering ray you are getting maxima but here you are getting minima. Here you are getting minima but in second case you are getting maxima.

So there is these two light the interfering light waves they are out of phase. The phase difference between them is odd multiple of pi. Here the phase difference is even multiple of pi.

Even multiple of pi means either the phase difference is zero. or that is 2 pi or that is 4 pi or that is 6 pi and so on. So, in this case you are going to get constructive interference but in the second case the phase difference is odd multiple of pi so that is 2n plus minus 1 pi. So, the phase difference that is of this order either pi or 3 pi or 5 pi or 7 pi and so on.

So, if you are getting this type of phase difference then you are going to get destructive interference. So in case of destructive interference the intensity that will be lesser than the sum of individual intensities right and if you are having two interfering rays with exactly same amplitude or with same intensity then in case of minima the result in case of destructive interference the resultant intensity will be 0. Why I am saying so? I will explain in next slide.

What I am saying here if you are having two interfering rays which are of exactly same amplitude and if the phase difference between those interfering rays at certain points is odd multiple of pi then at those points the resultant intensity will be zero and if you are having two interfering rays and if the phase difference at certain points between those light waves is even multiple of pi then you are going to get constructive interference at those points. So, this is your constructive interference and the second one is your destructive interference. Now, the conditions for interference means now we are going to write the conditions in terms of phase difference or we are also writing the intensity of constructive and destructive interference.

So, conditions for interference in this particular slide we can we will see these conditions. We know that intensity due to superposition of 2 wave is given by this. From where we are getting this result? So for this you have to remember, you have to just recall about your Young's double slit experiment. You have gone through that Young's double slit experiment in your class 12 and you know if you are having two light rays, sorry if you are having two light waves, one is of amplitude a1 and the second one is amplitude a2.

If a1 is the amplitude then I1 will be equal to a1 square and your I2 will be equal to a2 square. So these are the intensities of those two interfering rays right. So if delta is the phase difference between these two light waves then the resultant that according to the principle of superposition we have solved that and we have calculated the resultant intensity. So that resultant intensity is of this particular form. I hope you have recalled about this expression.

That resultant intensity because of the superposition of two light waves of intensity I1 and I2 and if delta is the phase difference between them is given by this equation. I is I1 plus I2 plus 2 root I1 I2 cos delta. So what is the meaning of all these terms? What is I1 here? That I1 signifies that I1 denote the intensity of first light wave.

I2 that represent the intensity of second light wave. And this delta that represent the phase difference between the two interfering rays. So between those two interfering light wave right. So where delta is a phase difference between them. So this delta actually the value of this delta is going to decide what you are going to get at a certain point.

Whether you will get maxima or minima that is completely decided by this delta. Why I am saying so? Because you can see in this equation I1 that is the intensity of first light wave constant. I2 that is again fixed. So only this particular factor only this cos delta is going to change the whole scenario.

So this cos delta will decide whether a point will be a point of maxima or that point will be a point of minima. So you know this cos delta. that will give you maximum value and that maximum value will be plus 1 and this cos delta will give you minimum value and that minimum value will be minus 1. So now this cos delta when it is taking plus 1 value you will get maxima and if this cos delta is taking minus 1 value then you are going to get minima because only these are the possibilities to get the maximum value of i or minimum value of i.

So now we are going to discuss the condition of maxima that is constructive interference. and condition of minima. So, the first one condition for constructive interference. So, when phase difference so for maxima what should be there cos delta should be equal to plus 1 and you know when this cos delta as you know cos 0 is 1, cos 2 pi is 1, cos 4 pi is 1, cos 6 pi is 1. It means when this delta will be equal to even multiple of pi. where this n that starts from 0. So, when n will take 0 value, delta will be 0. When n will take 1 value, delta will be equal to 2 pi.

When n will take 2 value, delta will be equal to 4 pi. So, when delta is 2 and pi means when this phase difference between the interfering light waves, if that phase difference is even multiple of pi, then delta is 2 and pi and starts from 0. and that is 0, 1, 2, 3 and so on means delta starting from 0 next value which it is taking that is 2 pi then 4 pi and 6 pi and so on. So, for those values you are going to get maxima. Why you are getting maxima? Because cos delta will be equal to plus 1 there.

So, if delta is taking 2 and plus sorry 2 and pi value then what will be the value of i? If delta is cos delta is 1 then I is I1 plus I2 plus 2 root I1 I2 right. So, this is the resultant intensity for maxima I is equal to I1 plus I2 this is for maxima I1 plus I2 plus 2 root I1 I2 right. It means now the resultant intensity of that constructive interference of that maxima is greater. Then the sum of individual intensities by this factor 2 root I1 I2.

Let me explain it. What is the condition of maxima in terms of phase difference? The condition of maxima is delta should be even multiple of pi.

And what is the value of this I? I for maxima is let me write it properly. It is I1 plus I2.

plus 2 root I1 I2. So, what is the final conclusion you can draw from here? The conclusion you can draw from here that when two light waves of intensity I1 and I2 when they are interfering then after their interference after the superimposition there are certain points where the intensity is greater than the sum of individual intensity.

and that is greater by this factor 2 root i1 i2 right. Now this is a condition in terms of phase difference. You can also write the condition in terms of part difference. So you know what is the relationship between phase difference and part difference?

The relationship is the part difference is lambda upon 2 pi times phase difference that is a relationship. So if you know the phase difference you can easily write the part difference. So the part difference will be equal to lambda upon 2 pi and what is the phase difference for maxima? It is 2 n pi.

So, this is this 2 pi will get cancelled and what you are getting? You are getting n lambda. So, what is n lambda?

This n lambda is the condition for maxima that is the condition for for part difference condition for constructive interference in terms of part difference. So, delta is n lambda and you know n starts from 0, 1, 2, 3. that n starts from 0 and it can take all the integers values like 1, 2, 3, 4 and so on. So, delta is 0, then lambda, 2 lambda.

When part difference is of this particular order, when part difference is 0, when part difference is lambda, when part difference is 2 lambda, when part difference is 3 lambda, for all those values, what you are going to get? You are going to get maxima. So, those points where the part difference is 0, lambda, 2 lambda, 3 lambda, 4 lambda and so on. Those points will be the points of maxima. So, I hope now the condition for maxima is condition for constructive interference is clear in each and every sense.

What is the condition of maxima in terms of phase difference? In terms of phase difference, you can say when the phase difference is even multiple of pi, then you are going to get a constructive interference. In terms of part difference, you can say when the part difference is n lambda.

then you are going to get maxima right. So, and the intensity of maxima is I1 plus I2 plus 2 root I1 I2. So, condition for maxima.

Now, we will discuss the condition for minima or you can say condition for destructive interference. So, you know that I what we have written there I is I1 plus I2 plus your 2 root I1 I2 cos delta. Now, from according to this equation this i will be minimum if cos delta is minus 1. So the condition of destructive interference should be cos delta that should be minus 1 and you know when this cos delta will be minus 1. If this delta is taking pi value if this delta is taking 3 pi value if this delta is taking 5 pi value then for all these odd multiple of pi's for cos pi, for cos 3 pi, for cos 5 pi, for cos 7 pi, for cos 9 pi.

For all these odd multiple of pi's, cos will take minus, cos odd multiple of pi will take minus 1 value. So, all these values or all these points will be the points of minima. So, when this delta is odd multiple of pi.

So, here this 2 and plus 1 is written. You can also write 2 and minus 1 here because you can write this. Whether you are writing 2n plus 1 or you are writing 2n minus 1 in both the cases this particular term is going to give you odd value.

But the difference is if you are writing 2n plus 1 then n will start from 0 as mentioned over here. But if you will write 2n minus 1 then n is going to start from 1 right. So the meaning is only this that when the phase difference between the super between the interfering light waves is odd multiple of pi then you are going to get the condition of minima at those points.

So the points are when delta is pi when delta is 3 pi 5 pi or you can say odd multiple of pi then you will get minima. So this is the condition of minima in terms of phase difference and you know the relationship between part difference and phase differences. The part difference sorry if you know the phase difference And the part difference is lambda upon 2 pi times of phase difference.

That is the relationship. So now if you want to write the part difference, it will be equal to lambda upon 2 pi times the phase difference. And phase difference is odd multiple of pi here. So this pi will get cancelled by this. So what you are getting?

You are getting this 2 and plus 1 lambda by 2. Or in other words, you can say the part difference is odd multiple of lambda by 2. So, here that power difference is written delta is 2n plus 1 lambda by 2. So, for minima or you can say for destructive interference phase difference should be odd multiple of pi and power difference should be odd multiple of lambda by 2 where n starts from 0 here. So, what are the values for values in terms of power difference for minima? The values are the first one is lambda by 2 when this 2n plus 1 when it is take one value. So, power difference will be lambda by 2. Next odd is 3. So, you will get 3 lambda by 2. Next one is 5. So, you will get 5 lambda by 2, 7 lambda by 2, 9 lambda by 2 and so on. So, the condition for minima is this, right.

So, what we have done? We have drawn the conditions for maxima and minima. So, I am just concluding these conditions here. So, on one side, I am writing the conditions of constructive interference. And on this side, I am writing the conditions of destructive interference.

So, first I am writing the intensity of constructive interference. So, that intensity is I1 plus I2 plus 2 root I1 I2. This is the intensity of maxima.

And if you want to write the intensity this is for maxima. And if you want to write the intensity of minima then what will be that? Intensity of minima that is equal to I1 plus I2.

And you know the last term is plus 2 I1 I2 cos delta. For destructive interference that cos delta is minus 1. So, what you are going to get here? You are going to get here minus 2 root I1 I2.

This is for minima and that one is for maxima. So, there is one important point and very interesting point here. Whatever is the gain for maxima? What is the gain for maxima with respect to sum of individual intensities that is I1 plus I2? The gain is in terms of 2 root I1 I2.

So whatever is the gain in case of maxima that is a loss for minima because in minima the intensity is less than the sum of individual intensities and the factor which is less that is 2 root I1 I2. So the loss for minima or destructive interference that will appear as gain for maxima means that light energy During the formation of interference that light energy will remain conserved. Energy that will remain that will follow the law of conservation of energy law of conservation means energy will remain conserved.

There is only redistribution of energy between maxima and minima. Why I am saying there is redistribution of energy between maxima and minima? Because whatever the laws for destructive interference that is appearing as gain for constructive interference right. Now, I am writing the conditions for maximas and minimas. So, in terms of phase difference, you know the condition for maxima is this phase difference is even multiple of pi where n starts from 0, right?

So, this is the condition for maxima here in terms of phase difference. And in terms of phase difference, the condition for minima is odd multiple of pi. Phase difference should be odd multiple of pi. And this n starts from 0, right?

So, what are the points? What should be the phase difference between the interfering rays? That should be 0 or 2 pi or 4 pi that is even multiple of pi and so on. And here this delta will be, this will be pi, this will be 3 pi, this will be 5 pi, this will be 7 pi. It means odd multiple of pi, right?

So, this is for constructive interference, this is for destructive interference. And in terms of power difference. What we have concluded for part difference in terms of part difference the condition is part differences n lambda right.

So, part difference for maxima should be 0 or lambda or 2 lambda or and so on. And in case of destructive interference part differences odd multiple of lambda by 2 that is 2 1 plus 1 lambda by 2. So, what is the meaning of this? The part difference between the interfering light wave for destructive interference that should be lambda by 2 or 3 lambda by 2 or 5 lambda by 2 and so on.

So, these are the conditions for constructive interference and these are the conditions for destructive interference. So, you have to remember all these conditions because in order to understand the interference in different different cases you should know for when you can get maxima at a certain point and you should know for when you can get maxima at a certain point and Under which condition you can get a minima at a certain point. So, all this is decided by the phase difference between the interfering light wave and the part difference between them, right.

So, these are the conditions. Then the next one is coherent sources. So, the coherent sources, this particular condition that the light sources from where you are getting the light that those light sources should be coherent in nature.

So, Now, we will try to understand what is the meaning of coherent sources. Two sources are said to be coherent if they emit light waves of same frequency, nearly the same amplitude and the most important condition is the phase difference that which is the phase difference between the light waves which are emitted from those sources, that phase difference should be either that should be 0 or that should be constant. So, that is the main condition to get coherent sources.

So, what are coherent sources? two sources are said to be coherent if they emit light wave of same frequency and nearly the same amplitude and the phase difference between the light waves which are emitted from those sources that phase difference either that should be zero or that should be constant so that is the main condition for the interference process the coherent sources are essential so whenever you want to get the interference of light then The main condition is the first condition is the first requirement is that you should have coherent sources. So, that is the first condition. two sources of light are said to be incoherent if they emit light wave whose phase changes with time if the phase changes with time then those light sources they are said to be incoherent in nature so we always require coherent light sources which can maintain a constant phase difference irrespective of time so those sources are said to be current in nature and there is a very important point that if you are having a single source and if you can divide that single source in two parts then those two sources which are generated from same source only the light waves which are emitted from those sources will be coherent in nature but if you are having two independent sources even the similar type of sources i'm taking an example here suppose you are having two sodium lamps two similar type of sodium lamps but those sodium lamps are independent of each other means to you are having two independent sodium lamps there and those two sodium lamps they are emitting light then the light which is emitted from those two independent sodium lamps that light will not be in co that light will not be coherent in nature you can get the coherent light only if the the light rays or the light waves are coming from a single source so two independent sources they can never give you coherent light because The initial phase of each and every source that is independent of each other.

So if there is any change in the phase of one source, you cannot maintain the similar type of change for the other source. So that is the reason whenever you want a coherent source, you have to divide a single source in two parts. So you know the main condition to get interference is you should have coherent sources of light only. Then you can get in sustained interference pattern right. So that is about coherent light.

Methods to obtain coherent sources in practice. So as I have mentioned that if you have two independent sources then you can never get coherent light. So now the question arises that how can you get coherent light.

So as I mentioned that you can get coherent light only by using a single source. So there are two methods through which you can get coherent light. from a single source.

One is division of wavefront. This is the first method through which you can get coherent light from a single source. You can divide a single light source in two parts and you can get coherent light from those two parts. So, division of wavefront is one example and the second one is division of amplitude.

So, using these two methods you can always divide a single source in two coherent sources. Now, the first one is division of wave front how you can get light there a coherent light there the in the incident wave front is divided into two parts these two parts travel and superimpose each other and produce interference pattern so the example one example one experiment through which you have study in your 12th class famous young's double set experiment so what you have in young's double set experiment you are having a single slit here which is illuminated by a monochromatic source of light. After that there are two slits which are placed at an equal distance from the single slit and there will be division of wave front at these two slits and these two that is S1 and S2.

Since they are eliminated from the single slit they are eliminated from the single source. So these two slits S1 and S2 which are eliminated from the single slit from the single source These two are known as coherent sources. From these two slits you will get the coherent light.

Reason is because they are illuminated from a same source. So by the division of wave front you are going to get coherent light. So the examples of division of wave front are the first one is Young's dubestate experiment and the second one is freshness by prism experiment.

There are some more experiments of more examples of this division of division of wave front like Lloyd's mirror matrix. method. So, all in all these examples by the division of wave front a single source is divided into two coherent sources.

So, always a single source is used to generate two coherent sources. So, this is one method and we have gone through the examples of this division of wave front in our previous classes. Now, in our syllabus here we have to discuss the division of amplitude that is the second type of.

method through which you can get coherent light and in this entire chapter we are going to discuss this division of amplitude right so the second one is sorry just in the exam experiment the setup of that young's double slit experiment what you have a source of light which will illuminate a single slit then two slits s1 and s2 which are placed at same distance from the single slit then these two slits this s1 let i'm writing it as let s1 and the second one s2 so s1 and s2 these two will act as current sources so there will be division of wave front and these two will act current sources and you will get the interference pattern over the screen. So, this is an example of division of wave trend. Now, the next one is division of amplitude. The intensity of incident light is divided into two parts, right?

These light waves superimpose afterwards and produce interference pattern. So, what will happen here? A single light ray, its amplitude is divided into two parts, its intensity is divided into two parts and two.

those two parts when a single ray is divided into two parts those two parts since they are generated from a single ray that is generated from a same source so if there is any change in the phase of that ray parent ray you will get the similar type of changes for the rays which are generated from that single ray it means you can maintain a constant phase difference here you can always maintain that constant phase difference for those um those divide those two rays which are obtained from a single ray. You have gone through this division of amplitude in Michelson model experiment. In Michelson model experiment there was an extended source of light from that extended source of light the light will incident over a glass plate which is inclined at 45 degree. So when light will incident over that glass plate which is inclined at 45 degree and that is also semi-silver some part will be reflected and some will be transmitted. So a single ray.

that will divide into two parts reflected and transmitted now that reflected and transmitted ray the amplitude of a single ray is divided into two parts reflected and transmitted now if there is any change in the phase of that parent ray you will get the similar type of changes for reflected and transmitted ray also it means you can maintain a constant phase difference here so those two rays will act as two current sources so that is the method to generate two current sources from a single source So, in division of amplitude the intensity of incident light is divided into two parts. As I have given you example some reflected portion, some transmitted portion. These light waves superimpose and then they produce interference pattern. Now the examples of division of amplitude Newton ring experiment, Michelson interferometer, interference and thin films of uniform thickness, non-uniform thickness, all these are the examples of division of amplitude. So, in this chapter where we are discussing this interference basically what we are going to discuss we are going to discuss Newtonian experiment here we are going to discuss interference and thin films of uniform thickness and non-uniform thickness in detail so that is our objective here right so the what are your current sources and what are the methods to generate those current sources so what I told you current sources are those sources from where we are getting the light waves of same frequency.

of a slight of nearly equal amplitude and the phase difference is either zero or constant between them. So, if you are having a two independent sources then those two independent sources can never give you coherent light. So, how you can get coherent light? In order to get coherent light what you have to do you have to divide you have to convert a single source into two parts and what are the methods to generate those.

those coherent sources the first method is by the division of wavefront and the second one is division of amplitude examples of division of wavefronts wavefront are young's double state experiment and frational biprism experiment and lloyd's mirror method all these are the examples of division of wavefront which you have gone through in your previous classes and the second one is division of amplitude which is the point of discussion in our chapter here So, division of amplitude means a single ray, the amplitude of that single ray or the intensity of that single ray is divided into two parts and then those two parts will act as two coherent sources right. And there are different examples of this division of amplitude. One is Michelson model experiment where the Michelson interferometer. In that interferometer the amplitude of single rays divided into two parts. Newton ray experiment that is an example of division of amplitude.

So, these are the examples. So, in this chapter we will discuss this division of amplitude. So, now a quick this review of Newton ring experiment here.

This is the experimental setup. So, in this experimental setup you can see here that this is a source of light, monochromatic source of light and this is a glass plate which is inclined at 45 degree. So, when light will incident over this glass plate the light will divide into two parts. So, there will be division of amplitude there right.

So, because of the division of amplitude you will get the interference pattern and the interference pattern in newtonian experiment that will be like this. That will be in the form of concentric circles or rings. So how you are getting this interference pattern in the form of concentric circles or rings, why the pattern is like this that we will try to understand when we will discuss this Newton ring experiment in detail.

Right so this is Newton ring experiment an example of division of amplitude. Now the conditions for sustained interference. To obtain a clear and stable interference pattern the following conditions must be fulfilled.

The light source must be monochromatic. means of single wavelength the light source must be current in nature you know the phase difference should be zero or constant the amplitude of both waves should must be nearly equal for good contrast between the fringes so as i mentioned that you should have the light waves which are having a slightly which are having a nearly equal amplitude the wave must propagate in the same direction so these are the conditions in to get sustained interference pattern so you That's all about today's lecture. These are the reference books which you can follow.

In next lecture we will start our division of amplitude with interference and thin films of uniform thickness. Thank you for today. Thank you very much.

Transcript for:Wave Optics: Interference and Diffraction

Transcript for:
Wave Optics: Interference and Diffraction