is my screen visible now yes ma'am yes so we were talking about parallel thin film and I said that your Optical path is nothing but it is refractive index times the geometrical path whatever is the path which is traveled by light which we call as geometrical path the actual path that is double by light if you multiply it by the refractive index of the medium through which your light is traveling through more career to Glass being equal to 1 your Optical path would be equal to geometrical path in case of your electromagnetic wave is moving through air so let's talk about the parallel thin fin let's say that you have a thin film here and let's say this thin film is probably of glass or anything and let's say that refractive index of that particular medium Is mu and the thickness of this film is d and mu is the refractive index of this medium on this side you have air and on this side you have air in both sides you have here and in between here you have something whose refractive index is given by mu now we'll talk about the interference pattern in the reflected light first and then we will talk about the transmitted light also so let's first talk about the a reflected light here let's say a ray was incident here at this point what would happen is that a part of this Ray will be reflected and a part of it will be refracted here when this Ray comes at this point here again a part of it will be reflected and a part of it will go to the next medium which is air again here but since we are not bothered about this portion so far so I'm not going to focus on that now at this point again a part of it will be reflected and a part of it will be transmitted on this side as I said before in this unit we are going to focus on interference due to division of amplitude so what we are saying is that whatever was the amplitude here a part of that is being reflected a part of it is being refracted and then after this refraction it's again reflected at this point and again reflected refracted at this point so what would be the interference pattern between Ray A1 and Ray 2. when you will superimpose them on the top of each other what kind of interference pattern do you expect at your screen so let's say the angle of incidence was I here and the angle of incidence this also will be I here now let me draw a perpendicular from this point on this Ray now if you talk about the ray one and Ray 2 up to this point let's call this as point a let me write a here up to this point a there was coming from a single Ray which was your incident Ray so up to point a both of them have traveled the similar distances from here onwards so after this line outwards are parallel to each other so that means after this line here they will again travel equal amount of distances so whatever is the path difference that is between Ray one and Ray 2 and path difference is going to decide if the intensity would be minimum or maximum right path difference multiple times Lambda oga for example if it's Lambda 2 Lambda 3 Lambda so up screen expected intensity maximum if it is some kind of or multiple of Lambda by 2 in that case you would expect to observe a minimum on the screen so we would like to know the path difference between Ray one and Ray 2. that is why I am saying up to point a they were coming from a single Ray and when you have grown perpendicular here after that both of these are parallel rays so they are going to travel equal distances up to the screen here so now we need to look at the whatever is going to happen between this point and this point for Ray 2 and this point and this point for the ray one let's call this as point B this is point C and this is point d Ray 2 up to this point Ray 2 has traveled distance a B plus BC and that distance has been traveled in a medium whose refractive index is Mu so now if I want to calculate the path difference if we want to calculate the path difference between Ray 1 and Ray 2 and let's call that as Delta this is equal to Mu times a B plus BC this is the distance which is traveled by Ray to additionally and by the ray one the distance that is traveled is 80 and that distance is traveled in air and we know that for air your refractive index is one so that is why here there is no mu because mu for air is equal to one so the path difference between Ray one and Ray 2 is whatever is the distance traveled by this Ray to up to line DC starting from a which is a B plus BC in a medium whose refractive index Is mu and from point A to line DC the distance that is traveled by wave 1 and that distance is a d here so this is now your path difference now what would you like to do is we would like to convert this path difference into our known quantities and that known quantity is more or less D which is the thickness of different we would like to associate this path difference with the thickness of the film and that is where your trigonometry comes into picture now from point B I can grow a perpendicular here and that perpendicular is let's call it e now if this is i r is your angle of refraction here then this is going to be R here if this is R then this is R because this line and let me write this f a f and EB are perpendicular to each other at this point R is the angle of incidence right that also means that is angle and this angle both of them are 90 minus r now let's talk about this line AF this line is perpendicular to the film here this angle here angle a d C so the angle D AC note ADC angle D AC is going to be 90 minus I am saying that this angle here is basically 90 minus and so your this angle would be then 90 minus sign first let's prove that a b is equal to BC here which you can very easily do because now we are saying that if you talk about triangle a b e and triangle with a I should write C and with b b e triangle to compare they have the corresponding angle equal is this is R this is R this is 90 minus r this is 90 minus r B is there are triangle there is a name for these kind of triangles this angles equal to each other I think congruent triangles right from here you can simply apply trigonometry and you can prove that a b is equal to BC if you talk about the triangle let me write here and then the famous rule if you talk about this triangle here a b e B in this case so if I write a b because I would like to know a b in terms of e b e b is nothing but the thickness of the film which is D here so then that means a b upon e b a b is your hypotenuse EB is the base so hypotenuse upon base would be equal to what would be that equal to secant of r so that means a b is equal to d e secant of r b e is nothing but the thickness of the film that means a b is equal to D times secant of r similarly in the second triangle CBE you can again use the trigonometry and your CB upon EB is again equal to secant of r which would imply that c b is also equal to D times secant of r so that means in this expression I can write 2 times mu D secant of r I can write that much minus a d now we need to write a d also in terms of D which is the thickness of the film how can we do that this line b e if I am saying that a e b and c e b these two triangles they are congruent congruent triangles all the three sides and all the three angles are equal to each other that means a is equal to e c your a would be equal to EC in that case right so that means you can write your AC being equal to 2 times a e foreign let's talk about triangle ADC is this is the triangle and this angle is now 90 minus r here so then I can say that a d upon AC a d is base AC is hypotenuse right and this is perpendicular so a d is base AC is hypotenuse that means based upon sorry about that I hope the screen is still there and it is not there again remember to see the blank screen yeah I think I lost the connection and I lost the previous recording also so that means I need to redraw the drawing again sorry for that so this was where we were and I'll draw it exactly with this similar kind of Expressions that we used previously so if this is I this is I this is r i can draw a perpendicular here this is R this is r and I think I call this as a b c and from here we drew a perpendicular which was D this was e if this is I this angle is 90 minus I and we said that the path difference is going to be equal to Mu times and the t is the thickness of the film mu is the refractive index then this is equal to a B plus BC minus a d is and a d Wala portion we wanted to solve so um let's come back to the triangle ADC sorry about the internet connection if you come back to this triangle again let me repeat that a d upon AC and let me write the formula I don't remember it anyway okay so uh a d is going to be your base AC is going to be your hypotenuse so this is going to be equal to cosine of 90 minus I this is how much it is going to be equal to and which is nothing but sine I so that means a d is equal to AC sine I but we would like to have now this AC in terms of the thickness of the film here I already said that because you are this triangle a e b and Triangle C E B these are congruent triangles so AE is basically equal to EC or in that case you can say that AC is going to be equal to 2 time a e so here then I can write AC being equal to 2 times a e sine of I now come back to this triangle a e b which is this triangle now let me change the color again to the blue is in this triangle I want a to be written in terms of the thickness of the film which is easier a e upon e b its perpendicular upon base this is equal to tangent of r which means that AE is nothing but e b tangent of r e b is nothing but the thickness of the film so this is equal to D times tangent of r now I can plug in the value of AE here and this I can write as 2 times a is D tangent of r sine of I right if you plug in this expression in the path difference you will get 2 times mu d secant of R minus 2 D tangent of R sine of I what I can do is that I convert this angle of incidence term into angle of refraction by using Snell slope right if you use Snell slope which here would basically say that we said that there is air on this side air on this side which will say that mu is equal to sine I upon sine r so that means sine I is equal to Mu times sine R so in that case and let me draw a line here so that we do not mix the Expressions so that means if I use this Snell slope I can write this expression for the path difference which we called as Delta being equal to 2 mu D secant of r minus 2 times sine R is Mu time sine r sorry sine I is Mu times sine R so I can write this also as Mu D tangent of R and sine of r in this form from here to Mu D you can take common out secant of R is 1 upon cosine of r two mood is multiplied by sine r is 1 minus sine Square r foreign this is your path difference in the simpler form mu refractive index materials now let's compare Ray 1 and Ray two this was the path difference which we have calculated from actually whatever was the difference in the length traveled by the whatever was the difference in the left length traveled by Ray one and Ray two a b that we haven't applied stock slow right let's look at Point a update reflect over here you are going from a air to something whose refractive index Is mu and which is going to be higher than air so 180 degrees remember your stock slow states that whenever you go from rarer to denser medium again then at the point of reflection there will be a sudden phase change of 180 degree but at point B you are going from a medium of higher refractive index to the lower refractive index so you are going from a denser to the rarer medium and we know that when you are moving from denser to rarer medium there is no phase change in the reflected light hence at this point there will be no phase change when your Ray will be reflected simply transmission from here to here hence there is no phase change so rate to make there is no phase change this path difference we have simply calculated based on the distances traveled by Ray one and Ray 2. but on the top of the distance is traveled by A1 and Ray 2 there is an additional phase difference of 180 degree in Ray 1 as compared to Ray 2 which is just because your ray is getting reflected from rarer to denser medium we have to take that also into account so that means the effective path difference or rather the word you can use total path difference would be whatever is the path difference geometrically or optically that you have calculated here plus minus Lambda by 2. either you can do plus Lam plus Lambda Y 2 you can do minus Lambda y that is totally up to you and we know that 360 degree 2 pi is equal to Lambda right after 2 pi Lambda K equivalent hence Pi is equivalent to Lambda by 2. this would be your relationship between the two and that is why there is this additional term plus minus Lambda by 2 you can either stick with Plus or you can stick with minus and I'm going to stick with the plus here for the rest of this unit but you might as well stick with minus as you will see consequently here now if this is your past difference let's now talk about the condition of minimum and maximum for Maxima what should be the path difference you would say that your path difference should be an integral multiple of Lambda right integral multiple of Lambda then on your screen you will see an intensity of maximum that means we are saying that we would like to have this which was our path difference being equal to n Lambda this means that we would like this to be equal to n Lambda minus Lambda by 2 n minus 1 Lambda by 2. looking at this expression here your n has to start from one own words rather than choosing here Plus that would have started from zero onwards so that is totally up to you plus Lambda of course it doesn't make any difference so this is now your condition for the maximum which is against our intuitions usually hamara the way we understand is this should be equal to right hand side usually we always assume to be equal to n Lambda and which is true in this case also but this extra Lambda by 2 is changing everything in this case in that case you would expect your path difference to be equal to 2 N plus 1 Lambda by 2 I am taking here minus sign so I'll start my counting from one onwards you start your kind counting from zero onwards and again this is simply because wave is a repetitive Behavior so that means what you would like is that your path difference which we defined by this expression here is equal to 2 N plus 1 Lambda by 2. Lambda by 2. this would basically means that two mu D cos R is equal to n Lambda where your n starts from 0 1 and so on so this will be a condition for Minima and this is for the reflected light remember we are talking about the ray one and Ray 2 which are reflected light now on the screen you will see alternatively bright dark bright dark bright dark are fringes have Observer okay these are called as fringes of equal inclination and I'll tell you why that is equal to and then condition of minimum and maximum is basically decided by this path difference you are saying that there is interference happening between Ray one and Ray 2 here and I'm saying at some infinite distance if you place a screen yeah okay and the fringes that you will see on this screen these are called as fringes of equal inclination why this is called fringes of equal inclination look at the expression for your past difference okay your path difference is dependent of cosine of r right angle of incidence the moment you change I automatically R will change and your path difference is deciding whether you are going to see on the screen a bright or a dark Fringe and this is the reason why these are called as fringes of equal inclination let me repeat that inclination basically means that what is the angle of inclination of Ray here the moment moment you change this angle of inclination which is angle of incidence here your angle of refraction will automatically change the angle of refraction change and your path difference decides if you are going to observe an intensity of minimum or maximum on the screen and hence depending on our Agra screen pay um that will be changed accordingly so that is why these are called as fringes of equal inclination now what about what about the transmitted light should draw it on the next page so far what we did was for the reflected light we did talk about the condition of minimum and maximum which are given here Maxima and Minima these conditions are conditions for the reflected light duty how many path difference calculations but what you can actually do is rather than looking for the interference pattern on this side let's say this was your day which was incident here part is refracted and then at this point part is reflected and at this point again a part is refracted reflected at this point when I said a part of light will be reflected but a part of light will be transmitted also or refracted right I can call it this as let's call this as Ray 3. and then at this point when a part of this incident light is refracted apart will be reflected also okay and at this point now again a path will be reflected and a part will be refracted on this side let's call this as Ray 4. the way you expected an interference with pattern between Raven and Ray 2 similarly we do expect an interference pattern between Ray 3 and Ray 4. are the transmitted light because if this was your incident light this is your reflected light or Joe light up case film K through transmit okay it appeared on the other side that will be basically cooled as your transmitted light here of course you can do the similar mathematics that we did previously that if this angle is R we said this is I this is I then this is R then this is R right if this is R this is going to be 90 minus r you can grow a perpendicular from here to here you can grow a perpendicular from here to here and then up to this point if you talk about Ray 3 and Ray 4 up to this point both of them were single Ray but is equal distance so mathematically now I can say that let's call A B C and this as D now I can again say that the path difference would be equal to Mu times a b plus BC minus a d you can again do the mathematics and trigonometry but rather than doing that you can argue physically also or conceptually also and that argument would be if you are saying that you are getting a maximum in the reflected light minimum and this is obvious energy have conserved here right that means [Music] or it could be other way around also minimum hoga in that case foreign now you are going from this medium whose refractive index is Mu which is the dense medium to a rarer medium reflect okay foreign this is the mathematics behind it will be equal to two n plus 1 Lambda by 2 where your n starts from 0 1 2 and so on up to maximum conditions of R is equal to n Lambda where your n starts from 0 1 2 which is opposite to the condition of Maximum and minimum in the reflected light could be reflected by maximum over transmitted by minimum hoga transmitted by maximum this is you can simply explain it without going into trigonometry and Mathematics simply by using the logic of energy conservation so this was all about the parallel thin film now we are going to talk about what if the thickness is not uniform throughout the film and those kind of films are called as wedge-shaped film ma'am please go back to previous slide once just one second thank you ma'am okay now previously we said those thickness uniformity now what if this thickness is not uniform s you place a Plano convex lens foreign if you want to imagine it this would look something like this foreign thin film again conditions of minimum and maximum will be opposite in both of the cases in the case of wed shape thin film we are again not going to do the derivation because the expression is going to be very much similar let's say that these two sides they make an angle Alpha which is basically the angle Alpha here this angle will be your Alpha angle or in the case of Newton ring experiment this is basically your angle Alpha here and this Alpha is usually very very small angle it's extremely extremely small angle now if you remember from the previous slides for the reflected light your condition of maximum was 2 mu D cosine of r is equal to 2N minus 1 L Alpha minus 1 multiplied by Lambda by 2 when your n starts from 1 onwards right this was your condition of Maxima in the parallel thin film the only difference you would get is here rather than having R you will have cosine of R plus Alpha that would be the only difference that you will get for the maximum in the reflected light for a well-shaped thin film will not do the mathematics derivation for this if you talk about the condition of minimum sorry not the transmitted light if you talk about condition of many mass here similar to the condition of Maxima the only difference would be you will have rather than having cosine of R you will have cosine of R plus Alpha being equal to n Lambda where your n can start from 0 1 2 and so on instead of yeah additional Plus in the case of wed shape thin field thin parallel thin film is now in these cases of course the path difference condition of minimum or maximum that means that is going to be equal to 2 mu D cosine of R plus Alpha plus Lambda biology difference so you will have exactly similar path difference only difference is that I have done the derivation in slides if you want I can provide you with the derivations of this but because it's not a part of your syllabus or expression drastically different honey so we are not going to do the derivation in this case now if you sangeeth it was by mistake okay so if this is your path difference now let's talk about D is the thickness of the film now right it is it will come different at different places so D is no longer a constant now but if these two edges if they are touching each other then at the point of this touch would you expect a bright ring or would you expect a dark ring can someone tell me that what would you expect where these two surfaces are just touching each other look at this yes dark look at this expression here when they are just touching each other that means you are talking about this point again your D is equal to zero foreign I landed here anyone has any doubts we'll start from this expression in the next class difference Maybe okay that shape Maybe mom please upload previous lectures I have already done that I just did it before the class the last one lecture I just did before the class and I uploaded the PPT for interference also and I have completed this lecture just within next 10-15 minutes thank you