In this lecture we are going to discuss anti reflection coating. Anti reflection coating is a direct application of the interference of light and this these anti reflection coatings are used in the lens of the camera in order to avoid the excess reflection of light from the lens surface. Now let us look at the construction of anti reflection coating. Usually the lens of camera is made up of glass. Suppose this is the surface of glass. For glass you know that refractory index mu is roughly about 1.55. Then over the surface of this glass a very thin layer of material called as magnesium fluoride MGF2. This deposited such that the refractory index of this MGF2 is about 1.38 which lies between the refractory index for air and glass. The refractory index for air is 1, while the refractory index for glass is 1.55 and the refractory index of MGF2 is 1.38 which lies between that of air and glass. Further the thickness of this film of MGF2 is selected in such a way that it is equal to lamina by 4, where lamina is the most sensitive wavelength to the human eye. It is about 5.500 extra minute divided by 4. Now with this as a normal let us assume that a beam of light AB is incident with some angle of incidence say eye. And on reaching at this point B part of the incident light is reflected and emerges out as ray 1. Part of it is reflected through the film of MGF2 with some angle of reflection R say along B C. On reaching at this point C again part of it is transmitted part of it is reflected say along C D. On reaching at this point D again part of it is transmitted say along B R 2 part of it is reflected and further it is transmitted. And hence we obtain rays in the reflector system as V R 1 and D R 2 these are the rays which seems to be such that they are reflected from the film while we obtain rays like C. And if I name these as CT1 and DG2 then CT1 and EG2 DG2 these are the 2 rays which seems to be such that they are transmitted through the film of MGF2. Now they will since the rays in the reflected system as well as transmitted system as they are obtained from the single incident source and therefore they are coherent in nature. So since they are coherent in nature they will interfere either constructively or destructively depending upon the part difference within them. And therefore in order to find out the part difference in case of difference we have done the procedures like before in case of we have discussed in case of thin panels added film we draw a perpendicular from this ray D R 2 on the path of ray D R 1. So that paths travelled by both the rays we on this were a particular and are same and therefore part difference is path BC plus CD minus path B M in air. But here there is no need to do all such procedures because the rays they emerge parallel to each other they emerge very close to each other and since the thickness of the film of MGF2 is very small which is lambda by 4 where lambda is 5 5 0 0 0 is the instrumented. So 5 5 0 is the instrumented means 5 5 0 0 into 10 raise to power minus 10 meter which is a very small thickness and therefore the optical part difference will be equal to path BC in film plus path CD in film. This is roughly the optical part difference for the rays in the reflected system and this path BC is nearly equal to the thickness of the film. So since path BC is nearly equal to thickness and you know that thickness is equal to lambda by 4. So path BC is equal to lambda by 4 path CD is equal to lambda by 4 and therefore optical part difference is equal to lambda by 4 plus lambda by 4 which is equal to lambda by 2. Now let us see that is there any additional part difference of lambda by 2 due to the rays which are reflected from the denser medium because if you recall that in previous lecture we have discussed some important results in optics and those important results are if the there is a phase if the light is reflected from the from the denser medium then there is a phase change of pi and that phase change of pi will always corresponds to a part difference of lambda by 2. So here if you look at the ray ray AB is incident from air and it is reflected from the film of MJF2 film of MJF2 is denser as compared to that of air. So this reflection takes place at the surface of denser medium it means that ray DR1 will undergo a phase change of pi and you know that a phase change of pi will always corresponds to a part difference of lambda by 2. But what about this ray that is DR2 here this reflection that is BC this reflection it also takes place at the surface of denser medium because this reflection takes place at the interface between the film of MJF2 and glass here the glass is denser as compared to that of MJF2 and therefore this reflection also takes place at the surface of denser medium so it will also undergo a phase change of pi and hence the phase difference between the rays in the reflected system being 0 as the phase difference is 0. So there is no corresponding part difference the corresponding part difference for this phase difference will also be 0 and therefore the effective part difference will remain only as optical part difference which is equal to lambda by 2. It means that the optical part difference or the effective part difference between the rays in the reflected system as it is equal to lambda by 2 which is odd integral part of the wavelength it means that it will satisfy the condition for destructive interference. If it satisfy condition for destructive interference it means that the rays in the reflected system will always interfere destructively to produce darkness it means that no light is reflected from the lens of the camera but maximum light is transmitted through the lens of the camera and that is what is our purpose of inter-reflection coating that we do not want that light to be reflected from the lens but maximum light should be transmitted through the lens of the camera which is necessary for the production of good quality film. So this is the direct application of interference of light and the rays in the reflected system they always interfere destructively to produce darkness just because of the two important facts that we have selected this material Mgf2 which has the reflected index which lie between that of air and glass and we have selected the thickness of the film such that it is equal to lambda by 4. Keep these two points in mind so that you can write down about the inter-reflection coating.