We will cover this interesting problem on Young's double slit experiment. In the normal problems you have air in between the slits and the screen. Here you have a medium with refractive index mu which is 1.3 and this is an air gap. Air gap here. Let the thickness be t and it is given that with this arrangement you get the third bright fringe here. Third bright fringe is here at the origin that is at y is equal to 0. We have to find the location of fourth bright fringe. So whether fourth fringe will be on this side, on this side first of all we should know that. And to know that let us just go back to the original Young's double slit experiment. We know that suppose I have this kind of an arrangement. This is S1, S2 and I place a glass slab here. Earlier without glass slab this was the, without glass slab this was the center. Means the central bright fringe was here. Now with glass slab what will happen is, without glass slab I was getting like this. Now suppose I place the glass slab here and this is air and this is denser medium. Please remember whenever you have medium as air and you place the denser medium in path of one of the slits the pattern shifts towards this side where you place the medium. This is the case where it is lighter and it is denser. So it is shifting towards the slab where you place and this is denser this is lighter. Here it is just the reverse. Here this was denser this is the lighter. So it will not shift towards this side. So the interference pattern here is actually shifting up. So here you have central fringe, central bright fringe and this is the third bright fringes here. So fourth fringe will be naturally down below here and you can do it by longer method also and shorter method also. Shorter method is very short. Just see what is the fringe width here. Lambda is D lambda upon D and take always lambda dash when you have a medium mu and lambda dash will be lambda upon mu into D is the fringe width. So what is the fringe width? Let us see in the problem we are given lambda s.78 into 10 to power minus 6 and D upon D is 1000 and this mu is 1.3. Let us place the value 1000 upon 1.3 and this is I can write 7.8 10 to power minus 7. It becomes 6. So it is 6 into 10 to power minus 4 or 0.6 into 10 to power minus 3 is the fringe width. So if this is the third fringe, the fourth fringe will be just add the fringe width. So it is at distance of minus 0.6 into 10 to power minus 3. This is one way of doing it. Let me do it by a slightly longer method also. The longer method is just to understand the concepts I will use that method. We know that path difference here suppose I come to this place. This place this is why and I draw it like this. This path S1 and this point be suppose A dash S1 A dash minus S2 A dash without this gap. What is the part difference is D sin theta. If this angle is theta, this is D sin theta. The standard formula I am not deriving that D sin theta is this part difference without this slab I am not presuming this. So it is S1 A dash minus S2 A dash is nothing but and what is sin theta can be written as D up Y upon D. This angle theta is same as this angle. Sin theta is Y upon D. So this is the condition of difference between these two paths without this air being there and they are travelling through a denser medium. Now when I consider this, let the path difference now be written as S2 A double dash. For this I will write instead of S2 A dash S2 A double dash because I put this and this will be equal to the actual S2 A dash. Here you are subtracting because optical path is decreasing. So earlier when you place the glass slab, when air is there and you place the glass slab, you are always adding a path which is mu minus 1 into T as the optical additional optical path here. You will write as mu minus 1 into T as minus. This path is effectively getting reduced. So what is the actual path difference S2 A double dash minus S1 A dash is the actual path difference here and that is equal to D Y upon D minus mu minus 1 into T is the path difference. So I write it here. Let me use this space. So the path difference is D Y upon D minus mu minus 1 into T is equal to the path is the path difference. We know that the third fringe is coming at Y is equal to 0. So and this has to be N lambda for bright fringe. Y is equal to 0 here. We are told that at center Y is equal to 0. So Y is equal to 0. So you are getting minus mu minus 1 into T is equal to 3 lambda. How it can be minus? Actually here N has to be taken as minus 3 because you are coming down. This is the central fringe here and is actually minus 3. Otherwise you cannot get lambda as negative. So actually N is minus 3. So you get mu minus 1 into T is equal to 3 lambda. This is one condition we have got. Now we have to find out the fourth bright fringe. That is N is equal to minus 4. You are coming down. So 4 N is equal to minus 4. Just write D Y upon D minus mu minus 1 into T is the path difference and is equal to minus 4 lambda. Put the value here D Y upon D. What is this value? mu minus 1 into T is how much? 3 lambda minus 3 lambda is equal to minus 4 lambda. So D Y upon D is equal to minus lambda. So D upon D is how much? 1000. At lambda, actually it is lambda. Dash we have to lambda. We have to take by dividing the lambda which is given by mu. It is actually lambda dash. Please note it will be lambda dash everywhere because it is the wavelength divided by the refractive index. So when you solve it, you get Y is equal to 1000 into 7.8 into 10 to power minus 7. So you get minus 0.6 millimeter as the answer.