Fringe width in Young's double slit experiment

we've already explored young's double-slit experiment in which we take a source and keep it behind two holes and that produces this nice pattern of alternating dark and bright spots and that happens due to constructive and destructive interference in this video we want to figure out an expression for the distance between two consecutive bright spots we want to figure out what is the expression for this distance why well one is because this is in our syllabus but more importantly we will see by figuring out this expression how young thomas young was able to find the wavelength of right that's right by using a piece of cardboard candle and a screen he was able to figure out the wavelength of light and we'll understand exactly how we did that so let's begin to start with this distance is often called beta and the name is fringe width that's the name we give to it fringe width and the name was confusing for me because i thought we were calculating the width of something but we are not see over here what we find is that the brightness is maximum then it becomes smaller and smaller then becomes minimum then the brightness increases and becomes maximum and then minimum and maximum minimum so fringe width is the distance between two consecutive brightest spots where the brightnesses are the maximum or distance between two consecutive darkest spots where the brightness is zero so it's not the width of anything it's the distance between two consecutive brightest spots or two consecutive darkest spots so how do we figure this out well there are a couple of ways to do that the way i like to do it is this way see when these arrays interfere over here these are the two rays of light interfering over here i know that they're undergoing constructive interference over here right all i need to figure out is what is this angle this angle if i know if i can figure out what this angle is then i can figure out this distance by using geometry see i can i know what this distance is experimentally i can the distance between the screen and the slits we can just call that as capital d and we also know what this distance is distant between the slits we can call that as small d so if i figure out what this angle is at which this interference constructive interference is happening then from geometry i can figure out what this distance is going to be so the next question is how do i figure out this angle the angle at which constructive interference is happening well for that we'll go back and ask ourselves what is the condition for constructive interference and this is something we have seen before in our previous videos you've seen for constructive interference the path difference between the two interfering waves that should equal an integral multiple of the wavelength lambda what is this equation saying well over here you can kind of see that this one is traveling slightly more distance compared to this wave right because this is slightly farther away from this source or this slit and so that extra distance is called the path difference if that equals an integral multiple of a wavelength then we'll always find constructive interference now if you're wondering why this is true and where this comes from we've talked about this in credit in our previous video so feel free to go back and check that out so when m equals zero we get the path difference to be zero and that happens right at the center so we get the central bright spot when m equals one we get the first constructive interference here or here when m equals two we get the second now since we are interested in uh the first constructive interference we'll put m equals one so we're writing first construction that's what i want and so we can say for us m equals one so over here i know the path difference is exactly equal to one lambda but from that how do i figure out what this angle is going to be well the path difference does depend on the angle think about it if the angle was zero then we are going to get zero part difference they're gonna travel exactly the same path but as the angle increases the lower wave has to travel more distance compared to the top wave to catch up and meet with it so clearly as the angle increases the path difference will increase so there is a relationship between them but what exactly is that how do we figure that relationship out and this is also something we've done before what we did just to recap is we zoomed in over here let me show you the zoomed in version of that and now to calculate the extra path traveled by this wave compared to this wave we said let's drop a perpendicular from here to here why why did we do that well we can now say hey the distance from here all the way to the screen and the distance from here all the way to the screen is the same because if you think about it if you drop a perpendicular here this becomes a giant isosceles triangle almost with these two sides 90 degrees these two angle 90 degrees and this angle is almost zero you have to imagine it that way and that means if these traditions are the same this part this is the extra distance traveled and from this geometry we can figure out what that is because i know what this length is this is d i know this is d and if this angle is theta you can find the angle inside this triangle and you can use trigonometry so great idea to pause and see if you can recollect what we did in the previous video or just use this geometry to figure out what this distance is going to be can you pause and quickly try and figure that out all right let me zoom in so we can see better all right so if this angle is theta the first thing we see is this angle should also be theta why because this whole angle is 90 right so if this is theta this should be 90 minus theta this is 90 so this should be theta as well and now i know this is the hypotenuse which is d and this is the opposite side so to figure out the opposite side i can use sine and if you do that you will now get the opposite side to be d sine theta and that is our path difference let me zoom out and so we can say the path difference is d sine theta and so for constructive interference for the first construction that should equal lambda and again if you need more clarity feel free to check out back our previous videos where we've discussed this in more detail but now comes the question from here how do i figure out what theta is well remember theta is incredibly small so when theta is small and it usually happens to be a few degrees one or two degrees then we can do another assumption and we can simplify this and from that we can figure out what theta is going to be and then we can use a little bit of geometry to figure out what beta is going to be so i want you to pause the video at this moment and see if you can do the rest of this yourself it's important to try it's okay if you go wrong but please try this first yourself all right now because theta is small the assumption we can do is sine theta equals theta for small angle signifies the same value as theta when you're dealing with radians so we can just say this is d times theta that equals lambda and from this we can immediately get what theta is going to be theta is going to be lambda divided by d so we found the angle that is subtended by these two bright spots over here now we found out the angle but how do i calculate this distance can again use trigonometry or you can again because the angle is very small we can use the arc angle formula we can say that hey this distance is almost the same as this distance and so this length is very small so you can assume this to be an arc and so for the arc angle formula i know s equals r theta oops sorry what is that okay s equals r theta so our arc is the beta over here beta the radius is the capital d times theta and therefore if i substitute i get beta to be alpha substitute theta that'll be lambda divided by d times capital d and if you had used trigonometry you would have gotten the same answer you could have used trigonometry and say i can use tan theta and then tan theta becomes the same thing as theta because the angle is small you get exactly the same result you can try that but anyways we have found the expression for fringeworth now before we talk about how thomas young calculated the wavelength one question i have is we found the angle between these two bright spots so the distance between these two bright spots would the answer be the same for distance between these two bright spots we get the same answer well it looks like that but can we prove it well yeah let's quickly try and do it again feel free to see if you can prove it yourself but now we'll consider the interference between uh for the second constructive interference let me use same yellow second constructive interference here and to calculate the angle the distance between these two bright spots i have to again calculate what this angle is going to be and again why don't you think about that you can do it very quickly over here all right so for the second constructive interference my m is going to be 2 so i'll get this to be 2 lambda so this will be 2 and this angle this total angle is going to be 2 lambda by d but since this angle is lambda by d total angle is 2 lambda by d that means this angle is also lambda by d so what we see is this angle is exactly the same as this angle and therefore this distance will turn out to be exactly the same as this distance again because we are assuming theta to be very small of course if i go very far away very high then eventually theta becomes big enough and then i can't do any of these assumptions and then it will fail but as long as the theta angles are very small we can use this assumption and so distance between any two consecutive bright spots or dark spots and you can prove this yourself you take any two dark spots also you will find the distance to be the same lambda d by small d so now let's come to the fun part see thomas young also did this calculation and he knew that beta is going to be lambda d by small d now in his experiment he experimentally found out using a ruler and maybe a microscope because this will be very small what was the distance between two consecutive bright spots so he experimentally calculated this and in his experiment he knew what the value of capital d and the value of small d was and so by plugging in he was able to figure out the wavelength of light which i feel is mind-blowing because you know if someone told me that you can use a candle light a cardboard a screen and a microscope to find the wavelength of light i wouldn't be able to believe that and that's why this is one of the most important experiments in history okay finally let's look at what this expression is telling us it's telling us that if the lambda is high if the wavelength of light is larger then beta will be larger so if you use red light which has a larger wavelength you would expect the beta to be larger the the bright spots will be farther apart if the distance increases if the distance between the source and the slit sorry slit and the screen increases then beta becomes larger which kind of makes sense right the angle does not depend on that but as you go further and further away you can see the they are diverging away and as a result you can see the bright spots go farther and farther away finally what we see is you tell me what do you think will happen if i bring the slits closer to each other do the bright spots will also come closer or do you think the spots will go farther away what do you think well from the equation you can see if the slits come closer to each other the beta value increases because in the denominator so the bright spots will end up going farther and farther from each other and we will be able to see a better interference pattern and this is important so to actually see a nice interference pattern you have to have the two slits close to each other if the slits are very far away if the distance is very large if let's say a few centimeters or meters that beta value will be so small you can hardly notice any interference and that's why if you have say two windows through which sunlight is coming in the windows are too far the distance is too far to see any interference on your wall