Let us continue with wave optics part 6. Consider two waves interfering at a point. We understand this by the principle of superposition of light waves. When two or more waves arrive at a point, simultaneously each wave produces its own displacement at that point, independent of the other. The result in this displacement at that point is equal to the vector sum of the displacement due to each wave at that instant. Why vector is equal to y1 vector plus y2 vector? If y1 and y2 are the amplitudes of each wave, either crest and crest can come together or a trough and a crest or a crest in a trough or a trough and a trough at that point. The amplitudes are as shown. So in the first one crest and crest gives you y1 plus y2, one trough and one crest gives you minus y1 plus y2, the crest and the trough gives plus y1 minus y2 and trough and the trough gives minus y1 minus y2. If y1 is equal to y2 is equal to y2, the amplitudes are same. These are all these are capital y. You get it as 2y, 0, 0 and minus 2y. The crest and the crest and the trough and the trough gives you what is known as the constructive interference and these zeros gives you the destructive interference. Constructive interference when two waves, two sources of light are in phase that is the part difference between them is an even integral multiple of lambda by 2. Then we get constructive interference. Part difference is equal to 2n, 2n implies even number, 2n lambda by 2. So 22 gets cancelled, you get any lambda, there is any equal to 0 plus minus 1 plus minus 2 plus minus 3 because they are on both the sides. Since phase difference is equal to 2 pi corresponds to the part difference lambda. Phase difference delta phi n is equal to n into 2 pi. See, we have replaced lambda by 2 pi, where n is equal to 0 plus minus 1 plus minus 2 and so on. The point is bright, destructive interference. When the two sources of light are in opposite phase and the part difference is odd integral multiple of lambda by 2, the part difference is equal to n minus half. See here it is odd number, n minus half times lambda, where n is equal to plus minus 1 plus minus 2 plus minus 3. Phase difference delta phi n is again n minus half 2 pi, where n is equal to plus minus 1 plus minus 2 plus minus 3 and so on. The point is dark. Young's double state experiment complies us of a source s which gives rise to s1 and s2 on a cardboard with two holes. This is how spherical wave runs which reaches the screen and gives you interference pattern. Thus the waveform is associated with it. Young, so this is the thing that is observed on the screen. Young's promise young was the first to demonstrate interference of light waves. A beam of light is allowed to pass through s. The light emerging from s is allowed to pass to two more pin holes s1 and s2 very close to each other. Thus the sources s1 and s2 are obtained on the primary source s. Hence the act as coherent sources and gives rise to spherical wave runs. This is the front view of the young's double state experiment. These are the interference frangers we call it, the alternate bright and the dark bands. Importance of young's experiment, it was the first experiment in which the interference of light was observed. It proves that light is propagated in the form of waves. It can be used to determine the wavelength of a monochromatic light. Young's experimental observation, interference frangers are produced in the region where the two waves overlap. If the source slits s1 and s2 are made narrow, the fringe width is unaffected by the intensity increases. If the distance between s1 and s2 is diminished, the bandwidth increases s is fixed. If the source slit s is widened, the fringe gradually disappear. The bandwidth s must be less than lambda d by d, where capital D is the distance of the screen from the two sources s1 and s2. If either s1 and s2 is closed with a finger, frangers will disappear. If the light is used, the central fringe is white and the fringe on either side is colored. Blue is the color near the central fringe and red is a wave. Let's treat the 9 interference bands analytically. So s1 and s2 are the two coherent sources separated by distance model, from which the wave starts and fades. Join s1p and s2p. These are the distance from the source to the screen. Why is the distance of p from the central, bright fringe? Part difference deltile is equal to s2p minus s1p. Now let's apply geometry to the double slit experiment. In the diagram, we apply Pythagoras theorem to the triangle s2p and s1ap. We get s2p square is equal to s2p square plus pp square. That is equal to capital D square plus y plus d by 2 the whole square. s1p square is equal to s1a square plus ap square. That is equal to capital D square plus y minus d by 2 the whole square. Subtract equation 2 from 1. s2p square minus s1p square is equal to capital D square plus y plus d by 2 the whole square minus capital D square minus y minus d by the 2 whole square. We get s2yd. So s2p minus s1p into s2p plus s1p we apply the function a square minus v square is equal to a plus v into a minus v that gives out to yd. s2p minus s1p is equal to yd upon s2p plus s1p. Since y is much less than capital D and the distance between the two states small day is much less than capital D, d by d is much less than 1. s2p and s1p are almost same so we can write it as 2d. Therefore s2p minus s1p is equal to 2yd upon 2d. s2p minus s1p is equal to 2yd by d. s2p minus s1p is a part difference between the two waves. For the end fringe the part difference delta l is equal to yn small d upon capital D. If point p is bright delta l is equal to yn d by d that is equal to n lambda where n is equal to 0 plus minus 1 plus minus 2 and so on. yn is equal to n lambda d by d. If the point p is dark then delta l is equal to yn d by d minus n. In mind is half lambda where n is equal to first minus 1 plus minus 2 and so on. yn is equal to n minus half lambda d by d. Now let us talk about the fringe width. The distance between the two consecutive bright or dark bands is called the fringe. For bright bands y is equal to yn plus 1. The consecutive yn plus 1 minus y that is equal to n plus 1 lambda d by d minus n lambda d by d. So that comes out to be w is equal to lambda d by small d. For dark bands w is equal to again yn plus 1 minus yn dark. That comes out to be n plus minus half lambda d by capital D minus n minus half lambda d by d. So w again comes out to be lambda d upon small d. From 5 and 6 the distance between two consecutive bright bands is equal to the distance between the two dark bands. Therefore the fringe width is equal to w lambda d by d whether it is bright or dark. The relation between the part difference delta n and the phase difference delta 5 is equal to delta 5 equal to 2 pi by lambda into delta n. Thus the phase difference between the two waves reaching p from s1 and s2 is delta 5 is equal to 2 pi by lambda yd by d. We substitute for delta l as yd by d. Now let us talk about the intensity distribution. i naught is directly proportional to e naught square, while i naught is the intensity of the incidence wave. The two waves coming from s1 and s2 mathematically follow the sign function with the amplitude e naught. Therefore, e1 is equal to e naught sin omega t and e2 is equal to e naught sin omega t plus 5. e is equal to e naught sin omega t plus e naught sin omega t plus 5. Well, that is equal to we use the arithmetic functions that gives you 2 e naught cos 5 by 2 into sin of omega t plus 5 by 2. Thus e is a sin wave having an amplitude 2 e naught cos 5 by 2. Intensity is proportional to the amplitude. The amplitude is 2 e naught cos 5 by 2 the whole square. So, intensity is proportional to 4 e naught square e square 5 by 2. That comes out to be 4 into i naught. We know that e naught and i naught have the relation as shown above that is equal to 4 i naught cos square 5 by 2. If e1 0 and e2 0 are the two amplitudes now of the two waves, then at the center of the bright fringe the intensity will be proportional to e1 0 plus e2 0 square. At the center of the top, fringe the intensity will be proportional to e1 0 minus e2 0.