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| Young's Double Slit Experiment |
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| Let A and B be two fine slits, a small distance 'd' apart. Let them be illuminated by a monochromatic light of wavelength l. |
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| MN in the screen is at a distance D from the slits AB. The waves from A and B superimpose upon each other and an interference pattern is obtained on the screen. The point C is equidistant from A and B and therefore the path difference between the waves will be zero and so the point C is of maximum intensity. It is called the central maximum. |
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| For another point P at a distance 'x' from C, the path difference at P = BP - AP. |
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| Now AB = EF = d, AE = BF = D |
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| \D BPF |
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| [Pythagoras theorem] |
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| Similarly in D APE |
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| (on expanding Binomially) |
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| For bright fringes (constructive wavelength) the path difference is integral multiple of wavelength i.e., path difference is nl. |
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| (x therefore represents distance of nth bright fringe from C) |
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| Now |
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| and so on. |
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| Therefore separation between the centers of two consecutive bright fringe is the width of a dark fringe. |
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| Similarly for dark fringes, |
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| The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe. |
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| The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe. |
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| All bright and dark fringes are of equal width as b1 = b2. |
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| Note: The intensity of all bright bands are the same. All dark bands also have same (zero) intensity. The intensity distribution Vs distance is shown as: |
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