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Bichromatic light is used in YDSE having wavelengths, lamda_1 = 400nm and lamda_2 = 700nm. Find minimum order of lamda_1which overlaps with lamda_2 |
Answer» Solution : LET `n_1` bright FRINGE of `lamda_1` overlaps with `n_2` bright fringe of `lamda_2` Then `(n_1 lamda_1 D)/(d) = (n_2lamda_2 D)/(d)` `n_1/n_2 = lamda_2/lamda_1 = 700/400 = 7/4` The ratio `n_1/n_2 = 7/4`implies that 7th bright fringe of`lamda_1`will OVERLAP with 4th of `lamda_2` . SIMILARLY 14th of `lamda_1` will over lap with 8th of `lamda_2`and so on. So the minimum order of `lamda_1` which overlaps with `lamda_2` is 7. |
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