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A beam of light considering of two wavelengths 6500Å and 5200Å is used to obtain interference fringes in a Young's double slit experiment. :) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 6500Å (ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide ? Distance a between the slits is 2 mm, distance between the slits and the screen D= 120 cm. |
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Answer» Solution : (i) The distance of the `m^(th)` BRIGHT fringe from the central MAXIMUM. `y_m = (m lamda L)/(d), y_3 = (3 lamda L)/(d) = (3 xx (6500 xx 10^(-10)) xx 1.20)/(2 xx 10^(-3))` = 1.17mm ii) Let the nth bright fringe of wavelength `lamda_n`and `m^th`bright fringe of wavelength a coincide at a distance `y_m`from the central maximum then `y = (m lamda_m L)/(d) THEREFORE m/n = (lamda_n)/(lamda_m) = 6500/5200 = 5/4` i.e., 5th bright fringe of wavelength 5200Å COINCIDES with the 4TH bright fringe of wavelength 6500Å ` therefore y_m = (m lamda_m L)/(d) = (5(5200 xx 10^(-10)) xx 1.20)/(2 xx 10^(-3)) = 1.56mm` |
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