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Monochromatic light of wavelength `lamda` passes through a very narrow slit S and then strikes a screen placed at a distance D = 1m in which there are two parallel slits `S_(1)` and `S_(2)` as shown. Slit `S_(1)` is directly in line with S while `S_(2)` is displaced a distance d to one side. The light is detected at point P on a second screen, equidistant from `S_(1)` and `S_(2)` When either slit `S_(1)` or `S_(2)` is open equal light intensities `I` are measured at point P. When both slits are open, the intensity is three times as large i.e. `3 I`. If the minimum possible wavelength is `Nd^(2)`, then find the value of `N (d lt lt D)` |
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Answer» Correct Answer - 3 `2I = I+I+2I cos phi` `phi =pi//3 rArr Delta =(lamda)/(6)` Also `Delta = sqrt(D^(2) +d^(2))-D` `=Dsqrt(1+(d^(2))/(D^(2)))-D=D(1+(d^(2))/(D^(2)))^((1)/(2))-D` `=D(1+(1)/(2)(d^(2))/(D^(2)))-D=(1)/(2)(d^(2))/(D)` `:. (1)/(2)(d^(2))/(D)=(lamda)/(6)` or `lamda=3 (d^(2))/(D)`. |
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