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)`

Answer

3

Explanation

`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)`.