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Two identical beam A and B of plane coherent waves of the sasme intensity and wavelength lambda fall on a plane screen. The direction of thebeam propagations make angles theta_(1) and theta_(2) with the normal to the screen and lie in the same plane as shown in the figure. Find the distance beta between adject interfrence fringes on the screen. |
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Answer» `(lambda)/(sin theta-sin theta_(2))`  Path difference between A 'P' and B'P' is `DELTA X =y sin theta_(1) - y sin theta_(2)` `Delta x = y(sin theta_(1) - sin theta_(2))` For bright fringes `Delta x =N lambda` `n lambda = y (sin theta_(1) - sin theta_(2))` `y = (n lambda)/(sin theta_(1) - sin theta_(2))` `:. beta = (lambda)/(sin theta_(1)-sin theta_(2))` |
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