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The intensity at a point where the path difference is (lambda)/(6)(lambda being the wavelength of the light used) is I. If I_(0) denotes the maximum intensity, (I)/(I_(0)) is equal to : |
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Answer» `(sqrt3)/(2)` `I_(0) =a^(2) + a^(2) + 2a.a cos theta = 4a^(2)` ....(1) When ` x = (lambda)/(6)` `phi = (2pi)/(lambda).x = (2pi)/(lambda)(lambda)/(6) = (pi)/(3) = 60^(@)` Then `I = a^(2) + a^(2) + 2a.acos 60^(@) = 3A^(2)` `THEREFORE (I)/(I_(0)) = (3a^(2))/(4a^(2)) = 3/4` |
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