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In Young's double slit experiment, using monochromatic light of wavelength lamda, the intensity of light at a point on the screen where path difference is lamda is K units. The intensity of light at a point where path difference is lamda/3 is |
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Answer» 4K `I = 4I_(0) cos^(2)(PHI)/2` where `I_(0)` is the intensity of either wave and `phi` is the phase difference between two waves. Phase difference, `phi = (2pi)/(lamda) xx` Path difference When path difference is `lamda` then `phi = (2pi)/(lamda) xx lamda = 2pi` `:. I = 4I_(0) cos^(2) ((2pi)/2) = 4I_(0) cos^(2) (pi) = 4I_(0) = K` ...(1) When path difference is `(lamda)/3` , then `phi = (2pi)/(lamda) xx (lamda)/3 = (2pi)/3` `:. I = 4I_(0) cos^(2) ((2pi)/(3)) = K/4`[Using (i)] |
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