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In Young.s double slit experiment intensity at a point is (1"/"4) of the maximum intensity. Angular position of this points is |
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Answer» Solution :`I= I_("MAX") COS^(2)(PHI/2) therefore (I_("max"))/(4)= I_("max") cos^(2)(phi/2)` `cos (phi)/(2)= (1)/(2) " or "(phi)/(2)= (pi)/(3)""therefore phi= (2pi)/(3)= ((2pi)/(LAMBDA))trianglex` where `trianglex = d SIN theta` `(lambda)/(3)= d sin theta, sin theta= (lambda)/(3d), theta= sin^(-1)((lambda)/(3d))`. |
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