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The maximum intensity in Young.s double slit experiment is I_0. Distance between slits is d= 5 lambda, where lambda is the wavelenght of the monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D= 10d. |
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Answer» SOLUTION :Path difference `trianglex = (yd)/(D)` Here, `y= (d)/(2)= (5 lambda)/(2)(" as "d= 5 lambda)" and "D= 10d= 50 lambda` So `trianglex = ((5lambda)/(2))((5lambda)/(50lambda))= (lambda)/(4)` CORRESPONDING PHASE difference `phi = (2lambda)/(lambda)XX (lambda)/(4)= (pi)/(2) I= I_(0) cos^(2) (lambda/4)= (I_0)/(2)`. |
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