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The young's slit experiment is donein a medium of refractive index 4//3 .A light of 600 nm wavelength is falling on the slits having 0.45 mm separation . The lower slit S_(2) is covered by a thin galss sheet of thickness 10.4 mum and r3efractive index 1.5 .The interferecnce pattern is observed on a screen placed 1.5 m form the slits as shown in figure .Find the light intensity at point O relative to the maximum fringle intensity (Given : cos3.4 =sqrt(0.75) |
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Answer» `1/4` `Delta = (mu-1)t` `"Phase difference" = (32pi)/(LAMBDA)xxDelta=6.8 rad` `THEREFORE (I)/(I_(0)) = cos^(2) (PHI)/(2)=cos^(2) (6.8)/(2) = 0.75` |
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