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A_n YDSE is carried out in a liquid of refractive index mu= 1.3 and thin film of air is formed in from of the lower slit as shown in the figure. In the above problem find the distance of fourth maxima from 0. |
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Answer» Solution :`beta= (lambda D)/( MU d)= (0.78)/(1.3)xx 10^(-6)xx 10^(3)= 0.6 MM` Shift in the fringe pattern will be upwards The refrective INDEX of the AIR film is less than the medium Find the maxima `implies : 0.6mm" below "0` `therefore -0.6mm` `therefore 7xx 0.6mm= 4.2mm` above 0 The FOURTH maxima are `4.2mm` above 0 and `0.6mm` below 0. |
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