InterviewSolution
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InterviewSolution
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A coherent beam light of intensity I and absolute wavelength lambda= 5000Å is being incident on the slits making an angle 30^@ with horizontal. If screen is placed at a distance D=1 m from the slits and the separation between- the slits is 3 xx 10^(-4) m. A thin film of glass mu_(g)= 3/2 and thickness 0.41 mm is placed near one slit as shown in the figure, find (a) the position of central maxima. (b) the intensity at'point O. |
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Answer» SOLUTION :Let central amxima lies at a point P at a distance x from the central line. Then optical path diffenrece `Delta`p at a point P is: `(S_(2)P - t) + tmu_(g) - mu_(w)dsintheta - S_(1)p` `implies S_(2)P - S_(1)P + (mu_(g) -1) t - mu_(w)dsintheta` `implies (xd)/D + (mu_(g) -1 )t - mu_(w)dsintheta` For central maxima `Deltap=0` `x=D/d[(mu_(w)dsintheta) - (mu_(g)-1)t]` `1/(3xx10^(-4)) [4/3 xx3 xx10^(-4) xx 1/2 -0.5 xx 0.41 xx10^(-3)]` `= (5xx10^(-6))/(3xx10^(-4)) = -1.66xx10^(-2)m=-1.66m` So central maxima lies at a distance 1.66 cm below the central line (b) At POINTO , optical path difference is `(mu_(g)-1)t - mu_(w)dsintheta=-5xx10^(-6)m` So intensity at 0, `I_(0) =I + I + 2sqrt(II) "cos" (2pi)/(5xx10^(-7))(-5xx10^(-6))=4I` 
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