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Question

# As shown in figure, a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given.If I0 is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

A
I08
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B
34I0
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C
I016
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D
25I0
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Solution

## The correct option is B I08Without P;A=A⊥+A11A1=A1⊥+A2⊥=A0⊥sin(kx−wt)+A0⊥sin(kx−wt+ϕ)A11=A(1)11+A(2)11A11=A011[sin(kx−wt)+sin(kx−wt+ϕ)where A0⊥,A011are the amplitudes of either of the beam in ⊥ and 11 polarizations.∴ Intensity ={|A0⊥|2+|A011|2}(sin2(kx−wt) (1+ cos2ϕ+2sinϕ)+sin2(kx−wt)sin2ϕ)average ={|A0⊥|2+|A011|2}(12).2(1+cosϕ)=2|A0⊥|2.(1+cosϕ)since|A0⊥|average=|A011|averageWith P:Assume A2⊥ is blocked:Intensity =(A111+A111)2+(A1⊥)2=|A0⊥|2(1+cosϕ)+|A0⊥|2.12Given: I0=4|A0bot|2= Intensity wothout polariser at principal maxima.intensity at principal maxima with polariser|A0⊥|2(2+12)=58I0Intensity at first minima with polariser=|A0⊥|2(1−1)+|A01|22=I08.

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