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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.
944138_a633981164d541a6b85f0482eb9ccc74.png

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 I08
Without P;
A=A+A11

A1=A1+A2=A0sin(kxwt)+A0sin(kxwt+ϕ)

A11=A(1)11+A(2)11

A11=A011[sin(kxwt)+sin(kxwt+ϕ)

where A0,A011

are the amplitudes of either of the beam in and 11 polarizations.

Intensity ={|A0|2+|A011|2}(sin2(kxwt)

(1+ cos2ϕ+2sinϕ)+sin2(kxwt)sin2ϕ)average

={|A0|2+|A011|2}(12).2(1+cosϕ)

=2|A0|2.(1+cosϕ)since|A0|average=|A011|average

With P:
Assume A2 is blocked:

Intensity =(A111+A111)2+(A1)2

=|A0|2(1+cosϕ)+|A0|2.12

Given: I0=4|A0bot|2

= Intensity wothout polariser at principal maxima.

intensity at principal maxima with polariser

|A0|2(2+12)

=58I0

Intensity at first minima with polariser

=|A0|2(11)+|A01|22

=I08.

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