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Question

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit double of that from other slit. If Im be the maximum intensity, the resultant intensity I. When they interfere at phase difference ϕ is given by

A
Im9(4+5cosϕ)
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B
Im3(1+2cos2ϕ2)
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C
Im5(1+4cos2ϕ2)
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D
Im9(1+8cos2ϕ2)
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Solution

The correct option is D Im9(1+8cos2ϕ2)
I=I0+4I0+2I0×4I0cosϕ
I=I0+4I0+4I0cosϕ
Im will be at cosϕ=1,
So, Im=9I0
IIm=5I0+4I0cosϕ9
IIm=1+8cos2ϕ29

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