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Question

In Young's double slit experiment, one of the slit is wider than another, so that amplitude of the light from one slit is 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)
Given, a1=2a2

I1=4I,I2=I

We know that, Im=(I1+I2)2

=(4I+I)2=9I

I=I1+I2+2I1I2cosϕ

=4I+I+24IIcosϕ

=4I+I+4Icosϕ

=5I+4Icosϕ

=I(5+4cosϕ)

=Im9[5+4(2cos2ϕ21]

=Im9[5+8cos2ϕ24]

=Im9[1+8cos2ϕ2].

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