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Question

Consider an YDSE that has different slit widths, as a result, amplitude of waves from two slits are A and 2A, respectively. If I0 be the maximum intensity of the interference pattern, then intensity of the pattern for phase difference ϕ is

A
I0cos2ϕ
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B
I03sin2ϕ2
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C
I09[5+4cosϕ]
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D
I09[5+8cosϕ]
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Solution

The correct option is C I09[5+4cosϕ]
As amplitude of the waves are A and 2A and,

We know that IA2

So intensities of the waves would be in the ratio 1:4, let us say I and 4I.

The intensity of the resultant wave is given by,

IR=I1+I2+2I1I2cosϕ

IRImax, when cosϕ=1,

Given that,

Imax=I0=I+4I+24I2=9I

I=I09

Intensity at a point where phase difference is ϕ any point,

I=I+4I+24I2cosϕ

I=5I+4Icosϕ=I09(5+4cosϕ)

Hence, option (C) is correct.

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