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 I∝A2 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+2√I1I2cosϕ IR→Imax, when cosϕ=1, Given that, Imax=I0=I+4I+2√4I2=9I ⇒I=I09 Intensity at a point where phase difference is ϕ any point, I′=I+4I+2√4I2cosϕ I′=5I+4Icosϕ=I09(5+4cosϕ) Hence, option (C) is correct.

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