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In a two-slit interference experiment the two slits have intensities…
Question
In a two-slit interference experiment the two slits have intensities in the ratio 4:1. At a point where the path difference equals one-third of a wavelength, the resultant intensity is what fraction of the maximum intensity? Give the numerical value of this fraction.
✓ Verified answer: 0.333333checked by our engine — not a guess
Step-by-step solution
Resultant intensity: I = I1 + I2 + 2*sqrt(I1 I2) cos(phi).
Path difference lambda/3 gives phase phi = (2pi/lambda)*(lambda/3) = 2pi/3, cos(2pi/3) = -1/2.
With I1=4, I2=1: I = 4 + 1 + 2*sqrt(4)*(-1/2) = 5 - 2 = 3.
Maximum intensity I_max = (sqrt I1 + sqrt I2)^2 = (2+1)^2 = 9.
Fraction = 3/9 = 1/3 = 0.333.
Final answer0.333
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