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Q.In Young's double slit experiment two coherent sources of intensity ratio 64 : 1 produce interference fringes. Calculate the ratio of maximum and minimum intensities.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2019Subjective· 3mImportance★★★★★
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Converting the given intensity ratio to an amplitude ratio and applying the standard two-source interference formula gives Imax:Imin=81:49I_{max}:I_{min} = 81:49.

In Young's double slit experiment, if two coherent sources have amplitudes a1a_1 and a2a_2 (with a1>a2a_1 > a_2), the resultant intensity at a point of path difference producing phase difference ϕ\phi is

I=a12+a22+2a1a2cos⁡ϕI = a_1^2 + a_2^2 + 2a_1a_2\cos\phi

which is maximum when ϕ=0\phi = 0 (constructive interference) and minimum when ϕ=π\phi=\pi (destructive interference):

Imax=(a1+a2)2,Imin=(a1−a2)2I_{max} = (a_1+a_2)^2, \qquad I_{min} = (a_1-a_2)^2

Since intensity is proportional to the square of amplitude (I∝a2I \propto a^2), the given intensity ratio I1:I2=64:1I_1 : I_2 = 64:1 corresponds to an amplitude ratio …

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