Interference fringes are produced on a screen by using two light sources of intensities I and 9 I . The phase difference between the beams is π 2 at the point P and π at the point Q on the screen. The

Question

Interference fringes are produced on a screen by using two light sources of intensities I and 9 I . The phase difference between the beams is π 2 at the point P and π at the point Q on the screen. The difference between the resultant intensities at point P and Q is

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At point P, resultant intensity of interfering wave is
${(I_{P})}_{r e s} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos θ$
For $θ = \frac{π}{2}, {(I_{P})}_{r e s} = I + 9 I = 10 I$
At point Q, resultant intensity of interfering wave is
${(I_{Q})}_{r e s} = 10 I - 2 . \sqrt{9 I^{2}} = 10 I - 6 I = 4 I$ for $θ = π$
$∴ ∆ I = {(I_{P})}_{r e s} - {(I_{g})}_{r e s} = 10 I - 10 I + 6 I = 6 I$

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