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Long Answer Questions · Q14

Q.Obtain the equation for resultant intensity due to interference of light.

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Step 1. Two waves of the same frequency, amplitudes a1,a2a_1,a_2, arrive at a point with phase difference ϕ\phi: y1=a1sin⁡ωty_1=a_1\sin\omega t, y2=a2sin⁡(ωt+ϕ)y_2=a_2\sin(\omega t+\phi).

Step 2. By the principle of superposition, the resultant displacement is y=y1+y2y=y_1+y_2; expanding y2y_2 using the angle-addition identity and collecting terms in sin⁡ωt\sin\omega t and cos⁡ωt\cos\omega t shows the sum can be written as y=Asin⁡(ωt+θ)y=A\sin(\omega t+\theta), a single sinusoid of the same frequency, with resultant amplitude AA given by A2=a12+a22+2a1a2cos⁡ϕA^2=a_1^2+a_2^2+2a_1a_2\cos\phi (from expanding and using cos⁡2+sin⁡2=1\cos^2+\sin^2=1).

Step 3. Since intensity is proportional to the square of amplitude, I∝A2I\propto A^2, so I∝a12+a22+2a1a2cos⁡ϕI\propto a_1^2+a_2^2+2a_1a_2\cos\phi; writing I1∝a12I_1\propto a_1^2 and I2∝a22I_2\propto a_2^2 (so a1a2∝I1I2a_1a_2\propto\sqrt{I_1I_2}), this becomes I∝I1+I2+2I1I2cos⁡ϕ\boxed{I\propto I_1+I_2+2\sqrt{I_1I_2}\cos\phi}. …

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