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NEET PHYSICSEasy

Two periodic waves of intensities I1I_1 and I2I_2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:

A

I1+I2I_1+I_2

B

(I1+I2)2(\sqrt{I_1}+\sqrt{I_2})^2

C

(I1−I2)2(\sqrt{I_1}-\sqrt{I_2})^2

D

2(I1+I2)2(I_1+I_2)

Step-by-Step Solution

  1. Identify the formulas for maximum and minimum intensity: When two waves of intensities I1I_1 and I2I_2 interfere, the resultant intensity is given by I=I1+I2+2I1I2cos⁡(Δϕ)I = I_1 + I_2 + 2\sqrt{I_1I_2} \cos(\Delta\phi) . The maximum intensity (constructive interference) is Imax⁡=(I1+I2)2=I1+I2+2I1I2I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = I_1 + I_2 + 2\sqrt{I_1I_2} . The minimum intensity (destructive interference) is Imin⁡=(I1−I2)2=I1+I2−2I1I2I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = I_1 + I_2 - 2\sqrt{I_1I_2} .
  2. Calculate the sum: Add the expressions for maximum and minimum intensities. Imax⁡+Imin⁡=(I1+I2+2I1I2)+(I1+I2−2I1I2)I_{\max} + I_{\min} = (I_1 + I_2 + 2\sqrt{I_1I_2}) + (I_1 + I_2 - 2\sqrt{I_1I_2}) Imax⁡+Imin⁡=2I1+2I2=2(I1+I2)I_{\max} + I_{\min} = 2I_1 + 2I_2 = 2(I_1 + I_2)
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