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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

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

B

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

C

(I1I2)2(\sqrt{I_1}-\sqrt{I_2})^2

D

2(I1I2)22(\sqrt{I_1}-\sqrt{I_2})^2

Step-by-Step Solution

When two periodic waves of intensities I1I_1 and I2I_2 superimpose, the resultant intensity II at a point is given by: I=I1+I2+2I1I2cosϕI = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi where ϕ\phi is the phase difference between the waves. For maximum intensity, cosϕ=1\cos \phi = 1: Imax=I1+I2+2I1I2=(I1+I2)2I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2 For minimum intensity, cosϕ=1\cos \phi = -1: Imin=I1+I22I1I2=(I1I2)2I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = (\sqrt{I_1} - \sqrt{I_2})^2 The sum of the maximum and minimum intensities is: Imax+Imin=(I1+I2+2I1I2)+(I1+I22I1I2)=2(I1+I2)I_{max} + I_{min} = (I_1 + I_2 + 2\sqrt{I_1 I_2}) + (I_1 + I_2 - 2\sqrt{I_1 I_2}) = 2(I_1 + I_2)

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