Question:
Three harmonic waves having equal frequency \(\nu\) and same intensity \(I_0\), have phase angles 0, \(\frac{\pi}{4}\) and \(-\frac{\pi}{4}\) respectively. When they are superimposed the intensity of the resultant wave is close to :
Three harmonic waves having equal frequency \(\nu\) and same intensity \(I_0\), have phase angles 0, \(\frac{\pi}{4}\) and \(-\frac{\pi}{4}\) respectively. When they are superimposed the intensity of the resultant wave is close to :
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Equal intensity means equal amplitude \(A\); add the three waves as phasors and use \(I\propto A^2\).
Updated On: Nov 5, 2026
- 5.8 \(I_0\)
- 3 \(I_0\)
- 0.2 \(I_0\)
- \(I_0\)
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The Correct Option is A
Solution and Explanation
Let each wave have amplitude \(A\), so \(I_0\propto A^2\).
Phasor sum along the reference direction: \(A\left[1+\cos\frac{\pi}{4}+\cos\left(-\frac{\pi}{4}\right)\right]=A(1+\sqrt{2})\).
Component perpendicular: \(A\left[\sin\frac{\pi}{4}+\sin\left(-\frac{\pi}{4}\right)\right]=0\).
Resultant amplitude \(A_R=A(1+\sqrt{2})\).
\(I=I_0(1+\sqrt{2})^2=I_0(3+2\sqrt{2})\approx 5.83\,I_0\).
Answer: (1) 5.8 \(I_0\)
Phasor sum along the reference direction: \(A\left[1+\cos\frac{\pi}{4}+\cos\left(-\frac{\pi}{4}\right)\right]=A(1+\sqrt{2})\).
Component perpendicular: \(A\left[\sin\frac{\pi}{4}+\sin\left(-\frac{\pi}{4}\right)\right]=0\).
Resultant amplitude \(A_R=A(1+\sqrt{2})\).
\(I=I_0(1+\sqrt{2})^2=I_0(3+2\sqrt{2})\approx 5.83\,I_0\).
Answer: (1) 5.8 \(I_0\)
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