Three harmonic waves having equal frequency $\mathrm{v}$ and same intensity $\mathrm{I}_{0}$, 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
JEE MAIN 2020, Medium
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Let amplitude of each wave is $A.$
Resultant wave equation
$=A \sin \omega t+A \sin \left(\omega t-\frac{\pi}{4}\right)+A \sin \left(\omega t+\frac{\pi}{4}\right)$
$=\mathrm{A} \sin \omega \mathrm{t}+\sqrt{2} \mathrm{A} \sin \omega \mathrm{t}$
$=(\sqrt{2}+1) \mathrm{A} \sin \omega \mathrm{t}$
Resultant wave amplitude $=(\sqrt{2}+1) \mathrm{A}$
as $I \propto A ^{2}$
so $\frac{\mathrm{I}}{\mathrm{I}_{0}}=(\sqrt{2}+1)^{2}$
$I=5.8 \mathrm{I}_{0}$
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