Two waves Y_1 = a sin omega t and Y_2 = a sin | vedclass

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(D) The resultant amplitude $R$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\delta$ is given by $R^2 = a_1^2 + a_2^2 + 2a_1 a_2

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$4a^2 \cos^2 \left( \frac{\delta}{2} \right)$

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(D) The resultant amplitude $R$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\delta$ is given by $R^2 = a_1^2 + a_2^2 + 2a_1 a_2 \cos \delta$.Here,$a_1 = a$ and $a_2 = a$.Substituting these values,we get $R^2 = a^2 + a^2 + 2(a)(a) \cos \delta$.$R^2 = 2a^2 + 2a^2 \cos \delta = 2a^2(1 + \cos \delta)$.Using the trigonometric identity $1 + \cos \delta = 2 \cos^2 \left( \frac{\delta}{2} ight)$,we get:$R^2 = 2a^2 	imes 2 \cos^2 \left( \frac{\delta}{2} ight) = 4a^2 \cos^2 \left( \frac{\delta}{2} ight)$.Since intensity $I \propto R^2$,the resultant intensity is proportional to $4a^2 \cos^2 \left( \frac{\delta}{2} ight)$.

Two waves $Y_1 = a \sin \omega t$ and $Y_2 = a \sin (\omega t + \delta)$ produce interference. The resultant intensity is ......

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