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(C) Given that the intensity of both waves is the same,let $I_1 = I_2 = I_0$.The phase difference between the waves is $\Delta \phi = \pi/3$.The formu

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$3I_0$

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(C) Given that the intensity of both waves is the same,let $I_1 = I_2 = I_0$.The phase difference between the waves is $\Delta \phi = \pi/3$.The formula for the resultant intensity $I_R$ of two superposing waves is given by:$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$Substituting the given values into the formula:$I_R = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos(\pi/3)$Since $\cos(\pi/3) = 1/2$,we get:$I_R = 2I_0 + 2I_0(1/2)$$I_R = 2I_0 + I_0 = 3I_0$Therefore,the resultant intensity at point $P$ is $3I_0$.

Two light waves of same intensity superpose at point $P$ with a phase difference of $\pi/3$. The resultant intensity at point $P$ will be?

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