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(C) The resultant intensity $I_R$ for two waves of intensity $I_0$ with a phase difference $\phi$ is given by the formula:$I_R = I_1 + I_2 + 2\sqrt{I_

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$\cos^2(\frac{\phi}{2})$

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(C) The resultant intensity $I_R$ for two waves of intensity $I_0$ with a phase difference $\phi$ is given by the formula:$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$Since $I_1 = I_2 = I_0$, we substitute these values:$I_R = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos \phi$$I_R = 2I_0 + 2I_0 \cos \phi = 2I_0(1 + \cos \phi)$Using the trigonometric identity $1 + \cos \phi = 2 \cos^2(\frac{\phi}{2})$:$I_R = 2I_0 \cdot 2 \cos^2(\frac{\phi}{2}) = 4I_0 \cos^2(\frac{\phi}{2})$Therefore, the resultant intensity $I_R$ is directly proportional to $\cos^2(\frac{\phi}{2})$.

Two waves of same intensity $I_0$ emitted from two sources having same phase difference $(\phi)$. Due to superposition of two waves, the intensity of resultant wave is directly proportional to . . . . . . .

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