Two identical light waves having phase differ | vedclass

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Question

(B) Let the intensity of each individual wave be $I$. When two identical waves with phase difference $\phi$ superpose,the resultant intensity $I_R$ is

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the intensity of each individual wave be $I$. When two identical waves with phase difference $\phi$ superpose,the resultant intensity $I_R$ is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$ Since the waves are identical,$I_1 = I_2 = I$. Substituting these values: $I_R = I + I + 2\sqrt{I \cdot I} \cos \phi$ $I_R = 2I + 2I \cos \phi$ $I_R = 2I(1 + \cos \phi)$ Using the trigonometric identity $1 + \cos \phi = 2 \cos^2 \frac{\phi}{2}$: $I_R = 2I(2 \cos^2 \frac{\phi}{2}) = 4I \cos^2 \frac{\phi}{2}$ Therefore,the resultant intensity is proportional to $\cos^2 \frac{\phi}{2}$.

Two identical light waves having phase difference $\phi$ propagate in the same direction. When they superpose,the intensity of the resultant wave is proportional to

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