Two identical light waves, propagating in the | vedclass

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Question

(C) Let the intensity of each individual wave be $I_0$. When two waves with the same frequency and amplitude superpose, the resultant intensity $I$ is

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$\cos ^2(\delta / 2)$

Vedclass · Answer

(C) Let the intensity of each individual wave be $I_0$. When two waves with the same frequency and amplitude superpose, the resultant intensity $I$ is given by the formula: $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \delta$ Since the waves are identical, $I_1 = I_2 = I_0$. Substituting these values: $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \delta$ $I = 2I_0 + 2I_0 \cos \delta$ $I = 2I_0 (1 + \cos \delta)$ Using the trigonometric identity $1 + \cos \delta = 2 \cos^2(\delta / 2)$: $I = 2I_0 (2 \cos^2(\delta / 2))$ $I = 4I_0 \cos^2(\delta / 2)$ Therefore, the resultant intensity is proportional to $\cos^2(\delta / 2)$.

Two identical light waves, propagating in the same direction, have a phase difference $\delta$. After they superpose, the intensity of the resulting wave will be proportional to

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