Two identical light waves having phase differ | vedclass

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(C) The resultant amplitude $A$ of two superposing waves with individual amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by: $A^2 = a_

Vedclass · Accepted Answer

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

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(C) The resultant amplitude $A$ of two superposing waves with individual amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by: $A^2 = a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi$.Since the waves are identical,$a_1 = a_2 = a$.Substituting this into the equation: $A^2 = a^2 + a^2 + 2 a^2 \cos \phi = 2 a^2 (1 + \cos \phi)$.Using the trigonometric identity $1 + \cos \phi = 2 \cos^2(\phi/2)$,we get: $A^2 = 2 a^2 (2 \cos^2(\phi/2)) = 4 a^2 \cos^2(\phi/2)$.Since intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$,we have $I \propto \cos^2(\phi/2)$.

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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