Question 286: Wave Optics

Question

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

Accepted Answer

To determine the intensity of the resultant wave formed by two identical light waves with a phase difference $\phi$ propagating in the same direction, we start with the formula for the amplitude $A$ of the resultant wave:
$  A^2 = a_1^2 + a_2^2 + 2a_1a_2 \cos \phi  $
Given that the amplitudes of the individual waves are equal, $a_1 = a_2 = a$, we can substitute to find:
$  A^2 = 2a^2(1 + \cos \phi)  $
Using the trigonometric identity, $\cos \phi = 2 \cos^2\left(\frac{\phi}{2}\right) - 1$, the equation becomes:
$ A^2 = 2a^2 \left(1 + 2 \cos^2 \frac{\phi}{2} - 1\right) $
Simplifying this, we obtain:
$ A^2 = 2a^2 \cdot 2 \cos^2 \frac{\phi}{2} = 4a^2 \cos^2 \frac{\phi}{2} $
Thus,
$ A^2 \propto \cos^2 \frac{\phi}{2} $
Since intensity $I$ is proportional to the square of the amplitude ($I \propto A^2$), we conclude:
$ I \propto \cos^2 \frac{\phi}{2} $
This shows that the intensity of the resultant wave is proportional to $\cos^2 \frac{\phi}{2}$.

Answer

$\cos ^2\left(\frac{\phi}{4}\right)$

Answer

$\cos ^2\left(\frac{\phi}{3}\right)$

Answer

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

Answer

$\cos ^2 \phi$

Wave Optics

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

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