Answer to Question #102918 in Optics for proloy

Question

two light waves propagating in the same direction are superposed .the frequency of the two light waves is the same and there is a constant phase difference between them obtain an expression for the intensity of the resultant wave

Answer 1

As per the given question,

The the two waves are getting superimposed,

Waves which are getting superimpose, have the same frequency

And, the phase difference between the wave is also constant

So, angular frequency also will be same.

Let the angular velocity is $\omega$ , a and b is the amplitude of the wave and t is the time.

$\phi$ is the phase difference.

Let the waves are

$y_1=a\sin\omega t$

$y_2=b\sin(\omega t+\phi)$

So, Resultant amplitude (A) =$\sqrt{a^2+b^2+2ab\cos\phi}$

We know that intensity is proportional to the amplitude of the wave

$I\propto A^2$

$I=kA^2$

$I=\sqrt{a^2+b^2+2ab\cos\phi}$