Question

Two waves, each of amplitude \( a \) and frequency \( \omega \) emanating from two coherent sources of light superpose at a point. If the phase difference between the two waves is \( \phi \), obtain an expression for the resultant intensity at that point.

Hint:

The intensity of two superimposed waves depends on the square of the cosine of half the phase difference.

Answer 1

Let the equations of the two waves be: \[ x_1 = a \cos(\omega t), \] \[ x_2 = a \cos(\omega t + \phi), \] where \( a \) is the amplitude, \( \omega \) is the frequency, and \( \phi \) is the phase difference. The resultant displacement \( x \) is: \[ x = x_1 + x_2 = a \cos(\omega t) + a \cos(\omega t + \phi) = a ( \cos(\omega t) + \cos(\omega t + \phi) ). \] Using the trigonometric identity: \[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right), \] we get: \[ x = 2a \cos\left(\frac{\phi}{2}\right) \cos\left(\omega t + \frac{\phi}{2}\right). \] The intensity \( I \) is proportional to the square of the amplitude: \[ I = K (\text{Amplitude})^2 = K \left(2a \cos\left(\frac{\phi}{2}\right)\right)^2 = 4K a^2 \cos^2\left(\frac{\phi}{2}\right). \] Let \( I_0 = Ka^2 \) be the intensity of each incident wave. Thus, the resultant intensity is: \[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right). \]