Question

The displacement of two light waves, each of amplitude 'a' and frequency \(\omega\), emanating from two coherent sources of light, are given by \(y_1 = a \cos(\omega t)\) and \(y_2 = a \cos(\omega t + \phi)\). \(\phi\) is the phase difference between the two waves. These light waves superpose at a point. Obtain the expression for the resultant intensity at that point.

Hint:

Remember that the maximum intensity occurs when \(\phi = 0\) (in phase) and the minimum intensity (possibly zero) occurs when \(\phi = \pi\) (out of phase), demonstrating constructive and destructive interference respectively.

Answer 1

Step 1: Formulate the resultant displacement. When two waves superpose, the net displacement \(y\) equals the sum of individual displacements: \[y = y_1 + y_2 = a \cos(\omega t) + a \cos(\omega t + \phi)\] Applying the trigonometric identity for the sum of cosines: \[y = 2a \cos\left(\frac{\phi}{2}\right) \cos\left(\omega t + \frac{\phi}{2}\right)\] Step 2: Calculate the resultant intensity. Intensity is directly proportional to the square of the wave's amplitude. The amplitude of the resultant wave is \(2a \cos\left(\frac{\phi}{2}\right)\), therefore, the intensity \(I\) is expressed as: \[I = k \left(2a \cos\left(\frac{\phi}{2}\right)\right)^2 = 4ka^2 \cos^2\left(\frac{\phi}{2}\right)\] with \(k\) representing the constant of proportionality.