Question

Find the intensity of light at a point on the screen when two interfering waves of the same intensity \((I_0)\) have a path difference of
(i) \(\frac{\lambda}{4}\)
(ii) \(\frac{\lambda}{3}\)

Hint:

For two coherent waves of equal intensity \(I_0\), \[ I=4I_0\cos^2\left(\frac{\phi}{2}\right) \] and \[ \phi=\frac{2\pi}{\lambda}\Delta x \] Always convert the given path difference into phase difference first, and then substitute into the intensity formula.

Answer 1

Concept: When two coherent light waves interfere, the resultant intensity depends upon the phase difference between the waves. The general expression for the resultant intensity is \[ I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi \] where
• \(I_1\) and \(I_2\) are the intensities of the two interfering waves,
• \(\phi\) is the phase difference between them. Since both waves have equal intensity \(I_0\), \[ I_1 = I_2 = I_0 \] Therefore, \[ I = I_0 + I_0 + 2\sqrt{I_0I_0}\cos\phi \] \[ I = 2I_0(1+\cos\phi) \] Using the trigonometric identity \[ 1+\cos\phi = 2\cos^2\left(\frac{\phi}{2}\right) \] we obtain \[ I = 4I_0\cos^2\left(\frac{\phi}{2}\right) \] Also, \[ \phi=\frac{2\pi}{\lambda}\Delta x \] where \(\Delta x\) is the path difference. Case (i): Path Difference \(=\dfrac{\lambda}{4}\)

Step 1: Calculate the phase difference. \[ \phi=\frac{2\pi}{\lambda}\left(\frac{\lambda}{4}\right) \] \[ \phi=\frac{\pi}{2} \]

Step 2: Substitute into the intensity formula. \[ I=2I_0(1+\cos\phi) \] \[ I=2I_0\left(1+\cos\frac{\pi}{2}\right) \] Since \[ \cos\frac{\pi}{2}=0 \] we get \[ I=2I_0(1+0) \] \[ I=2I_0 \] Result for case (i): \[ \boxed{I=2I_0} \] Case (ii): Path Difference \(=\dfrac{\lambda}{3}\)

Step 1: Calculate the phase difference. \[ \phi=\frac{2\pi}{\lambda}\left(\frac{\lambda}{3}\right) \] \[ \phi=\frac{2\pi}{3} \]

Step 2: Substitute into the intensity formula. \[ I=2I_0(1+\cos\phi) \] \[ I=2I_0\left(1+\cos\frac{2\pi}{3}\right) \] Since \[ \cos\frac{2\pi}{3}=-\frac12 \] we obtain \[ I=2I_0\left(1-\frac12\right) \] \[ I=2I_0\left(\frac12\right) \] \[ I=I_0 \] Result for case (ii): \[ \boxed{I=I_0} \] Final Answers: For path difference \[ \Delta x=\frac{\lambda}{4} \] \[ \boxed{I=2I_0} \] For path difference \[ \Delta x=\frac{\lambda}{3} \] \[ \boxed{I=I_0} \]