Question

In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.

Hint:

Intensity is proportional to the square of the amplitude (\(I \propto A^2\)). At maxima, amplitudes add (\(A+A=2A\)), so intensity becomes \((2A)^2 = 4A^2 = 4I\).

Answer 1

Step 1: Understanding the Concept:
Interference results from the superposition of wave amplitudes. The resultant intensity at any point depends on the phase difference between the two waves.

Step 2: Key Formula or Approach:
Resultant Intensity \(I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi\).

Step 3: Detailed Explanation:
Given \(I_1 = I_2 = I\). Substituting these into the formula: \[ I_R = I + I + 2\sqrt{I \cdot I} \cos \phi = 2I + 2I \cos \phi = 2I(1 + \cos \phi) \] Using the identity \(1 + \cos \phi = 2 \cos^2(\phi/2)\): \[ I_R = 4I \cos^2(\phi/2) \] 1. For Maxima (Constructive Interference): Phase difference \(\phi = 2n\pi\), so \(\cos(\phi/2) = \pm 1\). \[ I_{max} = 4I(1)^2 = 4I \] 2. For Minima (Destructive Interference): Phase difference \(\phi = (2n+1)\pi\), so \(\cos(\phi/2) = 0\). \[ I_{min} = 4I(0)^2 = 0 \]

Step 4: Final Answer:
The intensity at maxima is 4I and at minima is 0, as proved.