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: Basic principle of interference. 
When two coherent waves overlap, the resultant intensity depends on their phase difference due to superposition of amplitudes. 
Step 2: Use the intensity formula. 
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi \] 
Step 3: Substitute equal intensities. 
Given \( I_1 = I_2 = I \) 
\[ I_R = I + I + 2\sqrt{I \cdot I}\cos\phi \] 
\[ I_R = 2I + 2I\cos\phi \] 
\[ I_R = 2I(1 + \cos\phi) \] Using identity: \[ 1 + \cos\phi = 2\cos^2\left(\frac{\phi}{2}\right) \] 
\[ I_R = 4I \cos^2\left(\frac{\phi}{2}\right) \] 
Step 4: Condition for maxima. 
For constructive interference: \[ \phi = 2n\pi \] 
\[ \cos\left(\frac{\phi}{2}\right) = \pm 1 \] 
\[ I_{max} = 4I \] 
Step 5: Condition for minima. 
For destructive interference: \[ \phi = (2n+1)\pi \] 
\[ \cos\left(\frac{\phi}{2}\right) = 0 \] 
\[ I_{min} = 0 \] 
Final Answer: 
Maximum intensity \( = 4I \) 
Minimum intensity \( = 0 \)