Question

Two coherent light waves, each having amplitude ‘a’, superpose to produce an interference pattern on a screen. The intensity of light as seen on the screen varies between:

• A. 0 and \( 2a^2 \)
• B. 0 and \( 4a^2 \)
• C. \( a^2 \) and \( 2a^2 \)
• D. \( 2a^2 \) and \( 4a^2 \)

Hint:

The intensity in an interference pattern depends on the phase difference between the waves. Constructive interference leads to maximum intensity, while destructive interference leads to minimum intensity.

Answer 1

The Correct Option is B

To address this problem, we must apply the principles of superposition and interference for coherent light waves. When two coherent light waves, each with amplitude 'a', interfere, the resultant intensity \( I \) is determined by the equation:

$I=I$₁+I₂+2√I₁I₂cosϕ

Here, \( I_1 \) and \( I_2 \) represent the intensities of the individual waves, and \( ϕ \) is the phase difference between them. The intensity of a wave is directly proportional to the square of its amplitude:

$I=a^2$

Given that \( I_1 = a^2 \) and \( I_2 = a^2 \) for each wave, substituting these values into the interference intensity equation yields:

$I=a^2+a^2+2aacos\varphi$

This equation simplifies to:

$I=2a^2+2a^{2}cos\varphi$

The intensity's value fluctuates based on \( cosϕ \), which can range from -1 to 1. Consequently, the minimum intensity \( I_{min} \) is achieved when \( cosϕ = -1 \):

$I$_min=2a2-2a2=0

The maximum intensity \( I_{max} \) occurs when \( cosϕ = 1 \):

$I$_max=2a2+2a2=4a2

Therefore, the light intensity spans a range from 0 to \( 4a^2 \). The conclusive answer is:

0 and \( 4a^2 \)