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 solve this problem, we need to consider the principle of superposition and interference of coherent light waves. Two coherent light waves of amplitude ‘a’ are interfering with each other. The intensity \( I \) resulting from interference of waves is given by the equation:

$I=I$₁+I₂+2√I₁I₂cosϕ

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

$I=a^2$

For each of the waves, we have \( I_1 = a^2 \) and \( I_2 = a^2 \). Substituting these into the interference intensity equation gives:

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

which simplifies to:

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

The intensity varies depending on the value of \( cosϕ \), which ranges between -1 and 1. Thus, the minimum intensity \( I_{min} \) occurs when \( cosϕ = -1 \):

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

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

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

Therefore, the intensity of light varies between 0 and \( 4a^2 \). The correct answer is:

0 and \( 4a^2 \)