Question

Two light beams of intensities in the ratio of 9 : 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be:

• A. 2 : 3
• B. 16 : 81
• C. 25 : 169
• D. 25 : 1

Answer 1

The Correct Option is D

To find the ratio of the intensity of maxima and minima for two interfering light beams, we need to understand the concept of interference of light and how intensities are calculated for constructive and destructive interference.

Concept Explanation

When two coherent light waves interfere, the resulting intensity at any point depends on the phase difference between the waves. The intensity at any point is given by:

I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi)

where,

• I_1 and I_2 are the intensities of the two interfering waves.
• \phi is the phase difference between the waves.

Calculation of Maxima and Minima

Maxima: Occurs when the phase difference \phi is an even multiple of \pi (i.e., they are in phase), leading to constructive interference.

The intensity of maxima is given by:

I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2}

Minima: Occurs when the phase difference \phi is an odd multiple of \pi (i.e., they are out of phase), leading to destructive interference.

The intensity of minima is given by:

I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2}

Given Data and Solution

We are given that the intensities of the light beams are in the ratio 9:4. Let's denote:

• I_1 = 9k
• I_2 = 4k

Substituting in the formula for maxima:

I_{\text{max}} = 9k + 4k + 2\sqrt{9k \cdot 4k} = 13k + 12k = 25k

Substituting in the formula for minima:

I_{\text{min}} = 9k + 4k - 2\sqrt{9k \cdot 4k} = 13k - 12k = k

Ratio of Maxima to Minima

Therefore, the ratio of the intensity of maxima to minima is:

\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{25k}{k} = 25:1

Thus, the correct answer is 25:1.