Question

Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

• A. $(\sqrt{I_1}-\sqrt{I_2})^2$
• B. $2(I_1+I_2)$
• C. $I_1+I_2$
• D. $(\sqrt{I_1}+\sqrt{I_2})^2$

Answer 1

The Correct Option is B

To solve this problem, we need to understand the principle of interference of waves, where the resultant intensity of two interfering waves is derived by the superposition principle. When two waves interfere, they create points where the intensity is either maximum (constructive interference) or minimum (destructive interference).

The given intensities of the two waves are \(I_1\) and \(I_2\). The relevant formulae for calculating the resultant intensities are as follows:

• The maximum intensity occurs when the waves are in phase (constructive interference): I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2.
• The minimum intensity occurs when the waves are out of phase (destructive interference): I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2.

The problem asks for the sum of the maximum and minimum intensities:

I_{\text{max}} + I_{\text{min}} = (\sqrt{I_1} + \sqrt{I_2})^2 + (\sqrt{I_1} - \sqrt{I_2})^2

Expanding both terms:

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

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

Adding these, we get:

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

This simplifies to:

I_{\text{max}} + I_{\text{min}} = 2I_1 + 2I_2 = 2(I_1 + I_2)

Therefore, the correct answer to the problem is 2(I_1 + I_2), which matches with the provided correct option.