Question

Two coherent monochromatic light beams of intensities I and 4I are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is x I. The value of x is______.

Answer 1

Correct Answer: 8

The problem involves the superposition of two coherent monochromatic light beams with intensities I and 4I. We need to find the difference between the maximum and minimum possible intensities in the resulting interference pattern and express it as a multiple of I.

Concept Used:

The solution is based on the principle of superposition for coherent light waves. The resultant intensity \(I_R\) at a point where two waves with intensities \(I_1\) and \(I_2\) and a phase difference \(\phi\) superimpose is given by:

\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi \]

The intensity is maximum (\(I_{\text{max}}\)) for constructive interference, which occurs when the phase difference \(\phi = 2n\pi\) (where \(n\) is an integer), so \(\cos\phi = 1\). The formula for maximum intensity is:

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

The intensity is minimum (\(I_{\text{min}}\)) for destructive interference, which occurs when the phase difference \(\phi = (2n+1)\pi\), so \(\cos\phi = -1\). The formula for minimum intensity is:

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

Step-by-Step Solution:

Step 1: Identify the intensities of the two light beams.

The intensities of the two coherent beams are given as:

\[ I_1 = I \] \[ I_2 = 4I \]

Step 2: Calculate the maximum possible intensity (\(I_{\text{max}}\)).

Using the formula for constructive interference:

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

Substituting the given values:

\[ I_{\text{max}} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 \] \[ I_{\text{max}} = (3\sqrt{I})^2 = 9I \]

Step 3: Calculate the minimum possible intensity (\(I_{\text{min}}\)).

Using the formula for destructive interference:

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

Substituting the given values:

\[ I_{\text{min}} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 \] \[ I_{\text{min}} = (-\sqrt{I})^2 = I \]

Step 4: Find the difference between the maximum and minimum intensities.

\[ I_{\text{max}} - I_{\text{min}} = 9I - I = 8I \]

We are given that this difference is equal to \(xI\). Comparing our result with the given expression:

\[ xI = 8I \]

This implies that \(x = 8\).

The value of x is 8.

Answer 2

The maximum intensity \( I_{\text{max}} \) in the superimposed beam is given by:
\[I_{\text{max}} = \left( \sqrt{I} + \sqrt{4I} \right)^2 = \left( \sqrt{I} + 2\sqrt{I} \right)^2 = (3\sqrt{I})^2 = 9I\]
The minimum intensity \( I_{\text{min}} \) is given by:
\[I_{\text{min}} = \left( \sqrt{4I} - \sqrt{I} \right)^2 = \left( 2\sqrt{I} - \sqrt{I} \right)^2 = (\sqrt{I})^2 = I\]
Therefore, the difference \( I_{\text{max}} - I_{\text{min}} \) is:
\[x = I_{\text{max}} - I_{\text{min}} = 9I - I = 8I\]