Question

Two monochromatic light beams have intensities in the ratio 1:9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

• A. 8 : 1
• B. 9 : 1
• C. 3 : 1
• D. 4 : 1

Hint:

In an interference pattern formed by two sources of intensities \( I_1 \) and \( I_2 \), the maximum intensity is \( (\sqrt{I_1} + \sqrt{I_2})^2 \) and the minimum intensity is \( (\sqrt{I_1} - \sqrt{I_2})^2 \). Use the given ratio of intensities to express one intensity in terms of the other and then calculate the ratio of maximum to minimum intensity.

Answer 1

The Correct Option is D

Let the intensities of the two monochromatic light beams be \( I_1 \) and \( I_2 \). 
Given that the ratio of their intensities is 1:9, we can write \( \frac{I_1}{I_2} = \frac{1}{9} \). 
Let \( I_1 = I \) and \( I_2 = 9I \). The maximum intensity \( I_{max} \) in the interference pattern is given by: \[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \] Substituting the values of \( I_1 \) and \( I_2 \): \[ I_{max} = (\sqrt{I} + \sqrt{9I})^2 = (\sqrt{I} + 3\sqrt{I})^2 = (4\sqrt{I})^2 = 16I \] The minimum intensity \( I_{min} \) in the interference pattern is given by: \[ I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \] Substituting the values of \( I_1 \) and \( I_2 \): \[ I_{min} = (\sqrt{I} - \sqrt{9I})^2 = (\sqrt{I} - 3\sqrt{I})^2 = (-2\sqrt{I})^2 = 4I \] The ratio of the maximum to the minimum intensity is: \[ \frac{I_{max}}{I_{min}} = \frac{16I}{4I} = 4 \] So, the ratio of the intensities of maximum to minimum is 4:1.

Answer 2

To determine the ratio of the intensities of maximum to minimum in an interference pattern created by two monochromatic light beams, we need to apply the formula for the intensity in an interference pattern.

The formula for the intensity at any point in an interference pattern created by two interfering waves with intensities \(I_1\) and \(I_2\) is given by:

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

where \(\phi\) is the phase difference between the two waves.

The intensity at the maximum of the interference pattern (bright fringes) occurs when \(\phi = 0\) (constructive interference), so the intensity is:

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

The intensity at the minimum of the interference pattern (dark fringes) occurs when \(\phi = \pi\) (destructive interference), so the intensity is:

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

Given that the ratio of the intensities of the two beams is \(1:9\), we can let \(I_1 = I\) and \(I_2 = 9I\).

Substituting these into the formulas for \(I_{\text{max}}\) and \(I_{\text{min}}\), we get:

\(I_{\text{max}} = I + 9I + 2\sqrt{I \cdot 9I} = 10I + 6I = 16I\)

\(I_{\text{min}} = I + 9I - 2\sqrt{I \cdot 9I} = 10I - 6I = 4I\)

Therefore, the ratio of the intensities of maximum to minimum is:

\(\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{16I}{4I} = 4:1\)

Thus, the correct answer is 4:1.