Question

In Young's double slit interference experiment, using two coherent sources of different amplitudes, the intensity ratio between bright to dark fringes is \(5 : 1\). The value of the ratio of resultant amplitudes of bright fringe to dark fringe is

• A. \((\frac{\sqrt{5}+1}{\sqrt{5}-1})\)
• B. \(\sqrt{5} : 1\)
• C. \((\frac{\sqrt{5}-1}{\sqrt{5}+1})\)
• D. \(1 : \sqrt{5}\)

Hint:

Intensity \(\propto\) amplitude\(^2\); use sum and difference of amplitudes.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Intensity of light in interference is proportional to the square of the resultant amplitude (\(I \propto A_{\text{res}}^2\)).
Bright fringes correspond to maximum intensity and maximum amplitude (constructive interference).
Dark fringes correspond to minimum intensity and minimum amplitude (destructive interference).
Step 2: Key Formula or Approach:
Given: \(\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{5}{1}\).
Relation between intensity and amplitude: \(\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{A_{\text{max}}}{A_{\text{min}}} \right)^2\).
Step 3: Detailed Explanation:
The question asks for the ratio of resultant amplitudes of bright fringe (\(A_{\text{max}}\)) to dark fringe (\(A_{\text{min}}\)).
From the intensity relation: \[ \frac{I_{\text{bright}}}{I_{\text{dark}}} = \frac{I_{\text{max}}}{I_{\text{min}}} = 5 \] Taking the square root on both sides: \[ \sqrt{\frac{I_{\text{max}}}{I_{\text{min}}}} = \sqrt{5} \] Since \(\sqrt{\frac{I_{\text{max}}}{I_{\text{min}}}} = \frac{A_{\text{max}}}{A_{\text{min}}}\): \[ \frac{A_{\text{max}}}{A_{\text{min}}} = \frac{\sqrt{5}}{1} \] Thus, the ratio is \(\sqrt{5} : 1\).
Step 4: Final Answer:
The ratio of the resultant amplitudes of the bright fringe to the dark fringe is \(\sqrt{5} : 1\).