Question

In Young’s double slit experiment, light from two identical sources is superimposing on a screen. The path difference between the two lights reaching a point on the screen is \( \frac{7\lambda}{4} \). The ratio of the intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{4} \)

Answer 1

The Correct Option is A

To solve the problem, we need to understand how intensity varies in Young's double slit experiment where there is a path difference. 

1. The Basics of Young’s Double Slit Experiment:
   • The interference pattern consists of bright and dark fringes.
   • The intensity of the fringe observed at a point on the screen depends on the path difference between the light from the two slits.
   • The intensity is maximum when the path difference is a multiple of the wavelength \( \lambda \) (constructive interference).
2. Path Difference and Its Impact on Intensity:
   • Given: The path difference \( \Delta x = \frac{7\lambda}{4} \).
   • The intensity can be calculated using the formula for intensity in interference:
   • Here, \(\phi\) is the phase difference and is related to the path difference by the formula: \(\phi = \frac{2\pi}{\lambda} \times \Delta x\)
   • Substituting for \(\Delta x\), we get:
3. Calculation of Intensity Ratio:
   • To find intensity at given path difference, we use:
   • We know, \(\cos\left(\frac{7\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)
   • Thus, \(\cos^2\left(\frac{7\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\)
   • The ratio of this intensity with respect to the maximum intensity is:
4. Conclusion:

The correct option is \(\frac{1}{2}\), which matches the given correct answer.

Answer 2

Given:  
- Path difference \( \Delta x = \frac{7\lambda}{4} \).

Step 1. Calculate the phase difference \( \phi \):
  
  \(\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \times \frac{7\lambda}{4} = \frac{7\pi}{2}\)

Step 2. Determine the intensity at this point:
  The intensity \( I \) at a point with phase difference \( \phi \) is given by:  
 
  \(I = I_{\text{max}} \cos^2 \left(\frac{\phi}{2}\right)\)
Step 3. Calculate \( \frac{I}{I_{\text{max}}} \):
  
  \(\frac{I}{I_{\text{max}}} = \cos^2 \left(\frac{\phi}{2}\right) = \cos^2 \left(\frac{7\pi}{4}\right)\)
  
 Using \( \cos \frac{7\pi}{4} = \cos \left(2\pi - \frac{\pi}{4}\right) = \cos \frac{\pi}{4} \), we get:  

  \(\frac{I}{I_{\text{max}}} = \cos^2 \frac{\pi}{4} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\)
 

Thus, the ratio of intensity at this point to the maximum intensity is .\( \frac{1}{2} \)

The Correct Answer is: \( \frac{1}{2} \)