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

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: \( \frac{1}{2} \)