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 address the problem, an understanding of intensity variation in Young's double-slit experiment concerning path difference is required.

1. Young’s Double Slit Experiment Fundamentals:
   • The experiment produces an interference pattern characterized by alternating bright and dark fringes.
   • The intensity of a fringe at a specific point on the screen is contingent upon the path difference of light originating from the two slits.
   • Maximum intensity occurs when the path difference is an integer multiple of the wavelength \( \lambda \), indicating constructive interference.
2. Path Difference and its Influence on Intensity:
   • Provided path difference: \( \Delta x = \frac{7\lambda}{4} \).
   • Intensity is calculable using the standard interference intensity formula:
   • The phase difference, \(\phi\), is linked to the path difference by the equation: \(\phi = \frac{2\pi}{\lambda} \times \Delta x\)
   • Upon substitution of \( \Delta x \):
3. Intensity Ratio Calculation:
   • The intensity at the specified path difference is determined using:
   • The value of \( \cos\left(\frac{7\pi}{4}\right) \) is equivalent to \( \cos\left(\frac{\pi}{4}\right) \), which equals \( \frac{1}{\sqrt{2}} \).
   • Consequently, \( \cos^2\left(\frac{7\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \).
   • The ratio of this intensity to the maximum intensity is:
4. Conclusion:

The result of \( \frac{1}{2} \) is the correct option, aligning with the provided answer.