Question

A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of incident beam. At the first maximum of the diffraction pattern, the phase difference between the rays coming from the edges of the slit is

• A. \( \frac{\pi}{4} \, \text{rad} \)
• B. \( 2\pi \, \text{rad} \)
• C. \( \frac{\pi}{2} \, \text{rad} \)
• D. \( \pi \, \text{rad} \)

Hint:

For the first maximum in diffraction, the phase difference is \( 2\pi \) radians, corresponding to a path difference of one wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding diffraction.
For diffraction at the first maximum, the path difference between the rays coming from the edges of the slit is \( \lambda \). The phase difference corresponding to a path difference of \( \lambda \) is \( 2\pi \) radians.
Step 2: Conclusion.
Thus, the phase difference between the rays coming from the edges of the slit is \( 2\pi \) radians, corresponding to option (B).