Q96 / 110 · VITEEE 2023 · Physics
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 the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is
- A0
- B\(\pi\)
- C\(\frac{\pi}{2}\)
- Correct answerD\(2 \pi\)
Solution
In a single slit diffraction pattern, the path difference \((\Delta \mathrm{x})\) for \(n^{\text {th }}\) minima is given by \(\Delta \mathrm{x}=n \lambda\), where \(n= \pm 1, \pm 2, \pm 3, \ldots\)
or phase difference, \(\Delta \phi=n(2 \pi)\)
For \(1^{\text {st }}\) minimum, \(n=1\).
\(\therefore \Delta \phi=2 \pi\)
Hence, the correct answer is option: 'D'.