MHT CET · Physics · Wave Optics
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 maximum of the diffraction pattern, the phase difference between the rays coming from the edges of the slit is
- A \(\frac{\pi}{4}\)
- B \(\pi\)
- C \(\frac{\pi}{2}\)
- D \(3 \pi\)
Answer & Solution
Correct Answer
(D) \(3 \pi\)
Step-by-step Solution
Detailed explanation
For nth secondary maxima, the path difference \(x\) is the odd multiple of the half of the wavelength, i.e.,
\(x_n=(2 n+1) \frac{\lambda}{2}\)
So, for the first maxima, the phase difference would be:
\(\phi_1=\frac{2 \pi}{\lambda} x_1\)
\(\therefore \phi_1=\frac{2 \pi}{\lambda}\left(\frac{3 \lambda}{2}\right)=3 \pi\)
\(x_n=(2 n+1) \frac{\lambda}{2}\)
So, for the first maxima, the phase difference would be:
\(\phi_1=\frac{2 \pi}{\lambda} x_1\)
\(\therefore \phi_1=\frac{2 \pi}{\lambda}\left(\frac{3 \lambda}{2}\right)=3 \pi\)
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