Wave Optics - Result Question 32
34. A parallel beam of light of wavelength $\lambda$ 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 second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is
(a) $\pi \lambda$
(b) $2 \pi$
(c) $3 \pi$
(d) $4 \pi$
[NEET Kar. 2013]
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Answer:
Correct Answer: 34. (d)
Solution:
- (d) Conditions for diffraction minima are
Path diff. $\Delta x=n \lambda$ and Phase diff. $\delta \phi=2 n \pi$
Path diff. $=n \lambda=2 \lambda$
Phase diff. $=2 n \pi=4 \pi(\because n=2)$
35 .
(d) $\delta \phi=1.22 \frac{\lambda}{D}=1.22 \frac{5000 \times 10^{-10}}{10 \times 10^{-2}}=6.1 \times 10^{-6}$ $\therefore$ Order $=10^{-6}$
