Question

Consider fraunhofer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference (in radians) between the wavelets from the opposite edges of the slit is –

• A. p
• B. 2p
• C. p/4
• D. p/2

Answer 1

Answer: (B)

2p

Answer 2

For I^st minimum q = $\frac{\lambda}{a}$.

and f = $\frac{2\pi}{\lambda}$ . Dx = $\frac{2\pi}{\lambda}a\sin\theta$

f = $\frac{2\pi}{\lambda}$ . aq = $\frac{2\pi}{\lambda}a\frac{\lambda}{a}$

\f = 2p