Question

A parallel beam of monochromatic light of wavelength $\lambda$ falls normally on a narrow slit of width $a$. The diffraction pattern is observed on a screen placed perpendicular to the direction of the incident beam. The phase difference between the waves coming from the edges of the slit at the position of the first maximum of the diffraction pattern is

• A. $3\pi$
• B. $\frac{3\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $3\pi$

The path difference between the waves emerging from the edges of the slit on reaching the position of the first maximum is,

$\Delta x=\frac{3\lambda }{2}$

Therefore, the phase difference is given by

$\Delta \theta =\frac{2\pi }{\lambda }\times \Delta x$

$=\frac{2\pi }{\lambda }\times \frac{3\lambda }{2}=3\pi$

Hence, option $\left(A\right)$ is correct.