Diffraction

Question

At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

Difficult

Solution

The situation is shown in the figure.

In figure A and B represent the edges of the slit AB of width a and C represents the midpoint of the slit. For the first minimum at P,

For the first minimum at P,

$a\mathrm{sin}\theta =\lambda ....\left(i\right)$

$\text{where}\lambda \text{is the wavelength of light.}$

$\text{The path difference between the wavelets from}A\text{to}C\text{is}$

$\Delta x=\frac{a}{2}\mathrm{sin}\theta =\frac{1}{2}\left(a\mathrm{sin}\theta \right)$

$=\frac{\lambda}{2}\text{(using(i))}$

$\text{The corresponding phase difference}\Delta \varphi \text{is}$

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Download Now