Question

When a parallel beam of monochromatic source of light of wavelength is incident on a single slit of a width a, show how the diffraction pattern is formed at the screen by the interference of the wavelets from the slit. Show that, besides the central maximum at $\theta = 0$ secondary maxima are observed at $\theta = \left( {n + \dfrac{1}{2}} \right)\dfrac{\lambda }{a}$ and the minima at $\theta = \dfrac{{n\lambda }}{a}$ .Why do secondary maxima get weaker in intensity $n$ ? Explain.

Answer 1

We start by analysing the given data and applying the formulas wherever necessary. We draw appropriate figures to make it easier to access and manage the given data and formulas. Then we find the relation between the maxima and the first minima. That is, the distance between the two first positive and negative minima will be the width of the central maxima.

Formulas used:
The path difference for minima is given by the formula,
$\Delta x = \dfrac{\lambda }{2}$
The formula for finding the distance,
$y = \dfrac{{\lambda D}}{a}$
Formula for constructive interference,
$\theta = \left( {n + \dfrac{1}{2}} \right)\dfrac{\lambda }{a}$
Formula for destructive interference,
$\theta = \dfrac{{n\lambda }}{a}$

Complete step by step answer:


The diffraction design framed can be perceived by adding the commitment from different wavelets of the incident wavefront. At the central point, the slit is partitioned into two equal parts. The commitment of wavelets from those two parts are on stage. So, maxima is obtained at the centre.

We know that, for minima $\Delta x = \dfrac{\lambda }{2}$. We substitute the value of path difference from the figure,
$\dfrac{{a\sin \theta }}{2} = \dfrac{\lambda }{2}$
Theta is the angle between the line joining the first minima and the line joining the slit and the central maxima.Here, $\sin \theta = \tan \theta$ and the value of this will be $\dfrac{y}{D}$. The second minima will be formed at $y = \dfrac{{\lambda D}}{a}$ below the central minima, so the width will be twice as much as this value. That is, $y = \dfrac{{2\lambda D}}{a}$ will be the width of the central maxima.

Note: In a single slit test, monochromatic light is gone through one slit of limited width and a comparative pattern is seen on the screen. Dissimilar to the double slit diffraction design, the width and intensity in single slit diffraction design decrease as we move away from the central maxima.