Physics > Wave Optics > 3.0 Diffraction of light
Wave Optics
1.0 Introduction
1.1 Wavefronts
1.2 Huygens Principle
1.3 Interference of light
1.4 Intensity distribution
1.5 Phase difference
2.0 Young's double slit experiment
3.0 Diffraction of light
4.0 Polarisation
3.1 Diffraction due to a single slit
1.2 Huygens Principle
1.3 Interference of light
1.4 Intensity distribution
1.5 Phase difference
When the double slit in Young's experiment is replaced by a single narrow slit and it is illuminated by a monochromatic light, then the diffraction pattern is produced.
The diffraction pattern is produced. The diffraction pattern consists of a central maximum bright band with alternating bright and dark bands of decreasing intensity on both sides of the central maximum.
The essential condition for the diffraction of light is that the width of the slit should be comparable to the wavelength of the light. 
So,
$$a \approx \lambda $$
where,
$a$: Width of the slit
$\lambda $: Wavelength of light
In this experiment we use light having single wavelength which is commonly known as monochromatic light.
According to Huygen's principle, each portion of the slit acts as a source of light waves. Hence light from one portion of the slit can interfere with light from another portion and the resultant light intensity on a viewing screen depends on the direction $\theta $.
To analyse the diffraction pattern, let us divide the slit into two halves as shown in the figure.
For dark bands
Path difference between $BP$ & $MP$ and $AP$ & $MP$ is,
$$\Delta x = \frac{a}{2}\sin \theta \quad ...(i)$$
For destructive interference, phase difference $\phi $ should be equal to $180^\circ $ or $\pi $.
Relation between phase difference and path difference,
$$\phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x$$ For $\phi = \pi $, $$\begin{equation} \begin{aligned} \pi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x \\ \Delta x = \frac{\lambda }{2}\quad ...(ii) \\\end{aligned} \end{equation} $$
For destructive interference path difference should be equal to $\left( {\frac{\lambda }{2}} \right)$.
So, from equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \frac{a}{2}\sin \theta = \frac{\lambda }{2} \\ \sin \theta = \frac{\lambda }{a} \\\end{aligned} \end{equation} $$
When the slit is divided into 4 equal parts
Path difference between any two consecutive slits is given by,
$$\Delta x = \frac{a}{4}\sin \theta $$
For destructive interference path difference should be equal to $\left( {\frac{\lambda }{2}} \right)$.
So, $$\begin{equation} \begin{aligned} \frac{a}{4}\sin \theta = \frac{\lambda }{2} \\ \sin \theta = \frac{{2\lambda }}{a} \\\end{aligned} \end{equation} $$
When the slit is divided into 6 equal parts
Path difference between any two consecutive slits is given by,
$$\Delta x = \frac{a}{6}\sin \theta $$
For destructive interference path difference should be equal to $\left( {\frac{\lambda }{2}} \right)$.
So, $$\begin{equation} \begin{aligned} \frac{a}{6}\sin \theta = \frac{\lambda }{2} \\ \sin \theta = \frac{{3\lambda }}{a} \\\end{aligned} \end{equation} $$
Thus the general condition for destructive interference is, $$\sin \theta = \frac{{n\lambda }}{a}\quad {\text{where}}\quad n \in I$$
For bright bands
All bright bands are in between two dark band as shown in the figure.
