Describe diffraction of light at a single slit. Explain the formation of pattern of fringes on screen. Also, use the variation of intensity with diffraction angle $\theta$ to explain why the intensity of secondary maxima decreases with the order of maxima.
Let a diverging light from a monochromatic source S be made parallel after refraction through convex lens $L_1$. The refracted light forms a plane wavefront WW'. This plane wavefront WW' is incident on the slit AB of width $d$. According to Huygens' principle, each point of slit AB acts as a source of secondary disturbance or wavelets.
Convex lens $L_2$ helps in converging the parallel beam of light. Now consider a point O equidistant from points A and B on the screen which is placed at a distance D from the slit AB. The secondary wavelets from A and B reach the point O in the same phase covering the same distance so constructive interference takes place at O. In other words, O is the position of central maximum of diffraction pattern.
Now let the light be diffracted through an angle $\theta$, so the secondary wavelets will also be diffracted through an angle $\theta$. Let these wavelets meet the screen at point P after passing through the lens $L_2$. The point P will be of maximum or minimum intensity depending on the path difference between the secondary wavelets (reaching point P) originating from the corresponding points of the incident wavefront.
Path difference: To find the path difference between the secondary wavelets originating from corresponding points A and B of the plane wavefront, draw AN perpendicular on B$\theta$'. The path difference between these wavelets originating from A and B is BN.
From $\triangle$ABN,
$ \sin \theta = \frac{BN}{AB} $
$ BN = AB \sin \theta $
But BN is path difference and AB is the width $d$ of the slit
Path difference = $d \sin \theta $
(a) For minima
If path difference BN $= d \sin \theta$ is equal to one wavelength i.e., $\lambda$. Now the path difference between the secondary wavelets originating from A and C is equal to $\lambda/2$. So these wavelets will meet point P out of phase i.e., phase difference = $\pi$) and hence destructive interference will take place at P. Similarly, the path difference between the wavelets originating from C and B is $\lambda/2$ and hence these will also produce destructive interference at P. Thus, position P is of minimum intensity.
Hence, for first minimum
$ d \sin \theta_1 = \lambda \quad \dots (1) $
$ \sin \theta_1 = \frac{\lambda}{d} $
$\because \theta_1$ is very small
$ \theta_1 = \frac{\lambda}{d} \quad \dots (2)$
Similarly, if BN $= 2\lambda$, then the slit AB can be imagined to be divided into four equal halves. Then, the path difference between the secondary wavelets originating from the corresponding points of each half $= (2\lambda)/4 = \lambda/2$. Thus, these wavelets produce destructive interference and point P is of minimum intensity.
Thus, for second minimum
$ d \sin \theta_2 = 2\lambda $
$ \sin \theta_2 = \frac{2\lambda}{d}$
$ \theta_2 = \frac{2\lambda}{d} $
In general , for mth minima
$ d \sin \theta_m = m\lambda$
$ \sin \theta_m = \frac{m\lambda}{d} $
In general, for $m^{th}$ minima
$ \sin \theta_m = \frac{m\lambda}{d} $$
Since $\theta_m$ is small, so $\sin \theta_m \approx \theta_m$
$ \theta_m = \frac{m\lambda}{d} \quad \dots (3) $
where, $\theta_m$ is the angle giving the direction of the $m^{th}$ order minimum and $m = 1, 2, 3, \dots$ an integer.
(b) For secondary maxima :
If path difference, BN $= d \sin \theta$ is an odd multiple of $\lambda/2$, then the constructive interference takes place at P. Hence, point P is the position of secondary maxima.
$ d \sin \theta_m = (m + \frac{1}{2})\lambda$
$\sin \theta_m = (m + \frac{1}{2})\frac{\lambda}{d} $
Since $\theta_m$ is small, so $\sin \theta_m \approx \theta_m$
$ \theta_m = (m + \frac{1}{2})\frac{\lambda}{d} \quad \dots (4) $
where, $\theta_m$ is the angle giving the direction of the $m^{th}$ order secondary maximum and $m = 1, 2, 3, \dots$ an integer.
Diffraction pattern due to a single slit consists of a central maximum flanked by alternate minima and secondary maxima.} The variation of intensity of secondary maxima and minima with the angle of diffraction is shown in figure