A parallel beam of monochromatic light falls normally on a narrow slit of width 'a' to produce a diffraction pattern on the screen placed parallel to the plane of the slit. Use Huygens principle to explain that
(i) the central bright maxima is twice as wide as the other maxima.
(ii) the intensity falls as we move to successive maxima away from the centre on either side.

Dear Student,
Please find below the solution to the asked query:

The phenomenon of bending of light around the corners of an obstacle/aperture of the size of the wave length of light is called diffraction. The phenomenon resulting from the superposition of secondary wavelets originating from different parts of the same wave front is define as diffraction of light. Diffraction is the characteristic of all types of waves. Greater the wave length of wave higher will be it’s degree of diffraction.

If a parallel beam of monochromatic light falls on a narrow slit of slip width 'a', diffraction will be produced on the screen. Let the wavelength of the light incident is $\lambda$, distance between the screen and slit is D.

According to Huygens, every point on the slit acts like a secondary source of waves. So these waves produced by secondary source interact according to superposition principle and produces the diffraction pattern on the screen.

Diffraction pattern on the screen is as shown in the following figure,

Here, the central maxima will be having double angular width to that of the remaining fringes. The central maxima lies between the first minima on both sides.
For n^th secondary maxima at P on the screen.
Path difference $∆ = a \sin \theta = \left(2n+1\right)\frac{\lambda }{2}$; where n = 1, 2, 3 .....
(i) Angular position of n^th secondary maxima
      $\sin \theta \approx \theta \approx \frac{\left(2n+1\right)\lambda }{2b}$
(ii) Distance of n^th secondary maxima from central maxima
            $x_n=D.\theta =\frac{\left(2n+1\right)\lambda D}{2b}$
For central maxima,
$$\begin{gathered} Angular width = 2\theta = \frac{2\lambda }{b} \\ Linear Width = 2x = 2D\theta = \frac{2\lambda D}{b} \end{gathered}$$

Now let's discuss about Intensity,
If the intensity of the central maxima is I₀ then the intensity of the first and second secondary maxima are found to be $\frac{I_0}{22}$ and $\frac{I_0}{61}$. Thus diffraction fringes are of unequal width and unequal intensities. Intensity variation on the screen is as shown in the following figure,

The mathematical expression for in intensity distribution on the screen is given by,
 $I = I_0{\left(\frac{\sin \alpha }{\alpha }\right)}^2$
Where a is just a convenient connection between the angle q that locates a point on the viewing screening and light intensity I, f = Phase difference between the top and bottom ray from the slit width a.
Also $\alpha = \frac{1}{2}\varphi =\frac{\pi a}{\lambda } \sin \theta$
As the slit width increases (relative to wavelength) the width of the control diffraction maxima decreases; that is, the light undergoes less flaring by the slit. The secondary maxima also decreases in width (and becomes weaker).

 If $a>>\lambda$, the secondary maxima due to the slit disappear; we then no longer have single slit diffraction.

When the slit width is reduced by a factor of 2, the amplitude of the wave at the centre of the screen is reduced by a factor of 2, so the intensity at the centre is reduced by a factor of 4.

 

Hope this information will clear your doubts about the topic.

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