Question

Use Huygen's principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band?

Answer 1

Consider a parallel beam of monochromatic light is incident normally in a slit of width b as shown in the figure. Consider a particular point P on the screen receives waves from all the secondary sources. All these aves start from different places and give the resultant intensity at the point $P$. At $P_0$ all waves are travelling same optical path. As the all waves are in phase thus the interference of maximum intensity is observed.

The intensity at the point P is given by $I=I_0\frac{si{n}^2\alpha }{\alpha }$, where $\alpha =\frac{\pi bsin\theta }{\lambda }$

For central maxima $\alpha =0$ thus $I=I_0$

Width of central maxima is given by $\beta =\frac{2D\lambda }{b}$

D= Distance between the screen and slit; $\lambda =$ wavelength of the light and $b=$ size of the slit.

So with the increase in size of the slit the width of the central maxima decrease. Hence double the size of the slits would results in half the width of the central maxima.