Briefly explain how bright and dark fringes are formed on a screen due to the diffraction at a single slit. Hence, explain why the intensity at the…

Dividing the slit into smaller parts, and adding their contributions at P with the proper phase differences. The path difference NP - LP between the two edges of the slit, $\displaystyle \mathrm{NP}-\mathrm{LP}=\mathrm{NQ}=a \sin \theta \approx a \theta$ At the central point $\displaystyle C$ on the screen, the angle $\displaystyle \theta$ is zero. All path differences are zero and hence all the parts of the slit contribute in phase. This gives maximum intensity at C, Similarly other secondary maxima will be formed at $\displaystyle \theta=(n+1 / 2) \lambda / a$, For minima (zero intensity) the contributions from $\displaystyle \mathrm{M}_{1}$ and $\displaystyle \mathrm{M}_{2}$ are $\displaystyle 180$° out of phase and cancel in the direction $\displaystyle \theta=\lambda / a$. where $\displaystyle \lambda$ is the path difference. Similarly other minima are formed due to cancellation of contributions. At an angle $\displaystyle \theta=n \lambda / a$, where $\displaystyle n= \pm 1, \pm 2, \pm 3, \ldots$. Consider an angle $\displaystyle \theta=3 \lambda / 2 a$ which is midway between two of the dark fringes. Divide the slit into three equal parts. If we take the first two thirds of the slit, the path difference between the two ends would be $\displaystyle \lambda$. The first two-thirds of the slit can be divided into two halves which have a $\displaystyle \lambda / 2$ path difference. The contributions of these two halves cancel, only the remaining one-third of the slit contributes to the intensity at a point between the two minima. Which will be much weaker than the central maximum (where the entire slit contributes in phase). Similarly there are maxima at ( $\displaystyle n +1 / 2) \lambda / a$ with $\displaystyle n=2,3$, etc. These become weaker with increasing $\displaystyle n$, since only one-fifth, one-seventh, etc., of the slit contributes in these cases.