Briefly explain how bright and dark fringes are formed on the screen in Young’s double slit experiment. Hence, derive the expression for the fringe width.

The wave fronts from the two illuminated slits superimpose on the screen in Young's double slit experiment.
As a result of constructive and destructive interference, respectively, alternate bright and dark fringes
are formed. Central maxima refer to the region of the screen when light intensity is at its maximum.

Let S₁ and S₂ be two slits separated by a distance d. GG' is the screen at a distance D from the slits S₁ and S₂. Both the slits are equidistant from point C. The intensity of light will be maximum at this point due to the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits S1 and S2 is S2P−S1P.
Now, S₁S₂ = d, EF = d, and S₂F = D
$\begin{array}{l}\therefore \text{ In }{\mathrm{S}}_2\mathrm{P}={\left[{\mathrm{S}}_2{\mathrm{F}}_2+{\mathrm{PF}}_2\right]}^{1/2}\\ {\mathrm{S}}_2\mathrm{P}={\left[{\mathrm{D}}^2+{\left(\mathrm{x}+\frac{\mathrm{d}}{2}\right)}^2\right]}^{1/2}\\ =\mathrm{D}{\left[1+\frac{{\left(\mathrm{x}+\frac{\mathrm{d}}{2}\right)}^2}{{\mathrm{D}}^2}\right]}^{\frac{1}{2}}\end{array}$${\mathrm{S}}_2\mathrm{P}-{\mathrm{S}}_1\mathrm{P}=\frac{1}{2\mathrm{D}}\left[4\mathrm{x}\times \frac{\mathrm{d}}{2}\right]=\frac{\mathrm{xd}}{\mathrm{D}}$
Similarly, in ${\mathrm{\Delta S}}_1\mathrm{PE}$
$\begin{array}{l}{\mathrm{S}}_1\mathrm{P}=\mathrm{D}{\left[1+\frac{{\left(\mathrm{x}-\frac{\mathrm{d}}{2}\right)}^2}{{\mathrm{D}}^2}\right]}^{\frac{1}{2}}\\ {\mathrm{S}}_2\mathrm{P}-{\mathrm{S}}_1\mathrm{P}=\mathrm{D}{\left[1+\frac{{\left(\mathrm{x}+\frac{\mathrm{d}}{2}\right)}^2}{{\mathrm{D}}^2}\right]}^{\frac{1}{2}}-\mathrm{D}{\left[1+\frac{{\left(\mathrm{x}-\frac{\mathrm{d}}{2}\right)}^2}{{\mathrm{D}}^2}\right]}^{\frac{1}{2}}\end{array}$
On expanding it binomially,
For bright fringes (constructive interference), the path difference is an integral multiple of wavelengths, i.e., path difference is nλ.
$\therefore \mathrm{n\lambda }=\frac{\mathrm{xd}}{\mathrm{D}}$
$\mathrm{x}=\frac{\mathrm{n\lambda D}}{\mathrm{d}},$where n=0, 1, 2, 3, 4,...........
For n = 0,         x₀ = 0

$\begin{array}{l}\mathrm{n}=1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_1=\frac{\mathrm{\lambda D}}{\mathrm{d}}\\ \mathrm{n}=2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_2=\frac{2\mathrm{\lambda D}}{\mathrm{d}}\\ \mathrm{n}=3,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_3=\frac{3\mathrm{\lambda D}}{\mathrm{d}}\\ ⋮\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \mathrm{n}=\mathrm{n},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_{\mathrm{n}}=\frac{\mathrm{n\lambda D}}{\mathrm{d}}\end{array}$
The separation between the centres of two consecutive bright interference fringes is the width of a
dark fringe.
$\therefore {\mathrm{\beta }}_1={\mathrm{x}}_{\mathrm{n}}-{\mathrm{x}}_{\mathrm{n}-1}=\frac{\mathrm{\lambda D}}{\mathrm{d}}$
Similarly, for dark fringes, ${\mathrm{x}}_{\mathrm{n}}=(2\mathrm{n}-1)\frac{\mathrm{\lambda D}}{\mathrm{d}}$
For n = 1, ${\mathrm{x}}_1=\frac{\mathrm{\lambda D}}{2\mathrm{d}}$
For n = 2,  ${\mathrm{X}}_2=\frac{3\mathrm{\lambda D}}{2\mathrm{d}}$
The separation between the centres of two consecutive dark interference fringes is the width of a 
bright fringe.
$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\beta }}_2={\mathrm{x}}_{\mathrm{n}}-{\mathrm{x}}_{\mathrm{n}-1}=\frac{\mathrm{\lambda D}}{\mathrm{d}}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\beta }}_1={\mathrm{\beta }}_2\end{array}$
All the bright and dark fringes are of equal width