Question

Why the fringes are of equal width?

Answer 1

Fringe width

While performing Young's double-slit experiment we get a pattern of Bright and dark patterns, The length of each pattern is known as fringe width

Explanation

The formula for the position of $n^{th}$ bright fringe is $y_n=\frac{nD\lambda }{d}$ and for $m^{th}$ dark fringe is $y_m=(m+\frac{1}{2})\frac{D\lambda }{d}$

where $D$ is the distance between the source and the screen

$\lambda$ is the wavelength of the incident light

$d$ is the distance between the two slits

Distance between any two bright fringes: $y_{n+1}-y_n=\frac{(n+1-n)D\lambda }{d}$ ⇒ $\frac{D\lambda }{d}$

Distance between any two dark fringes: $y_{m+1}-y_m=(m+\frac{3}{2}-m-\frac{1}{2})\frac{D\lambda }{d}$ ⇒ $\frac{D\lambda }{d}$

So, the distance between any two bright fringes or any two dark fringes is constant

Hence, the fringes are of equal width