Expression for fringe's width:
Let us consider two wave from slit $S_{1}$ and $S_{2}$ superimpose on each other and form interfernece patteren on the screen. The distance between the two slits is $d$ and distance between slit to screen is $D$. Now take a $n^{th}$ fringe from the centre $O$ of the screen which is at distance $y_{n}$.
So the path difference between the rays
$\Delta x = S_{2}P- S_{1}P \quad(1)$
In $\Delta S_{1}PM$
$S_{1}P^{2}=S_{1}M^{2}+PM^{2} \quad(2)$
From figure:
$S_{1}M =D$
$PM= y_{n}- \left(\frac{d}{2}\right) $
Now subtitute these values in equation $(2)$, then
$S_{1}P^{2}=D^{2}+ \left( y_{n}- \frac{d}{2} \right)^{2} \quad(3)$
In $\Delta S_{2}PN$
$S_{2}P^{2}=S_{2}N^{2}+PN^{2} \quad(4)$
From figure:
$S_{2}N =D$
$PN= y_{n} + \left(\frac{d}{2}\right) $
Now subtitute these values in equation $(4)$, then
$S_{2}P^{2}=D^{2}+ \left( y_{n} + \frac{d}{2} \right)^{2} \quad(5)$
Now subtract the equation $(3)$ in equation $(5)$
$S_{2}P^{2} - S_{1}P^{2} = \left[D^{2}+ \left( y_{n} + \frac{d}{2} \right)^{2}\right] - \left[ D^{2}+ \left( y_{n}- \frac{d}{2} \right)^{2} \right] $
$S_{2}P^{2} - S_{1}P^{2} = D^{2}+ \left( y_{n} + \frac{d}{2} \right)^{2} - D^{2} - \left( y_{n}- \frac{d}{2} \right)^{2} $
$S_{2}P^{2} - S_{1}P^{2} = \left( y_{n} + \frac{d}{2} \right)^{2} - \left( y_{n}- \frac{d}{2} \right)^{2} $
$S_{2}P^{2} - S_{1}P^{2} = y^{2}_{n} + \left(\frac{d}{2}\right)^{2} +2 y_{n} \left(\frac{d}{2}\right) - y^{2}_{n} - \left(\frac{d}{2}\right)^{2} +2 y_{n} \left(\frac{d}{2}\right) $
$S_{2}P^{2} - S_{1}P^{2} = y^{2}_{n} + \left(\frac{d}{2}\right)^{2} + y_{n} d - y^{2}_{n} - \left(\frac{d}{2}\right)^{2} + y_{n} d $
$S_{2}P^{2} - S_{1}P^{2} = y_{n} d + y_{n} d $
$ \left( S_{2}P + S_{1}P \right) \left( S_{2}P - S_{1}P \right) = 2y_{n} d$
$ \left( S_{2}P + S_{1}P \right) \Delta x = 2y_{n} d \quad \left\{from \: equation\: (1)\right\} \qquad(6)$
In this experiment, the distance between slit and screen i.e. $D$ is much greater than the distance between the slit i.e. $d$ so angle $\theta$ will be very small. So from figure
$S_{1}P = S_{2}P = D$
Substitute this value in the equation $(6)$, therefore equation $(6)$ can be written as
$ 2D \Delta x = 2y_{n} d $
$D \Delta x = y_{n}$
$\Delta x = \frac{y_{n} d}{D} \quad (7) $
This is equation of path differnce between two wave from slit $S_{1}$ and slit $S_{2}$.
Condition For Bright Fringe:
The path difference for bright fringe is
$\Delta x= n \lambda \qquad(8)$
From equation $(7)$ and equation $(8)$
$n \lambda = \frac{ d}{D} $
$y_{n} = \frac{n \lambda D}{d} \quad(9)$
This is the equation for distance of $n^{th}$ bright fringe. Now the distance of $(n+1)^{th}$ bright fringe.
$y_{n+1} = \frac{\left( n+1 \right) \lambda D}{d} \quad(10)$
The dark fringe lie between the two consecutive bright fringe. So width of the dark fringe
$\beta = y_{n+1} - y_{n}$
Now subtitute the value of $y_{n+1}$ and $y_{n}$ in the above equation $(9)$ and equation $(10)$, then
$\beta = \frac{\left( n+1 \right) \lambda D}{d} - \frac{n \lambda D}{d}$
$\beta = \frac{\lambda D}{d} \quad(11)$
This is the equation of width of dark fringe.
Condition For Dark Fringe:
The path difference for dark fringe is
$\Delta x = (2n-1) \lambda \quad(12)$
From equation $(7)$ and equation $(12)$
$ (2n-1) \lambda = \frac{y_{n} d}{D} $
$y_{n} = \frac{(2n-1) \lambda D}{d} \quad(13)$
This is the equation for distance of $n^{th}$ dark fringe. Now the distance of $(n+1)^{th}$ dark fringe.
$y_{n+1} = \frac{\left[ 2 (n+1)-1) \right] \lambda D}{d} $
$y_{n+1} = \frac{(2n+1)\lambda D}{d} \quad(14)$
The bright fringe lie between the two consecutive dark fringe. So width of the bright fringe
$\beta = y_{n+1} - y_{n}$
Now subtitute the value of $y_{n+1}$ and $y_{n}$ in the above equation $(13)$ and equation $(14)$, then
$\beta = \frac{(2n+1)\lambda D}{d} - \frac{(2n-1) \lambda D}{d}$
$\beta = \frac{\lambda D}{d} \quad(15)$
This is the equation of width of bright fringe.
The equation $(11)$ and equation $(15)$ shows that the width of bright fring and dark fringe is same.
Alternative method to find the path differnce of two wave in Young's double-slit Experiment
Let us consider: Two wave from slit $S_{1}$ and $S_{2}$ superimpose on each other and form interfernece patteren on the screen.
The distance between the two slits is $\rightarrow d$
The distance between slit to screen is $\rightarrow D$
The distance of $n^{th}$ fringe from the centre $O$ of the screen is $\rightarrow y_{n}$.
So the path difference between the two rays is
$\Delta x = S_{2}M \quad(1)$
Now from figure, In $\Delta S_{1}S_{2}M$
$sin \theta = \frac{S_{2}M}{S_{1}S_{2}} \quad(2)$
In $\Delta PNO$
$tan \theta = \frac{PO}{NO} \quad(3)$
The distance between slit and screen i.e. $D$ is much greater than the distance between the slit i.e. $d$ so angle $\theta$ will be very small i.e. $sin \theta \approx tan \theta \approx \theta $. Now equate the equation $(2)$ and equation $(3)$
$\frac{PO}{NO} = \frac{S_{2}M}{S_{1}S_{2}} \quad(4)$
Now the value from the above figure
$\frac{y_{n}}{D} = \frac{\Delta x}{d} $
$\Delta x = \frac{y_{n} d}{D} $
This is the equation of the path difference between two wave from slit $S_{1}$ and $S_{2}$.
Now the derivation for the condition for Bright and Dark fringe can be used here as it is given above.
