Derive an expression for fringe width in Young's double slit experiment. State the factors on which fringe width depends.What is the shape of interference fringes obtained in Young's double slit experiment
Dark and bright bands in the interference pattern are called interference fringes.
Consider two coherent sources $S_1$ and $S_2$ separated by a distance $d$. Let $D$ be the distance between the screen and the plane of slits $S_1$ and $S_2$.
Light waves emitted from $S_1$ and $S_2$ reach point O on the screen after travelling equal distances. So, path difference and hence phase difference between these waves is zero. Therefore, $S_1$ and $S_2$ meet at O in phase and hence constructive interference takes place at O. Thus, O is the position of the central bright fringe.
Let the waves emitted by $S_1$ and $S_2$ meet at point P on the screen at a distance $y$ from the central bright fringe. The path difference between these waves at P is given by
$ \Delta x = S_2P - S_1P \quad \cdots (i) $
From right angled triangle $S_2BP$, we get, $S_2P = [S_2B^2 + PB^2]^{1/2}$
$S_2P = [D^2 + (y + \frac{d}{2})^2]^{1/2}$
$S_2P= D \left[ 1 + \frac{(y + d/2)^2}{D^2} \right]^{1/2}$.
Using Binomial theorem, and neglecting terms containing higher powers, we get,
$ S_2P = D \left( 1 + \frac{(y + d/2)^2}{2D^2} \right)$
$S_2P= D + \frac{(y + d/2)^2}{2D} \quad \cdots (ii)$
Similarly, from right angled triangle $\Delta S_1AP$, we get
$S_1P = D + \frac{(y - d/2)^2}{2D} \quad \cdots (iii)$
Substituting the values of (ii) and (iii) in (i),
we get,
$\Delta x = \left( D + \frac{(y + d/2)^2}{2D} \right) - \left( D + \frac{(y - d/2)^2}{2D} \right)$
$\Delta x = \frac{(y + d/2)^2 - (y - d/2)^2}{2D}$
$\Delta x = \frac{(y^2 + yd + d^2/4) - (y^2 - yd + d^2/4)}{2D}$
$\Delta x = \frac{2yd}{2D}$
$\Delta x = \frac{yd}{D} \quad \cdots (iv)$
Bright Fringes (Constructive Interference /Maxima)
If path difference is an integral multiple of $\lambda$, then bright fringe will be formed at P, i.e.
$ \Delta x = m \lambda $
$ \frac{yd}{D} = m \lambda $
$y = \frac{m \lambda D}{d} \quad \cdots (v) $
Where $m = 0, \pm 1 , \pm 2 , \pm 3 \cdots$
Here, $y$ is the position of $m^{th}$ bright fringe from the central bright fringe.
If $m = 1$, then $y_1 = \frac{1 \cdot \lambda D}{d} = \frac{\lambda D}{d}$. It is the position of first bright fringe from the central bright fringe.
If $m = 2$, then $y_2 = \frac{2 \lambda D}{d}$. It is the position of second bright fringe from the central bright fringe and so on.
If $m = (m-1)$, then $y_{m-1} = \frac{(m-1) \lambda D}{d}$. It is the position of $(m-1)^{th}$ bright fringe from the central bright fringe.
Bright Fringe Width ($\beta$). The distance between any two successive bright fringes is called fringe width. i.e.
$ \beta = y_2 - y_1 = \frac{2 \lambda D}{d} - \frac{\lambda D}{d} = \frac{\lambda D}{d}$$\beta = \frac{\lambda D}{d} \quad \cdots (vi)$
Dark Fringes (Destructive Interference /Minima).
If path difference is odd multiple of $\lambda/2$, then dark fringe is formed at P. i.e.
$ \frac{yd}{D} = (m + \frac{1}{2}) \lambda$
$ y = \frac{(m + \frac{1}{2}) \lambda D}{d} \quad \cdots (vii) $
where $m = 0, \pm 1, \pm 2, \pm 3 \cdots $
Here $y$ is the position of $m^{th}$ dark fringe from the central bright fringe.
If $m = 0$, then $y_0 = \frac{(0 + \frac{1}{2}) \lambda D}{d} = \frac{\lambda D}{2d}$.
It is the position of first dark fringe from the central bright fringe.
If $m = 1$, then $y_1 = \frac{(1 + \frac{1}{2}) \lambda D}{d} = \frac{3 \lambda D}{2d}$.
It is the position of second dark fringe from the central bright fringe.
If $m = (m-1)$, then $y_{m-1} = \frac{(m - 1 + \frac{1}{2}) \lambda D}{d} = \frac{(m - \frac{1}{2}) \lambda D}{d}$.
It is the position of $(m - 1)^{th}$ dark fringe from the central bright fringe.
Dark Fringes Width ($\beta$) The distance between any two successive dark fringes is also called fringe width.
$ \beta = y_1 - y_0 = \frac{3 \lambda D}{2d} - \frac{\lambda D}{2d}$
$ \beta= \frac{2 \lambda D}{2d} = \frac{\lambda D}{d}$
$\beta = \frac{\lambda D}{d} \quad \cdots (viii) $
Factors on which Fringe Width Depends :
We know that, fringe width
$\beta = \frac{\lambda D}{d}$
It means, fringe width
(i) $\beta \propto \lambda$, larger the wave length of light, larger will be the fringe width and vice-versa. It means, fringe width will be greater for red colour than for violet colour.
(ii) $\beta \propto D$, larger the distance of the screen from the slits, larger will be fringe width.
(iii) $\beta \propto \frac{1}{d}$, smaller the distance between the coherent sources (i.e. slits $S_1$ and $S_2$), larger will be the fringe width.
Shape of Interference Fringes :
We know, for bright fringes,
$S_2P - S_1P = m \lambda$
and for dark fringes,
$S_2P - S_1P = (m + \frac{1}{2}) \lambda$
Since $m$ and $\lambda$ are constants, hence $S_2P - S_1P = \text{constant}$.
Therefore, above equations tell that the locus of point P is a hyperbola with $S_1$ and $S_2$ as foci. However, if the screen distance (D) from the plane of slits $S_1$ and $S_2$ is very large as compared to the fringe width ($\beta$), then the fringes in the interference pattern are nearly straight lines.
