Wave Optics
2.0 Young's double slit experiment
2.0 Young's double slit experiment
Young's double slit experiment was the first to demonstrate the phenomenon of interference of light in the year 1801. This experiment is named after the scientist Thomas Young.
In this experiment monochromatic light (single wavelength) from a narrow vertical slit $S$ falls on two other narrow slits $S_1$ and $S_2$, which are very close and almost parallel to $S$.
$S_1$ and $S_2$ acts as two coherent sources as they are both derived from a single source $S$.
By using two slits $S_1$ & $S_2$ separated by small distance $d$ and a monochromatic light source, an alternate bright and dark fringes of equal width are observed. These fringes are known as interference fringes.
The interference fringes are obtained on a screen placed at a distance $D$ as shown in the figure.
The light wave from $S_1$ and $S_2$ meet at point $P$ on the screen as shown in the figure.
Path difference $\left( {\Delta x} \right)$ between the two light ray, $$\Delta x = {S_2}P - {S_1}P$$
Let ${n^{th}}$ fringe is formed at point $P$ which is at a distance ${Y_n}$ from the centre.
As we know the distance $d$ between the two slits $S_1$ and $S_2$ is very small as compared to the distance $D$ between the source and screen. So,
$$d < < D$$
From the figure we can write, $$\begin{equation} \begin{aligned} \sin \theta = \frac{{\Delta x}}{d} \\ \tan \theta = \frac{{{Y_n}}}{D} \\\end{aligned} \end{equation} $$
As $\theta $ is very small, $$\begin{equation} \begin{aligned} \tan \theta \approx \sin \theta \\ \frac{{{Y_n}}}{D} = \frac{{\Delta x}}{d} \\ {Y_n} = \frac{{\Delta xD}}{d}\quad ...(i) \\\end{aligned} \end{equation} $$
At point $P$, resultant of two waves of amplitude $A_1$ & $A_2$ having phase difference $\phi $ is given by,
$$R = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $$
Based on the interference of waves, two cases are possible.
1. Bright fringe is formed at point $P$ due to constructive interference
2. Dark fringe is formed at point $P$ due to destructive interference
Bright fringe is formed at point $P$ due to constructive interference
For constructive interference, $$\cos \phi = 1$$ or $$\phi = 2n\pi $$
Relation between path difference and phase difference is given by, $$\phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x$$ or $$\begin{equation} \begin{aligned} \phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x \\ \Delta x = \phi \left( {\frac{\lambda }{{2\pi }}} \right) \\ \Delta x = \left( {2n\pi } \right)\left( {\frac{\lambda }{{2\pi }}} \right) \\ \Delta x = n\lambda \quad {\text{where }}n \in I\quad ...(ii) \\\end{aligned} \end{equation} $$
From equation $(i)$ & $(ii)$ we can write the position of ${n^{th}}$ bright fringe as, $${Y_n} = \frac{{n\lambda D}}{d}$$
where,
$n = 0,\;{Y_0} = 0$ (Central bright fringe)
$n = 1,\;{Y_1} = \frac{{\lambda D}}{d}$ (First bright fringe)
$n = 2,\;{Y_2} = \frac{{2\lambda D}}{d}$ (Second bright fringe)
$d$: Distance between two slits
$D$: Distance of slits from the screen
$Y_n$: Position of ${n^{th}}$ bright fringe from the centre
Dark fringe is formed at point $P$ due to destructive interference
For constructive interference, $$\cos \phi = -1$$ or $$\phi = \left( {2n + 1} \right)\pi $$
Relation between path difference and phase difference is given by, $$\phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x$$ or $$\begin{equation} \begin{aligned} \phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x \\ \Delta x = \phi \left( {\frac{\lambda }{{2\pi }}} \right) \\ \Delta x = \left\{ {\left( {2n + 1} \right)\pi } \right\}\left( {\frac{\lambda }{{2\pi }}} \right) \\ \Delta x = \left( {2n + 1} \right)\frac{\lambda }{2}\quad {\text{where }}n \in I\quad ...(iii) \\\end{aligned} \end{equation} $$
From equation $(i)$ & $(iii)$ we can write the position of ${n^{th}}$ dark fringe as, $${Y_n} = (2n + 1)\frac{{\lambda D}}{{2d}}$$
where,
$n = 0,\;{Y_0} = \frac{{\lambda D}}{{2d}}$ (First dark fringe)
$n = 1,\;{Y_1} = \frac{{3\lambda D}}{2d}$ (Second dark fringe)
$d$: Distance between two slits
$D$: Distance of slits from the screen
$Y_n$: Position of ${n^{th}}$ dark fringe from the centre
