2.1 Fringe width $(w)$

  Wave Optics

The distance between any two consecutive bright or dark fringes is known as fringe width.

Case 1: For bright fringes

$$\begin{equation} \begin{aligned} w = {Y_{n + 1}} - {Y_n} \\ w = \frac{{(n + 1)\lambda D}}{d} - \frac{{n\lambda D}}{d} \\ w = \frac{{\lambda D}}{d} \\\end{aligned} \end{equation} $$

Case 2: For dark fringes

$$\begin{equation} \begin{aligned} w = {Y_{n + 1}} - {Y_n} \\ w = \frac{{\{ 2(n + 1) + 1\} \lambda D}}{{2d}} - \frac{{(2n + 1)\lambda D}}{{2d}} \\ w = \frac{{(2n + 3 - 2n - 1)\lambda D}}{{2d}} \\ w = \frac{{\lambda D}}{{d}} \\\end{aligned} \end{equation} $$

where,

$w$: Fringe width

$\lambda $: Wavelength of monochromatic light

$D$: Distance between slits and the screen

$d$: Distance between the two slits

So, the fringe width $(w)$ is same for two consecutive bright fringes and two consecutive dark fringes.

Note:

• In Young's double slit experiment (YDSE), we are interested in finding the resultant intensity at an arbitrary point $P$ on the screen.

As $d < < D$, the intensity at point $P$ due to individual sources $S_1$ and $S_2$ are almost equal.

$${I_1} \approx {I_2} \approx {I_0}\quad ...(i)$$

Resultant intensity at point $P$ is given by, $$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi \quad ...(ii)$$

From equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} I = 2{I_0} + 2{I_0}\cos \phi \\ I = 4{I_0}{\cos ^2}\left( {\frac{\phi }{2}} \right) \\\end{aligned} \end{equation} $$

	Maximum Intensity	Minimum Intensity
Phase difference $\left( \phi \right)$	$$\begin{equation} \begin{aligned} {\cos ^2}\left( {\frac{\phi }{2}} \right) = 1 \\ \cos \left( {\frac{\phi }{2}} \right) = \pm 1 \\ \left( {\frac{\phi }{2}} \right) = n\pi \\ \phi = 2n\pi \\\end{aligned} \end{equation} $$	$$\begin{equation} \begin{aligned} {\cos ^2}\left( {\frac{\phi }{2}} \right) = 0 \\ \cos \left( {\frac{\phi }{2}} \right) = 0 \\ \left( {\frac{\phi }{2}} \right) = \left( {2n + 1} \right)\frac{\pi }{2} \\ \phi = \left( {2n + 1} \right)\pi \\\end{aligned} \end{equation} $$
Path difference $\left( {\Delta x} \right)$	$$\begin{equation} \begin{aligned} \phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x \\ \Delta x = \frac{{2n\pi }}{{2\pi }}\lambda \\ \Delta x = n\lambda \\\end{aligned} \end{equation} $$	$$\begin{equation} \begin{aligned} \phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x \\ \Delta x = \frac{{\left( {2m + 1} \right)\pi }}{{2\pi }}\lambda \\ \Delta x = \left( {2n + 1} \right)\frac{\lambda }{2} \\\end{aligned} \end{equation} $$
Intensity	$${I_{\max }} = 4{I_0}$$	$${I_{\min }} = 0$$
Ratio of ${I_{\max }}$ and ${I_{\min }}$	$$\frac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\frac{{\sqrt {{I_1}} + \sqrt {{I_2}} }}{{\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}$$
Fringe visibility	$$V = \left( {\frac{{\sqrt {{I_{\max }}} - \sqrt {{I_{\min }}} }}{{\sqrt {{I_{\max }}} + \sqrt {{I_{\min }}} }}} \right)$$
Variation of intensity with phase difference	$$I = {I_{\max }}{\cos ^2}\left( {\frac{\phi }{2}} \right)$$

• Relation between refractive index and the wavelength is given by, $$\frac{{{\mu _1}}}{{{\mu _2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}}$$ or $$\begin{equation} \begin{aligned} \frac{1}{\mu } = \frac{{{\lambda _2}}}{\lambda } \\ {\lambda _2} = \frac{\lambda }{\mu } \\\end{aligned} \end{equation} $$

The position and width of fringe depends on the wavelength of monochromatic light. If the YDSE apparatus is immersed in a liquid of refractive index $\mu $ as shown in the figure, then the position and width of the fringe is decreased by $\mu $ times.

• Optical path length: Optical path length is defined as the product of the geometrical path length of light and the refractive index $\mu $ of the medium through which light propagates.

Path difference produced by a slab of refractive index $\mu $ and thickness $t$.

Original path length: ${L_1} + t + {L_2}$

Changed path length: ${L_1} + \mu t + {L_2}$

Path difference $\left( {\Delta x} \right)$, $$\begin{equation} \begin{aligned} \Delta x = {\text{Changed path length }} - {\text{ Original path length}} \\ \Delta x = \left( {{L_1} + \mu t + {L_2}} \right) - \left( {{L_1} + t + {L_2}} \right) \\ \Delta x = \left( {\mu - 1} \right)t \\\end{aligned} \end{equation} $$

• Shifting of fringe in YDSE

Let a glass slab of thickness $t$ and refractive index $\mu $ is placed in the path of source $S_1$ as shown in the figure.

$\begin{equation} \begin{aligned} {S_1}P = x + \left( {\mu - 1} \right)t \\ {S_2}P = x + d\sin \theta \\\end{aligned} \end{equation} $

Path difference can be written as, $$\begin{equation} \begin{aligned} {S_2}P - {S_1}P = \left( {x + d\sin \theta } \right) - \left\{ {x + \left( {\mu - 1} \right)t} \right\} \\ \Delta x = d\sin \theta - \left( {\mu - 1} \right)t \\\end{aligned} \end{equation} $$

Also, $$\tan \theta = \frac{{Y{'_n}}}{D}$$ As, $\theta $ is very small. So, $$\tan \theta \approx \sin \theta $$ Therefore, $$\Delta x = \frac{{dY{'_n}}}{D} - \left( {\mu - 1} \right)t$$

Relation between phase difference and path difference is given by, $$\begin{equation} \begin{aligned} \phi = \left( {\frac{{2\pi }}{\lambda }} \right)\left( {\Delta x} \right) \\ \phi = \frac{{2\pi }}{\lambda }\left[ {\frac{{dY{'_n}}}{D} - \left( {\mu - 1} \right)t} \right]\quad ...(i) \\\end{aligned} \end{equation} $$

For constructive interference or bright fringe

$$\phi = 2n\pi \quad n \in I\quad ...(ii)$$

From equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} 2n\pi = \left( {\frac{{2\pi }}{\lambda }} \right)\left[ {\frac{{dY{'_n}}}{D} - \left( {\mu - 1} \right)t} \right] \\ Y{'_n} = \frac{{n\lambda D}}{d} + \frac{{\left( {\mu - 1} \right)tD}}{d} \\\end{aligned} \end{equation} $$

Bright fringe is shifted by, $$\begin{equation} \begin{aligned} \Delta {Y_n} = Y{'_n} - {Y_n} \\ \Delta {Y_n} = \frac{{\left( {\mu - 1} \right)tD}}{d} \\\end{aligned} \end{equation} $$

For destructive interference or dark fringe

$$\phi = \left( {2n + 1} \right)\pi \quad n \in I\quad ...(iii)$$

From equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \left( {2n + 1} \right)\pi = \frac{{2\pi }}{\lambda }\left[ {\frac{{dY{'_n}}}{D} - \left( {\mu - 1} \right)t} \right] \\ Y{'_n} = \frac{{\left( {2n + 1} \right)\lambda D}}{{2d}} + \frac{{\left( {\mu - 1} \right)tD}}{d} \\\end{aligned} \end{equation} $$

Dark fringe is shifted by, $$\begin{equation} \begin{aligned} \Delta {Y_n} = Y{'_n} - {Y_n} \\ \Delta {Y_n} = \frac{{\left( {\mu - 1} \right)tD}}{d} \\\end{aligned} \end{equation} $$

The fringe is shifted towards the side in which the slab is introduced.

• Lloyd's mirror

In this experiment, a light from a slit $S_1$ is illuminated. The incident light forms a virtual image $S_2$ after reflection.

Now, $S_1$ and $S_2$ acts as two coherent sources as shown in the figure.

The interference pattern at point $P$.

In $\Delta {S_1}OM$ and $\Delta {S_2}OM$, $$\begin{equation} \begin{aligned} {S_1}M = {S_2}M\quad \left( {M{\text{ is midpoint}}} \right) \\ \angle {S_1}OM = \angle {S_2}OM \\ \angle M{S_1}O = \angle M{S_2}O \\\end{aligned} \end{equation} $$ So, $$\Delta {S_1}OM \cong \Delta {S_2}OM$$ Therefore, $${S_1}O = {S_2}O$$

Path difference $\left( {\Delta x} \right)$ is given by, $$\begin{equation} \begin{aligned} \Delta x = {S_2}P - {S_1}P \\ \Delta x = d\sin \theta \\\end{aligned} \end{equation} $$ Also, $$\tan \theta = \frac{{{Y_n}}}{D}$$ As, $\theta $ is very small. So, $$\tan \theta \approx \sin \theta $$ or $$\Delta x = \frac{{d{Y_n}}}{D}$$

Relation between phase difference and path difference is given by, $$\begin{equation} \begin{aligned} \phi = \left( {\frac{{2\pi }}{\lambda }} \right)\Delta x \\

\phi = \left( {\frac{{2\pi }}{\lambda }} \right)\left( {\frac{{d{Y_n}}}{D}} \right)\quad ...(i) \\\end{aligned} \end{equation} $$

After reflection the phase is changed by $180^\circ $ or $\pi $ as it is reflected from the denser medium. So, now the condition of constructive and destructive interference is interchanged.

For constructive interference or bright fringe

$$\phi = \left( {2n + 1} \right)\pi \quad ...(ii)$$

From equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \left( {2n + 1} \right)\pi = \frac{{2\pi }}{\lambda }\left( {\frac{{d{Y_n}}}{D}} \right) \\ {Y_n} = \frac{{\left( {2n + 1} \right)\lambda D}}{{2d}} \\\end{aligned} \end{equation} $$

For destructive interference or dark fringe

$$\phi = 2n\pi \quad ...(iii)$$

From equation $(i)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} 2n\pi = \left( {\frac{{2\pi }}{\lambda }} \right)\left( {\frac{{d{Y_n}}}{D}} \right) \\ {Y_n} = \frac{{n\lambda D}}{d} \\\end{aligned} \end{equation} $$

• Colour of thin films

A soap film or a thin film of oil spread over water surface when seen in white light appears coloured. This effect can be explained in terms of phenomenon of interference.

Consider a film of uniform thickness $t$ and refractive index $\mu $ as shown in the figure.

Let us assume that the light rays traveling in air are nearly normal to the surfaces of the film.

The wavelength of light in a medium whose refractive index is $\mu $. $${\lambda _\mu } = \frac{\lambda }{\mu }$$

where $\lambda $ is the wavelength of light in vacuum or air.

Light ray 1 is reflected from denser medium. So, it undergoes a phase change of $\pi $. However, there is no phase change in case of light ray 2.

Path difference $\left( {\Delta x} \right)$, $$\Delta x = 2t\cos r$$

Relation between path difference and phase difference is given by, $$\begin{equation} \begin{aligned} \phi = \left( {\frac{{2\pi }}{{{\lambda _\mu }}}} \right)\left( {\Delta x} \right) \\ \phi = \left( {\frac{{2\pi }}{{{\lambda _\mu }}}} \right)\left( {2t\cos r} \right)\quad ...(i) \\\end{aligned} \end{equation} $$

For constructive interference or bright fringe

$$\phi = \left( {2n + 1} \right)\pi \quad ...(ii)$$

From equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \left( {2n + 1} \right)\pi = \frac{{2\pi }}{{{\lambda _\mu }}}\left( {2t\cos r} \right) \\ \left( {\frac{{2n + 1}}{2}} \right){\lambda _\mu } = 2t\cos r \\\end{aligned} \end{equation} $$

As, ${\lambda _\mu } = \frac{\lambda }{\mu }$

$$\begin{equation} \begin{aligned} 2t\cos r = \frac{{\left( {2n + 1} \right)\lambda }}{{2\mu }} \\ 2\mu t\cos r = \left( {n + \frac{1}{2}} \right)\lambda \\\end{aligned} \end{equation} $$

Light rays traveling in air are nearly normal to the surface of the film.

$$r \approx 0$$ or $$\cos r = 1$$ So, $$2\mu t = \left( {n + \frac{1}{2}} \right)\lambda $$

For destructive interference or dark fringe

$$\phi = 2n\lambda \quad ...(iii)$$

From equation $(i)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} 2n\lambda = \left( {\frac{{2\pi }}{{{\lambda _\mu }}}} \right)\left( {2t\cos r} \right) \\ 2t\cos r = n{\lambda _\mu } \\\end{aligned} \end{equation} $$

As, ${\lambda _\mu } = \frac{\lambda }{\mu }$

$$\begin{equation} \begin{aligned} 2t\cos r = \frac{{n\lambda }}{\mu } \\ 2\mu t\cos r = n\lambda \\\end{aligned} \end{equation} $$

Light rays traveling in air are nearly normal to the surface of the film.

$$r \approx 0$$ or $$\cos r = 1$$ So, $$2\mu t = n\lambda $$