Young Double Slit Experiment Class 12 Notes, Definition, Working Principle, Formula & Real-Life Applications

Thomas Young came up with a technique in 1802, which created a stationary interference pattern. A single wavefront was divided into two, and these two wavefronts appeared to be coming from two sources with a stable phase connection. This resulted in a steady interference pattern when they were let to interfere and called Young's Double Slit Experiment.

Figure : 4.1 : Young's Arrangement to produce stationary interference pattern by division of wave front $S_0$  into $S_1$ and $S_2$

This is an important concepts considering JEE Mains exam and NEET exam since 2 questions were asked in the last year paper carrying a total weightage of 8 marks.

In the configuration above, we have guaranteed that the light waves traveling through $S_1$ and $S_2$ are in phase. However, since the wave from $P$ must take a longer route to reach P than the wave from $S_2$ , the waves may not be in phase when they arrive at $P$ from $S_1$ . The path difference-induced phase difference has already been covered. The arriving waves experience constructive interference and are precisely in phase if the path difference is zero or an integral multiple of wavelengths. Arriving waves that have a route difference that is an odd multiple of half a wavelength are out of phase and experience completely destructive interference. Therefore, the intensity at a point $P$ is determined by the path difference Δx.

Path difference $\mathrm{\Delta }p=S_{1}P-S_{2}P=\sqrt[]{{\left(y+\frac{d}{2}\right)}^2+D^2}-\sqrt[]{{\left(y-\frac{d}{2}\right)}^2+D^2}$
Approximation I : For D >> d, we can approximate rays ${}_1$  and 2 as being approximately parallel, at angle $\theta$  to the principle axis.
Now, ${\mathrm{S}}_1\mathrm{P}-{\mathrm{S}}_2\mathrm{P}={\mathrm{S}}_1\mathrm{A}={\mathrm{S}}_1{\mathrm{S}}_2\mathrm{s}\mathrm{i}\mathrm{n}\theta$
$\Rightarrow$ path difference $=d\mathrm{s}\mathrm{i}\mathrm{n}\theta$

Approximation II : Further if $\theta$  is small, i.e., $y\ll D,\mathrm{s}\mathrm{i}\mathrm{n}\theta =\mathrm{t}\mathrm{a}\mathrm{n}\theta =\frac{y}{D}$
and hence, path difference $=\frac{dy}{D}$
for maxima (constructive interference),

$\begin{array}{c}\Delta p=\frac{\mathrm{d}.\mathrm{y}}{\mathrm{D}}=n\lambda \Rightarrow y=\frac{\mathrm{n}\lambda \mathrm{D}}{\mathrm{d}},n=0,\pm 1,\pm 2,\pm 3\#\left(4.4\right)\end{array}$

Here $\mathrm{n}=0$  corresponds to the central maxima
$\mathrm{n}=\pm 1$  correspond to the 1st maxima
$\mathrm{n}=\pm 2$ correspond to the 2 nd maxima and so on.
for minima (destructive interference).
$\mathrm{\Delta }\mathrm{p}=\pm \frac{\lambda }{2},\pm \frac{3\lambda }{2}\pm \frac{5\lambda }{2}\mathrm{}\Rightarrow \mathrm{}\mathrm{\Delta }\mathrm{p}=\left\{\begin{array}{ll}(2\mathrm{n}-1)\frac{\lambda }{2}& \mathrm{n}=\mathrm{1,2},3\dots \dots \dots \dots \\ (2\mathrm{n}+1)\frac{\lambda }{2}& \mathrm{n}=-1,-2,-3\dots \dots .\end{array}\right.$ consequently,

$y=\left\{\begin{array}{ll}(2n-1)\frac{\lambda D}{2d}& n=\mathrm{1,2},3\dots \dots \dots \dots \\ (2n+1)\frac{\lambda D}{2d}& n=-1,-2,-3\dots \dots \end{array}\right.$
Here $\mathrm{n}=\pm 1$  corresponds to first minima,
$\mathrm{n}=\pm 2$  corresponds to second minima and so on.

It is the distance between two maxima of successive order on one side of the central maxima. This is also equal to distance between two successive minima.
fringe width $\beta =\frac{\lambda D}{d}$

Notice that it is directly proportional to wavelength and inversely proportional to the distance between the two slits.

In section 4.1 we obtained, $y=\frac{n\lambda D}{d},n=0,\pm 1,\pm 2\dots$ .
for interference maxima, but n cannot take infinitely large values, as that would violate the approximation (II) i.e., $\theta$  is small or $\mathrm{y}\ll \mathrm{D}$
$\Rightarrow \frac{\mathrm{y}}{\mathrm{D}}=\frac{\mathrm{n}\lambda }{\mathrm{d}}\ll 1$ Hence the above formula ( $4.4\&4.5$ ) for interference maxima/minima are applicable when

$\mathrm{n}\ll \frac{\mathrm{d}}{\lambda }$

when $n$  becomes comparable to $\frac{d}{\lambda }$  path difference can no longer be given by equation (4.3) but by (4.2) Hence for maxima

$\begin{array}{c}\Delta p=n\lambda \Rightarrow dsin\theta =n\lambda \Rightarrow n=\frac{\mathrm{d}\mathrm{s}\mathrm{i}\mathrm{n}\theta }{\lambda }\#\left(4.7\right)\end{array}$

Hence highest order of interference maxima, $\mathrm{}{n}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\left[\frac{d}{\lambda }\right]$
where [ ] represents the greatest integer function.
Similarly highest order of interference minima,

$\begin{array}{c}{\mathrm{n}}_{\text{min}}=\left[\frac{\mathrm{d}}{\lambda }+\frac{1}{2}\right]\#\left(4.8\right)\end{array}$

Alter
$\mathrm{\Delta }\mathrm{p}={\mathrm{S}}_1\mathrm{P}-{\mathrm{S}}_2\mathrm{P}$ (3rd side of a triangle is always greater than the difference in

Suppose the electric field components of the light waves arriving at point $\mathrm{\Delta }\mathrm{p}\le \mathrm{d}\Rightarrow \mathrm{\Delta }{\mathrm{p}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathrm{d}$  (in the Figure : 4.3) from the two slits $P$  and $S_1$  vary with time as
$S_2$ and $E_1=E_0\mathrm{s}\mathrm{i}\mathrm{n}\omega t$
Here $E_2=E_0\mathrm{s}\mathrm{i}\mathrm{n}(\omega t+\varphi )$
and we have assumed that intensity of the two slits $\varphi =k\mathrm{\Delta }x=\frac{2\pi }{\lambda }\mathrm{\Delta }x$  and $S_1$  are same (say $S_2$  ); hence waves have same amplitude E.
then the resultant electric field at point $I_0$  is given by,

$P$ where $E=E_1+E_2=E_0\mathrm{s}\mathrm{i}\mathrm{n}\omega t+E_0\mathrm{s}\mathrm{i}\mathrm{n}(\omega t+\varphi )=E_0^{\text{'}}\mathrm{s}\mathrm{i}\mathrm{n}\left(\omega t+{\varphi }^{\text{'}}\right)$
Hence the resultant intensity at point $\mathrm{}{\mathrm{E}}_0{}^2={\mathrm{E}}_0{}^2+{\mathrm{E}}_0{}^2+2{\mathrm{E}}_0.{\mathrm{E}}_0\mathrm{c}\mathrm{o}\mathrm{s}\varphi =4{\mathrm{E}}_0{}^2{\mathrm{c}\mathrm{o}\mathrm{s}}^2\varphi /2$ ,

$P$ $\begin{array}{c}I=4I{\mathrm{c}\mathrm{o}\mathrm{s}}^2\frac{\varphi }{2}\#\left(4.9\right)\end{array}$ when $I_{\mathrm{m}\mathrm{a}\mathrm{x}}=4{\mathrm{I}}_0$ .

$\frac{\varphi }{2}=\mathrm{n}\pi ,\mathrm{}\mathrm{}n=0,\pm 1,\pm 2,\dots \dots$

Here

$I_{\mathrm{m}\mathrm{i}\mathrm{n}}=0\text{when}\frac{\varphi }{2}=\left(n-\frac{1}{2}\right)\pi \mathrm{}n=0,\pm 1,\pm 2\dots \dots \dots$

If $\varphi =\mathrm{k}\mathrm{\Delta }\mathrm{x}=\frac{2\pi }{\lambda }\mathrm{}\mathrm{\Delta }\mathrm{x}$
If $D\gg d,\mathrm{}\varphi =\frac{2\pi }{\lambda }d\mathrm{s}\mathrm{i}\mathrm{n}\theta$
However if the two slits were of different intensities $D\gg d\&y\ll D,\varphi =\frac{2\pi }{\lambda }d\frac{y}{D}$ and ${\mathrm{I}}_1$ ,
say ${\mathrm{I}}_2$  and $E_1=E_{01}\mathrm{s}\mathrm{i}\mathrm{n}\omega t$
then resultant field at point P ,

$E_2=E_{02}\mathrm{s}\mathrm{i}\mathrm{n}(\omega t+\varphi )$

where $E=E_1+E_2=E_0\mathrm{s}\mathrm{i}\mathrm{n}(\omega t+\varphi )$
Hence resultant intensity at point P ,

$\mathrm{}{E}_0^2=E_{01}{}^2+E_{02}{}^2+2{\mathrm{E}}_{01}{\mathrm{E}}_{02}\mathrm{c}\mathrm{o}\mathrm{s}\varphi$

Suppose there is a glass slab that has a refractive index μ and t thickness. Say, this glass is placed in the way of one of the interfering waves. In this case, the optical length of this path will be μ, which will rise by (t-1) μ. Through this, the path difference will be   Δx = μt – t = (μ-1)t 

In case, the present location of fringe is  y =D/d (Δx); then new position of the fringe will be 

 y’ = D/d(Δx + (μ-1)t).

where

D: The distance from the lens to the image plane (or screen).

d: The distance from the lens to the object (or the source plane).

Therefore, Fringe lateral shift will be  y₀ = y’ – y

 y₀  = D/d(μ-1)t

Fringe width, w = D/dλ 

So, we will get

 y₀ = β/λ (μ-1)t