## What isÂ Youngâ€™s Double Slit Experiment?

Youngâ€™s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than theÂ wavelength of lightÂ is used. Young’s double-slit experiment helped in understanding the wave theory of light which is explained with the help of a diagram. A screen or photodetector is placed at a large distance â€™Dâ€™ away from the slits as shown.

The original Youngâ€™s double-slit experiment used diffracted light from a single source passed into two more slits to be used as coherent sources. Lasers are commonly used asÂ coherent sourcesÂ in the modern-day experiments.

### Table of Content

- Derivation
- Position of Fringes
- Shape of Fringes
- Intensity of Fringes
- Special Cases
- Displacement of Fringes
- Constructive and Destructive Interference

Each source can be considered as a source of coherent light waves. At any point on the screen at a distance â€˜yâ€™ from the centre, the waves travel distancesÂ l_{1}Â andÂ l_{2}Â to create a path difference of Î”l at the point in question. The point approximately subtends an angle of Î¸ at the sources (since the distance D is large there is only a very small difference of the angles subtended at sources).

## Derivation of Youngâ€™s Double Slit Experiment

Consider a monochromatic light source â€˜sâ€™ kept at a considerable distance from two slits s_{1} and s_{2}. S is equidistant from s_{1} and s_{2}. s_{1} and s_{2} behave as two coherent sources, as both bring derived from S.

The light passes through these slits and falls on a screen which is at a distance â€˜Dâ€™ from the position of slits s_{1} and s_{2}. â€˜dâ€™ be the separation between two slits.

If s_{1} is open and s_{2} is closed, the screen opposite to s_{1} is closed, only the screen opposite to s_{2} is illuminated. The interference patterns appear only when both slits s_{1} and s_{2} are open.

When the slit separation (d) and the screen distance (D) are kept unchanged, to reach *P* the light waves from s_{1} and s_{2} must travel different distances. It implies that there is a path difference in Youngâ€™s double slit experiment between the two light waves from s_{1} and s_{2}.

**Approximations inÂ Youngâ€™s double slit experimentÂ **

**ApproximationÂ****1:****D > > d:**Since D > > d, the two light rays are assumed to be parallel, then the path difference,**ApproximationÂ****2:**Â**d/Î» >> 1:Â**Often, d is a fraction of a millimetre and Î» is a fraction of a micrometre for visible light.

Under these conditionsÂ Î¸ is small, thus we can use the approximationÂ sin Î¸ approx tan Î¸ = Î³/D.

âˆ´ path difference, Î”z = Î³/D

This is the path difference between two waves meeting at a point on the screen. Due to this path difference in Youngâ€™s double slit experiment, some points on the screen are bright and some points are dark.

Now, we will discuss the position of these light, dark fringes and fringe width.

## Position of Fringes InÂ Young’s Double Slit Experiment

### Position of Bright Fringes

For maximum intensity at P

Î”z = nÎ» (n = 0, Â±1, Â±2, . . . .)

Or d sin Î¸ = nÎ» (n = 0, Â±1, Â±2, . . . .)

The bright fringe for n = 0 is known as the central fringe. Higher order fringes are situated symmetrically about the central fringe. The position of n^{th} bright fringe is given by

y (bright) = (nÎ»\d)D (n = 0, Â±1, Â±2, . . . .)

### Position of Dark Fringes

For minimum intensity at P,

\(\Delta z=\left( 2n-1 \right)\frac{n\lambda }{2}\left( n=\pm 1,\pm 2,….. \right)\) \(d\sin \theta =\left( 2n-1 \right)\frac{n\lambda }{2}\)The first minima are adjacent to the central maximum on either side. We will obtain the position of dark fringe as

\({{y}_{dark}}=\frac{\left( 2n-1 \right)\lambda D}{2d}\left( n=\pm 1,\pm 2,….. \right)\)### Fringe Width

Distance between two adjacent bright (or dark) fringes is called the fringe width.

\(\beta z\frac{n\lambda D}{d}-\frac{\left( n-1 \right)\lambda D}{d}=\frac{\lambda D}{d}\)â‡’ \(\beta \propto \lambda\)

If the apparatus of Youngâ€™s double slit experiment is immersed in a liquid of refractive indexÂ (u), then wavelength of light and hence fringe width decreases â€˜uâ€™ times.

\({{\beta }^{1}}=\frac{\beta }{\mu }\)If a white light is used in place of monochromatic light, then coloured fringes are obtained on the screen with red fringes larger in size that of violet.

### Angular width of Fringes

Let the angular position of n^{th} bright fringe is \({{\theta }_{n}}\) and because of its small value \(\tan {{\theta }_{n}}\approx {{\theta }_{n}}\)
\(\tan {{\theta }_{n}}=\frac{\gamma n}{D}\approx {{\theta }_{n}}=\frac{\gamma n}{D}\)

â‡’ \({{\theta }_{n}}=\frac{n\gamma D/d}{D}=\frac{n\lambda }{d}\)

Similarly, the angular position of (n+1)^{th} bright fringe is \({{\theta }_{n+1}},\) then

âˆ´ Angular width of a fringe inÂ Youngs double slit experiment is given by,

\(\theta ={{\theta }_{n+1}}-{{\theta }_{n}}=\frac{\left( n+1 \right)\lambda }{d}-\frac{n\lambda }{d}=\frac{\lambda }{d}\) \(\theta =\frac{\lambda }{d}\)We know that \(\beta =\frac{\lambda D}{d}\)

â‡’ \(\theta =\frac{\lambda }{d}=\frac{\beta }{D}\)

Angular width is independent of â€˜nâ€™ i.e angular width of all fringes are same.

### Maximum Order of Interference Fringes

The position of n^{th} order maxima on the screen is \(\gamma =\frac{n\lambda D}{d};n=0,\pm 1,\pm 2, . . \)

But â€˜nâ€™ values cannot take infinitely large values as it would violate 2^{nd} approximation.

i.eÂ Î¸ is small (or) y < < D

â‡’ \(\frac{\gamma }{D}=\frac{n\lambda }{d}<<1\)

Hence, the above formula for interference maxima is applicable when \(n<<\frac{d}{\lambda }\)

When â€˜nâ€™ value becomes comparable to \(\frac{d}{\lambda },\) path difference can no longer be given by \(\frac{d\gamma }{D},\) Hence for maxima, path difference = nÎ»

â‡’ \(d\sin \theta =n\lambda\)

â‡’ \(n=\frac{d\sin \theta }{\lambda }\) \({{n}_{\max }}=\left[ \frac{d}{\lambda } \right]\)

The above represents box function or greatest integer function.

Similarly, the highest order of interference minima

\({{n}_{\min }}=\left[ \frac{d}{\lambda }+\frac{1}{2} \right]\)## Shape of Interference Fringes in YDSE

From the given YDSE diagram, the path difference from the two slits is given by

\({{s}_{2}}p-{{s}_{1}}p\approx d\sin \theta\) (constant)The above equation represents a hyperbola with its two foci as s_{1} and s_{2}.

The interference pattern we get on the screen is a section of a hyperbola when we revolve hyperbola about the axis s_{1}s_{2}.

If the fringe will represent 1^{st} minima, the fringe will represent 1^{st} maxima, it represents central maxima.

If the screen is yz plane, fringes are hyperbolic with a straight central section.

If the screen is xy plane, the fringes are hyperbolic with a straight central section.

## Intensity of Fringes In Youngâ€™s Double Slit Experiment

For two coherent sources s_{1} and s_{2}, the resultant intensity at point *p* is given by

I = I1 + I2 + 2 âˆš(I1 . I2) cosÂ Ï†

Putting I1 = I2 = I0Â (Since, d<<<D)

I = I0 + I0 + 2 âˆš(I0.I0) cos Ï†

I = 2I0 + 2 (I0) cos Ï†

I = 2I0 (H cos s Ï†)

I = \(4{{I}_{0}}{{\cos }^{2}}\left( \frac{\phi }{2} \right)\).

#### For maximum intensity

\(\cos \frac{\phi }{2}=\pm 1\) \(\frac{\phi }{2}=n\pi ,n=0,\pm 1,\pm 2,……\)or Ï† = 2nÏ€

**phase difference**Â **Ï† = 2nÏ€**

then, **path difference** \(\Delta x=\frac{\lambda }{{2}{\pi }}\left( {2}n{\pi } \right)\) = nÎ»

The intensity of bright points are maximum and given by

Imax = 4I0

#### For Minimum Intensity

\(\cos \frac{\phi }{2}=0\) \(\frac{\phi }{2}=\left( n-\frac{1}{2} \right)\pi \,\,\,\,\,where\,\left( n=\pm 1,\pm 2,\pm 3,….. \right)\)Ï† = (2n – 1) Ï€

Phase difference Ï† = (2n – 1)Ï€

\(\frac{2\pi }{\lambda }\Delta x=\left( 2n-1 \right)\pi\) \(\Delta x=\left( 2n-1 \right)\frac{\lambda }{2}\)Thus, **intensity of minima** is given by

Imin = 0

If \({{I}_{1}}\ne {{I}_{2}},{{I}_{\min }}\ne 0.\)

## Special Cases

### Rays Not Parallel to Principal Axis:

From the above diagram,

Path difference \(\Delta x=\left( A{{S}_{1}}+{{S}_{1}}P \right)-{{S}_{2}}P\) \(\Delta x=A{{S}_{1}}-\left( {{S}_{2}}P-{{S}_{1}}P \right)\) \(\Delta x=d\sin \theta -\frac{4d}{D}\)

For maxima \(\Delta x=n\lambda\)

For minima \(\Delta x=\left( 2n-1 \right)\frac{\lambda }{2}\)

Using this we can calculate different positions of maxima and minima.

### Source Placed Beyond the Central Line:

If the source is placed a little above or below this centre line, the wave interaction with S_{1} and S_{2} has a path difference at a point *P* on the screen,

Î” x= (distance of ray 2) â€“ (distance of ray 1)

= \(\left( S\,{{S}_{2}}+{{S}_{2}}P \right)-\left( S\,{{S}_{1}}+{{S}_{1}}P \right)\)

= \(\left( S\,{{S}_{2}}+S\,{{S}_{1}} \right)+\left( {{S}_{2}}P-{{S}_{1}}P \right)\)

= bd/a + yd/D â†’ (*)

We knowÂ Î”x = nÎ» for maximum

Î”x = (2n – 1) Î»/2 for minimum

By knowing the value ofÂ Î”x from (*) we can calculate different positions of maxima and minima.

## Displacement of Fringes inÂ YDSE

When a thin transparent plate of thickness â€˜tâ€™ is introduced in front of one of the slits in Youngâ€™s double slit experiment, the fringe pattern shifts toward the side where the plate is present.

The dotted lines denote the path of the light before introducing the transparent plate. The solid lines denote the path of the light after introducing a transparent plate.

Path difference before introducing the plate \(\Delta x={{S}_{1}}P-{{S}_{2}}P\)

Path difference after introducing the plate \(\Delta {{x}_{new}}={{S}_{1}}{{P}^{1}}-{{S}_{2}}{{P}^{1}}\)

The path length \({{S}_{2}}{{P}^{1}}={{\left( {{S}_{2}}{{P}^{1}}-t \right)}_{air}}+{{t}_{plate}}\) \(={{\left( {{S}_{2}}{{P}^{1}}-t \right)}_{air}}+{{\left( \mu t \right)}_{plate}}\)

where \(’\mu t'\) is optical path

\(={{S}_{2}}{{P}^{1}}_{air}+\left( \mu -1 \right)t\)Then we get

\({{\left( \Delta x \right)}_{new}}={{S}_{1}}{{P}^{1}}_{air}-\left( {{S}_{2}}{{P}^{1}}_{air}+\left( \mu -1 \right)t \right)\) \(={{\left( {{S}_{1}}{{P}^{1}}-{{S}_{2}}{{P}^{1}} \right)}_{air}}-\left( \mu -1 \right)t\) \({{\left( \Delta x \right)}_{new}}=d\sin \theta -\left( \mu -1 \right)t\) \({{\left( \Delta x \right)}_{new}}=\frac{\gamma d}{D}-\left( \mu -1 \right)t\)Then,

\(\begin{matrix} y=\frac{\Delta xD}{d}+\frac{D}{d}\left[ \left( \mu -1 \right)t \right] \\ \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \, \\ \left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\,\,\,\, \\ \end{matrix}\)Term (1) defines the position of a bright or dark fringe, the term (2) defines the shift occurred in the particular fringe due to the introduction of a transparent plate.

## Constructive and Destructive Interference

For constructive interference, the path difference must be an integral multiple of the wavelength.

Thus for a bright fringe to be at ‘y’,

nÎ»Â =Â yÂ dD

Or, ynthÂ =Â nÎ»Â Dd

WhereÂ n =Â Â±0,1,2,3â€¦..

TheÂ 0th fringe represents the central bright fringe.

Similarly, the expression for a dark fringe in Youngâ€™s double slit experiment can be found by setting the path difference as:

Î”lÂ =Â (n+12)Î»

This simplifies toÂ yn =Â (n+12)Î»Dd

Note that these expressions require that Î¸ be very small. HenceÂ yDÂ needs to be very small. This implies D should be very large and y should be small. This, in turn, requires that the formula works best for fringes close to the central maxima. In general, for best results,Â dDÂ must be kept as small as possible for a good interference pattern.

The Youngâ€™s double slit experiment was a watershed moment in scientific history because it firmly established that light indeed behaved as a wave.

The Double Slit Experiment was later conducted using electrons, and to everyoneâ€™s surprise, the pattern generated was similar as expected with light. This would forever change our understanding of matter and particles, forcing us to accept that matter like light also behaves like a wave.