One of the first demonstration of the intererference of light waves was given by Young – an English physicist in 1801. We have learnt that two essential conditions to obtain an interference phenomenon are :

- Two sources should be coherent and
- Two coherent sources must be placed close to each other as the wavelength of light is very small.

**Young’s double slit experiment derivation**

Young placed a monochromatic source (S) of light in front of a narrow slit S_{0} and arranged two very narrow slits S₁ and S₂ close to each other in front of slit S_{0} **young’s double slit experiment derivation diagram** below. Slits S₁ and S₂ are equidistant from S_{0}, so the spherical wavefronts emitted by slit S_{0} reach the slits S₁ and S₂ in equal time.

These wavefronts after arriving at S₁ and S₂ spread out of these slits. Thus the emerging waves are of the same amplitude and wavelength and are in phase. Hence slits S₁ and S₂ behave as coherent sources.

The wavefronts emitted by coherent sources S₁ and S₂ superpose and give rise to interference. When these wavefronts are received on the screen, interference fringes are seen as shown in **young’s double slit experiment diagram** below.

The points where the destructive interference takes place, we get minima or dark fringe and where the constructive interference takes place, maxima or bright fringe is obtained. The pattern of these dark and bright fringes obtained on the screen is called interference pattern.

Young had used sun light as source of light and circular slits in his experiment.

**Theory of the Experiment**

Suppose S is the monochromatic source of light. S_{0} is the slit through which the light passes and illuminates the slits S₁ and S₂. The waves emitted by slits S₁ and S₂ are the part of the same wavefront, so these waves have the same frequency and the same phase.

Hence slits S_{1} and S_{2} behave as two coherent sources. Interference takes place on the screen. If we consider a point O on the perpendicular bisector of S₁S_{2}, the waves traveling along S₁O and S₂O have traveled equal distances. Hence they will arrive at O in phase and interfere constructively to make O the centre of a bright fringe or maxima.

## Derivation of Young’s double slit experiment

To locate the position of the maxima and minima on both sides of O, consider any point P at a distance x from O. Join S_{1}P and S_{2}P. Now draw S_{1}N normal on S_{2}P. Then the path difference between S_{2}P and S_{1}P

Now from **△**S_{1}PL,

and from **△**S_{2}PM,

Since the distance of screen from slits S_{1} and S_{2} is very large, so S_{2}P ≈S_{1}P ≈D

Path difference,

**Maxima or Bright fringes**

If the path difference (S_{2}P-S_{1}P) = xd/D is an integral multiple of λ, then the point P will be the position of bright fringe or maxima.

That is for bright fringe,

Eqn. (1) gives the position of different bright fringes.

IF

**P = 0, x =0**, i.e., the central fringe at O will be bright.

IF

This is the position of first bright fringe w.r.t. point O.

IF

This is the position of second bright fringe w.r.t. point O.

………………………………………………………………………………………

………………………………………………………………………………………

IF

This is the position of **pth** bright fringe w.r.t. point O.

IF

This is the position of **(p+1)** bright fringe w.r.t. point O.

The distance between two successive bright fringes is called **fringe width** and is given by

**Minima or Dark fringes**

If the path difference** (S _{2}P-S_{1}P)=xd/D** is an odd multiple of

**λ/2**, then the point P will be the position of dark fringes or minima.

Thus for dark fringes,

Eqn. (3) gives the position of different dark fringes.

IF

This is the position of first dark fringe w.r.t. point O.

IF

This is the position of second dark fringe w.r.t. point O.

IF

This is the position of third dark fringe w.r.t. point O.

…………………………………………………………………………….

…………………………………………………………………………….

IF

This is the position of **pth** dark fringe w.r.t. point O.

If

This is the position of **(p+1)** dark fringe w.r.t. point O.

The distance between two successive dark fringes is called fringe width **(β)** of the dark fringes which is given by

This eqn. (4), ‘**β**=**λD/d’** is called **Fringe width formula in Young’s experiment**.

From eqns. (2) and (4), it is evident that the fringes width of bright fringe and dark fringe is the same.

If we know the value of “D” and “d” then the measurement of the fringe width **( β)** gives a direct determination of the wavelength of light.

**FAQ on Young’s double slit experiment derivation**

### Double slit experiment formula?

In a double-slit experiment, **λ= xd / L** is the formula for the calculation of wavelength.

### Fringe width formula in Young’s experiment?

If we know the value of “D” and “d” then the measurement of the fringe width **( β)** gives a direct determination of the wavelength of light. Fringe width formula in Young’s experiment is given by:

**β**=

**λD/d**