Young’s double slit experiment derivation

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 :

1. Two sources should be coherent and
2. 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 S0 and arranged two very narrow slits S₁ and S₂ close to each other in front of slit S0 young’s double slit experiment derivation diagram below. Slits S₁ and S₂ are equidistant from S0, so the spherical wavefronts emitted by slit S0 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. S0 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 S1 and S2 behave as two coherent sources. Interference takes place on the screen. If we consider a point O on the perpendicular bisector of S₁S2, 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 S1P and S2P. Now draw S1N normal on S2P. Then the path difference between S2P and S1P

Now from S1PL,

and from S2PM,

Since the distance of screen from slits S1 and S2 is very large, so S2P ≈S1P ≈D

Path difference,

Maxima or Bright fringes

If the path difference (S2P-S1P) = 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 (S2P-S1P)=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