In 1801, an English physician and physicist established the principle of interference of light, where he made a pinhole camera in cardboard and allowed sunlight to pass through it.

This white light was then allowed to fall upon another cardboard having two pin holes placed together symmetrically.

The emerging light was received on a plane screen placed at some distance.

At a given point on screen the waves emerging from two holes had different phases, interfering to give a pattern of bright and dark areas.

Such a variation of intensity on the plane screen demonstrated the light waves emerging from the two holes.

Therefore, this pattern of bright (constructive fringe) and dark (destructive fringe) areas can be sharply defined only if the light of a single wavelength is used.

(Image to be added soon)

The two waves interfering at P have covered different distances.

(Image to be added soon)

Let the waves from two coherent sources of light be represented as

y1 = a sinωt…(1)

y2 = b Sin(ωt + Φ)...(2)

Here, a and b are amplitudes of the two waves resp.

Φ is the constant phase angle by which the second wave leads the first wave.

Applying the superposition principle, the displacement(y) of the resultant wave at time (t) would be:

y = y1 + y2 = a sinωt + b sin(ωt + Φ)

Expanding sin(ωt + Φ) = sin ωt cosΦ + cosωt . sinΦ

= sinωt (a+ b cosΦ ) + cosωt . b sinΦ

Put = a + b cosΦ = R cosӨ ….(3)

and b sinΦ = R sinӨ ….(4)

y = sinωt. R cosӨ + cosωt . R sinӨ

We get,

y = R sin(ωt + Ө)......(4)

The wave equation (4) represents the harmonic wave of amplitude R.

Now, squaring (3) and (4) and adding, we get

R2 (cos2Ө + sin2Ө) = (a + b cosΦ)2+ (b sinΦ)2

R2.1 = a2+ b2 Cos2Φ + 2ab cosΦ + b2Sin2 Φ

a2+ b2(Sin2 Φ+Cos2Φ)+2abcosΦ

R = √a2+ b2 + 2ab cosΦ

Since, intensity I α amplitude R2.

I α a2+ b2 + 2ab cosΦ

For constructive fringe:

I should be maximum for which cosΦ = max or +1; Φ = 0, 2π, 4π…

I.e.,

Φ = 2nπ

(n = integer such that n = 0,1,2…)

If x is the path difference between the two waves reaching point P (in Fig.2) corresponding to phase difference Φ, then,

x = λ/2π Φ = λ/2π ( 2nπ)

x = nλ….(p)

For destructive interference:

I should be minimum i.e., CosΦ= minimum when Φ = -1 or π, 3π, 5π….

So,

Φ = (2n - 1)π

The path difference will be:

x = λ/2π Φ = λ/2π (2n - 1)π

x = λ/2(2n - 1)...(q)

(Image to be added soon)

Let S1 and S2 be two slits separated by a distance d, and the center O equidistant from S1 and S2.

Let the slits be illuminated by a monochromatic source S of light of wavelength λ.

Consider a point P at a distance y from C.

Here, O is the midpoint of S1 and S2, and

D = OP₀ and D >> d

As S1S2 are perpendicular to OP₀ and S1A nearly perpendicular to O., we have

∠ S₁ S₂A = ∠ POP₀ = Ө

The path difference between two waves approaching at P is,

Δ x = S₂P - S₁P = S₂P - PA (Since D>>d)

= S₂A = d SinӨ

= d tanӨ = dy/D

Δ x = dy/D = nλ (where n is an integer)

So,

y = nDλ/d…..(a)

At n = 0, y = 0

n = 1, y = Dλ/d

n = 2, y = 2Dλ/d

n =3, y = Dλ/d,...etc

The centers of the dark fringes will be obtained when

Δ x = dy/D = (2n - 1)λ

y = (2n -1)Dλ/d….(b)

At n =1, y = Dλ/2d

n =2, y = 3Dλ/2d,

n = 3, y = 5Dλ/2d,...etc

Now, to find the fringe width, subtracting equation (b) from (a), we get

Fringe width, w = (2n -1)Dλ/d - nDλ/d = Dλ/d

If a glass slab of refractive index μ and thickness t is introduced on one of the paths of interfering waves, the optical length of this path will become μ instead of t, increasing by (t-1)μ.

This generates a path difference, given by,

Δx = μt - t = (t-1)μ

This path difference comes due to the glass slab.

If current position of fringe is y =D/d (Δx ), the new position will be

y’ = D/d(Δx + (t-1)μ)

Lateral shift of fringe:

y_{₀ } = y’ - y

= D/d(t-1)μ

As, fringe width, w = D/dλ

We get,

y_{₀} = β/λ (t-1)μ

FAQ (Frequently Asked Questions)

1. What is Fringe Width?

Fringe width is the distance between two consecutive dark and bright fringes and is denoted by a symbol, β.

Fringe width depends on the following factors that are outlined below:

The wavelength of light.

The distance between the slits and the screen or slit separation.

What is Fringe Width in YDSE?

The distance between any two consecutive dark or bright fringes and all the fringes are of equal lengths. Fringe width is given by, β = D/dλ.

The angular width, Ө = λd = βD.

3. How is Fringe Width Calculated?

Let’s say the wavelength of the light is 6000 Å. The distance between the two slits is d = 0.8 x 10^{-3} m . Distance (D) between slit and screen is 1.2 m. The fringe width will be calculated by the formula: β = Dλ/d = 1.2 x 6 x 10^{-7}/0.8 x 10^{-3} ( 1 Å = 10^{-10}m)

On calculating, we get

β = 9 x 10^{-4}m

4. What is The Ratio of Fringe Width For Bright And Dark Fringes?

In Young’s double slit experiment, dark and bright fringes are equally spaced. Therefore, the ratio of fringe width for dark to bright fringes is 1.