Question

Using the geometry of the double slit experiment, derive the expression for fringe width of interference bands.

Answer 1

Let S₁ and S₂ be the two sources of light separated by distance d. A screen is placed at a distance of D from the two light sources. Let 'A be the wavelength of light.

Draw S₁ S'₁ and S₂ S'₂ perpendicular to the screen. Let O' be the midpoint of S'₁ S'₂. 

∴ `O"'"S"'"_1 = O"'"S"'"_1 = d/2`

Let 'P' be the point at distance y from point 'O' on screen.

Finding whether point P is bright or dark depends upon the path difference. The path difference is S₂P − S₁P.

In ΔS₁S'₁P,

(S₁P)² = (S₁S'₁)² + (S'₁P)²

`(S_1P)^2 = D^2 + (y - d/2)^2`

`(S_1P)^2 = D^2 + y^2 - yd + d^2/4`    ...(i)

In ΔS₂S'₂P,

(S₂P)² = (S₂S'₂)² + (S'₂P)²

`(S_2P)^2 = D^2 + (y + d/2)^2`

`(S_2P)^2 = D^2 + y^2 + yd + d^2/4`    ...(ii)

Subtracting equation (i) from (ii),

(S₂P)² − (S₁P)²

= `(D^2 + y^2 + yd + d^2/4) - (D^2 + y^2 - yd + d^2/4)`

∴ (S₂P + S₁P)(S₂P − S₁P) = 2yd

We know (S₂P − S₁P) = Path difference

∴ Path difference = `(2yd)/((S_P + S_1P))`

Now, D is very large as compared to y and

i.e. D >> y and D >> d.

∴ S₂P ≈ S₁P ≈ D

∴ Path difference = `(2yd)/(D + D) = (2yd)/(2D)`

∴ Path difference = `(yd)/D`    ...(iii)

Case-I:

The point P will be bright if the path difference = nλ

`therefore (yd)/D = n lambda`

`y = (D n lambda)/d`

Equation for position of n^th bright band is

`therefore y_n = (D n lambda)/d`    ...(iv)

Case-II:

The point P will be dark if path difference = `(2n - 1)lambda/2`

`therefore (yd)/D = (2n - 1)lambda/2`

`therefore y = (D(2n - 1)lambda)/(2 d)`

The equation for position of the n^th dark band is

`therefore y_n = (D(2n - 1)lambda)/(2d)`    ...(v)

Fringe width (W): The distance between two successive bright or dark bands in an interference pattern is called fringe width or bandwidth.

∴ Fringe width (W) = y_{n + 1} − y_n

From equation (iv),

Equation of n^th bright band

`y_n = (D n lambda)/d`

The equation for (n + 1)^th bright band

`y_(n + 1) = (D(n + 1)lambda)/d`

∴ Fringe width (W) = y_{n + 1} − y_n

= `(D(n + 1)lambda)/d - (D n lambda)/d`

= `(D lambda)/d [(n + 1) - n]`

= `(D lambda)/d(1)`

∴ Fringe width (W) = `(lambda D)/d`