Question

Derive the expression for the fringe width of interference pattern in Young’s double-slip experiment.

Answer 1

In the figure, A and B represent the two slits through which the light passes. A screen is placed at a distance D. When monochromatic light passes through the slits, the spherical wavefronts will interfere. As a result, we will get a pattern on the screen.

We know that there is a difference in the paths traveled by the two rays of light AC and BC. This difference in the path traveled by the rays is called the path difference and it can be obtained by considering the two triangles ΔABE and ΔACE.

Here the path difference is given by, BE = ABsinθ \((\because \sin\theta = \frac{BE}{AB})\)

We know that AB is the distance between the two slits and it can be taken as d .

Let us denote the path difference as Δx

Now we can write the path difference as,

⇒ Δx = dsinθ   … equation

For small values of θ we can write sinθ ≈ tanθ

From the figure we can write, sinθ ≈ tanθ = \(\frac YD\)

Substituting in the equation, we get

⇒ \(Δx = d(\frac YD)\)

⇒ \(\frac{Δx}d = \frac YD\)

We assume that Y is the distance between the centre of the screen and the point where the fringe is formed.

Then,

\(Y =\frac{\Delta xD}d\)

The separation between two consecutive dark fringes or bright fringes is called the fringe width.

We can write the path difference as, Δx = nλ

Where n is an integer called the order of the fringe and λ is the wavelength of the monochromatic source of light.

Now we can find the separation between two fringes using the expression

⇒ \(Y = \frac {n\lambda D}d\)

For an n^th order fringe, \(Y_n = \frac {n\lambda D}d\)

And the (n+1)^th order fringe, Y_n+1 = \(\frac{(n+1)\lambda D}d\)

Let β be the separation between two consecutive bright or dark fringe then

The fringe width can be written as,

⇒ \(\beta = \frac{(n+ 1)\lambda D}d - \frac{n\lambda D}d\)

⇒ \(\frac{\lambda D}d (n + 1 - n)\)

Therefore we get,

⇒ \(\beta = \frac{\lambda D}d\)

Answer 2

S₁ & S₂ are two coherent sources. GG – screen at a distance D from the sources. 

Let d = distance between the slits (two sources). 

P = a point at a distance from the middle of the screen where a bright fringe is formed. 

For constructive interference path difference = nλ 

S₂P – S₁P = nλ where n = 0,1,2,3…….

From the diagram

(S₂P)² – (S₁P)² = 2xd 

(S₂P – S₁P)(S₂P + S₁P) = 2xd 

S₂P ≅ S₁P ≅ OP = D 

S₂P – S₁P = \(\frac{xd}{D}\) = nλ 

X_n = \(\frac{n\lambda D}{2}\)

Where n = 0, +1, ±2, ± 3 for bright fringes ±1, ±2, ±3 for dark fringes 

Since the fringes are equally spaced the distance between two consecutive bright or consecutive dark fringes gives fringe width.