Wave Optics - Young's Experiment

Description:

It was carried out in 1802 by the English scientist Thomas Young to prove the wave nature of light.

Two slits S₁ and S₂ are made in an opaque screen, parallel and very close to each other. These two are illuminated by another narrow slit S and light falls on both S₁ and S₂ which behave like coherent sources.

Note that the coherent sources are derived from the same sources. In this way, any phase which occurs in S will occur in both S₁ and S₂. The phase difference (∅₁ - ∅₂) between S₁ and S₂ is unaffected and remains constant.

Light now spreads out from both S₁ and S₂ and falls on a screen. It is essential that the waves from the two sources overlap on the same part of the screen. If one screen is covered up, the other produces a wide smoothly illuminated patch on the screen. But when both slits are open there is formation of interference fringes.

Where,

• Φ = Phase difference

• I = Resultant intensity

• I₁,I₂ = Intensity of individual waves.

Condition for bright fringes or maxima:

Φ = 2nΠ or p = nλ path difference

Where,

n = 0, 1, 2, …………

Condition for dark fringes or minima:

Φ = (2n - 1)Π or p = [n - 1/2]λ, path difference

where,

n = 1, 2, 3, …………

The relation between phase difference (Φ) and the path difference (p) is given by

How to find the position of the n^th Maxima or Minima on the screen?

Let P be the position of the n^th Maxima on the screen. The two waves arriving at P follow the path S₁P and S₂P, thus the path difference between the two waves is

p = S₁P - S₂P = dsinθ

From experimental conditions, we know that D >> d, therefore, the angle θ is small,

sinθ = tanθ = Y_n/D

Thus,

p = dsinθ = dtanθ = d(Y_n/D) ⇒ Y_n = pD/d

For n^th maxima,

p = nλ

For n^th minima

p = (n - 1/2)λ

Note − The n^th minima comes before the n^th maxima.