Young’s Double slit experiment:
Coherent sources are those which have exactly the same frequency and are in this same phase or have a constant difference in phase.
Conditions: (i) The sources should be monochromatic and originating from common single source.
(ii) The amplitudes of the waves should be equal.
Expression for Fringe Width: Let S1 S2 and be two coherent sources separated by a distance d. Let the distance of the screen from the coherent sources be D. Let M be the foot of the perpendicular drawn from O, the midpoint of S1 and S2 on the screen. Obviously point M is equidistant from S1 and S2. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity. Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2 P
The path difference between two waves reaching at P from S1 and S2 is Δ =S2 P -S1 P ≈ S2N As D > > d, therefore ∠ S2 S1N = θ is very small
(i) Positions of bright fringes (or maxima): For bright fringe or maximum intensity at P, the path difference must be an integral multiple of wavelength (λ) of light used. i.e. Δ = nλ
∴ yd/D = nλ, n = 0,1,2,3, ....
∴ y = ndλ/d.
This equation gives the distance of nth bright fringe from the point M. Therefore writing yn for y, we get
yn = nDλ/d ...(ii)
(ii) Positions of dark fringes (or minima): For dark fringe or minimum intensity at P, the path difference must be an odd number multiple of half wavelength. i.e. Δ =(2n -1)λ/2
This equation gives the distance of nth dark fringe from point M. Therefore writing yn for y, we get
(iii) Fringe Width β: The distance between any two consecutive bright fringes or any two consecutive dark fringes is called the fringe width. It is denoted by ω.
For Bright Fringes: If yn + 1 and yn denote the distances of two consecutive bright fringes from M, then we have
For Dark Fringes: If yn + 1 and yn are the distances of two consecutive dark fringes from M, then we have
Thus, fringe width is the same for bright and dark fringes equal to
β = Dλ/d
The condition for the interference fringes to be seen is
s/b < λ/d