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Describe Young’s double slit experiment to produce interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width.

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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 Son 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   

y= 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 ydenote 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

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