Let S1 and S2 are two coherent sources separated by a distance d and made from a monochromatic source 5. Let a screen be placed at a distance from the coherent source as shown in Figure 4.1 point 0. On the screen is equidistance from 51 and 52.
. . As such path difference at 0 will be zero and the intensity at 0 will be maximum.
Consider a point Pat a distance y from 0. The wave reaches the point P from S1 and S2.
From Figure 4.1, PQ = (y- d/2) and PR = (y + d/2)
(S 2 p- S 1 P)” [ D + (y +d/2)2/2d’) – [ D + (y +d/2) ]2
∆.= yd/ D
Corresponding phase difference will be
∆ θ = 2π /λ ∆ =2π /λ (yd /d)
Position of bright fringes:
Path difference = even multiple of (‘λ/2)
Yd/d = 2nλ/2 = nλ
which gives position of different maximum at
y 0 = 0, y1 =λd/d ,y2= 2λd/d ,y3 =3λd/d,…
∆y =y1 –y0 =y2 –y1
=…=yn –yn-1 =λd/d
Position of dark fringes:
Path difference odd multiple of (λ../2)
yd /d = (2n + 1) λ/2
y n =(2n +1)λd/2d
which gives the position of different minima
Y0= λd/2d, Y1 = 3λd/2d, y3 = 5λd/2d
∆y = Y1-Y0= Y2-Y1 = Y3-Y2 = …… = Yn-Yn-1- = λd/d … (9)
Fringe width: The distance between any two consecutive bright or dark fringes is called
fringe width and is denoted by β.
(β) right = (∆y)min =λd/d
(β)dark = (∆Y)max =λd/d
(β) = (β)bright = (β)dark =λd/d
It is found that
(β) r λ: β r d and β r 1/d
Angular fringe width e11: It is given by
θβ =θ n+1 –θ =y n-1/d –yn /d =β/d
1. At the centre we get bright fringe.
2. Then alternately dark, bright, dark …… fringe on both sides of the central maximum.
3. Fringe width of dark and bright fringe are same.
4. Width of fringe depends upon wavelength, separation of sources and source to screen separation.
If white light is used each wavelength will give its own fringe pattern upto a short distance from 0 the fringes of different colour would stand side by side but beyond that a
general illumination is observed. The central fringe with white light is white as for all colours
the central fringe is formed at 0 and they together produce white light.