# Describe Young's double-slit interference experiment and derive conditions for occurence of dark and bright fringes on the screen. Define fringe width and derive a formula for it. - Physics

Describe Young's double-slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.

#### Solution

Description of Young's double-slit interference experiment:

1. A plane wavefront is obtained by placing a linear source S of monochromatic light at the focus of a convex lens. It is then made to pass through an opaque screen AB having two narrow and similar slits S1 and S2. S1 and S2 are equidistant from S so that the wavefronts starting simultaneously from S and reaching
S1 and S2 at the same time are in phase. A screen PQ is placed at some distance from screen AB as shown in following figure.

Young's double-slit experiment
2. S1 and S2 act as secondary sources. The crests/- troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in the above figure. The point where these lines meet the screen have high intensity and is bright.
3. Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference. These dark and bright regions are called fringes or bands and the whole pattern is called an interference pattern.

Conditions for occurrence of dark and bright lunges on the screen:

Consider Young's double-slit experimental set up. Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength λ emerges out of two narrow and closely spaced, parallel slits S1 and S2 of equal widths. The separation S1 S2 = d is very small. The interference pattern is observed on a screen placed parallel to the plane of S1S2 and at considerable distance.D (D » d) from the slits. OO' is the perpendicular bisector of a segment S1S2

Geometry of the double-slit experiment

Consider, a point P on the screen at a distance y from O' (y « 0). The two light waves from S1 and S2 reach P along paths S1P and S2P, respectively. If the path difference (Δl) between S1P and S2P is an integral multiple of λ, the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between S1P and S2P is a half-integral multiple of λ, there will be destructive interference and a dark fringe will be produced at P.

From above figure,

(S2P)2 = (S2S2')2 + (PS2')2

= (S2S2')2 + (PO' + O'S2')2

= "D"^2 + ("y" + "d"/2)^2       ....(1)

and (S1P)2 = (S1S1')2 + (PS1')2

= (S1S1')2 + (PQ' - Q'S1)2

= "D"^2 + ("y" - "d"/2)^2     .....(2)

(S2P)2 - (S1P)2 = {"D"^2 + ("y" + "d"/2)^2} - {"D"^2 + ("y" - "d"/2)^2}

∴ (S2P + S1P)(S2P - S1P)

= ["D"^2 + "y"^2 + "d"^2/4 + "yd"] - ["D"^2 + "y"^2 + "d"^2/4 - "yd"] = 2"yd"

∴ S2P + S1P = Δ l = 2yd/S2P + S1P

In practice, D » y and D » d,

∴ S2P + S1P ≅ 2D

∴ Path difference,

Δ l = S2P + S1P ≅ 2 "yd"/"2D" = "y" "d"/"D"    ....(3)

The expression for the fringe width (or band width) : The distance between consecutive bright (or dark) fringes is called the fringe width (or bandwidth) W. Point P will be bright (maximum intensity), if the

path difference, Δ l = "y"_"n" "d"/"D" = "n" lambda where n = 0, 1, 2, 3, .....

Point P will be dark (minimum intensity equal to zero), if "y"_"m" "d"/"D" = ("2m" - 1) lambda/2, where, m = 1,2,3...,

Thus, for bright fringes (or bands),

"y"_"n" = 0, lambda "D"/"d", (2lambda"D")/"d" ...

and for dark fringes (or bands),

"y"_"n" = lambda/2 "D"/"d", 3 lambda/2 "D"/"d", 5lambda/2 "D"/"d" ....

These conditions show that the bright and dark fringes (or bands) occur alternately and are equally spaced. For Point O', the path difference (S2O' - S1O') = 0. Hence, point O' will be bright. It corresponds to the centre of the central bright fringe (or band). On both sides of O', the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit.

Let "y"_"n" and "y"_"n + 1", be the distances of the nth and (m + 1)th bright fringes from the central bright fringe.

∴ ("y"_"n""d")/"D" = "n" lambda

∴ "y"_"n" = ("n" lambda "D")/"d"    .....(4)

and ("y"_("n + 1")"d")/"D" = ("n + 1")lambda

∴ ("y"_("n + 1")) = (("n + 1") lambda "D")/"d"   .....(5)

The distance between consecutive bright fringes

= "y"_("n + 1") - "y"_"n" = (lambda "D")/"d" [("n + 1") - "n"] = (lambda"D")/"d"    ....(6)

Hence, the fringe width,

∴ W = triangle "y" = "y"_("n + 1") - "y"_"n" = (lambda"D")/"d" (for bright fringes) ... (7)

Alternately, let "y"_"m" and "y"_"m + 1" be the distances of the m th and (m + 1)th  dark fringes respectively from the central bright fringe.

∴ ("y"_"m""d")/"D" = (2"m" - 1) lambda/2 and

("y"_("m+1")"d")/"D" = [2("m + 1") - 1] lambda/2 = (2"m" + 1) lambda/2    ....(8)

∴ "y"_"m" = (2"m - 1") (lambda"D")/"2d" and

"y"_"m + 1" = (2"m" + 1) (lambda"D")/"2d"    .....(9)

∴ The distance between consecutive dark fringes,

"y"_"m + 1" - "y"_"m" = (lambda"D")/"2d" [(2"m" + 1) - (2"m" - 1)] = (lambda"D")/"d"    ....(10)

∴ W = "y"_"m + 1" - "y"_"m"

= (lambda"D")/"d" (for dark fringes)         .....(11)

Eqs. (7) and (11) show that the fringe width is the same for bright and dark fringes.

Concept: Interference
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 7 Wave Optics
Exercises | Q 6 | Page 184