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Question
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.
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Solution
Description of Young's double-slit interference experiment:
- a plane wavefront is made to fall on an opaque screen AB having two similar narrow slits S1 and S2.
- The plane wavefront can be either obtained by placing a linear source S far away from the screen or by placing it at the focus of a convex lens kept close to AB.
- The rays coming out of the lens will be parallel rays and the wavefront will be a plane wave front as shown in Figure.
- The figure shows a cross-section of the experimental set up and the slits have their lengths perpendicular to the plane of the paper. For better results, the slits should be about 2-4 mm apart from each other. An observing screen PQ is placed behind of AB.
- For simplicity, we assume that the slits S1 and S2 are equidistant from the S so that the wavefronts starting from S and reaching the S1 and S2 at every instant of time are in phase.
Young's double-slit experiment - 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.
- 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 the occurrence of dark and bright lunges on the screen:
Consider Young's double-slit experimental set up. Wavefront splitting produces two narrow coherent light sources as monochromatic light of wavelength emerges from 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 a 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 the 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"` ....
The bright and dark fringes (or bands) alternate and are evenly spaced in these situations. 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.
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Answer in brief:
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