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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.
RELATED QUESTIONS
A narrow slit S transmitting light of wavelength λ is placed a distance d above a large plane mirror, as shown in the following figure. The light coming directly from the slit and that coming after the reflection interfere at a screen ∑ placed at a distance D from the slit. (a) What will be the intensity at a point just above the mirror, i.e. just above O? (b) At what distance from O does the first maximum occur?
A long narrow horizontal slit is paced 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit. Find the fringe-width if the light used has a wavelength of 700 nm.
Answer in brief:
In Young's double-slit experiment what will we observe on the screen when white light is incident on the slits but one slit is covered with a red filter and the other with a violet filter? Give reasons for your answer.
Why two light sources must be of equal intensity to obtain a well-defined interference pattern?
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to ______.
What is interference of light?
How does wavefront division provide coherent sources?
How do source and images behave as coherent sources?
Explain Young’s double-slit experimental setup and obtain the equation for path difference.
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio `("I"_"max" - "I"_"min")/("I"_"max" + "I"_"min")` will be ______
In Young's double-slit experiment, in an interference pattern, a second minimum is observed exactly in front of one slit. The distance between the two coherent sources is 'd' and the distance between source and screen is 'D'. The wavelength of the light source used is ______
A thin transparent sheet is placed in front of a slit in Young's double slit experiment. The fringe width will ____________.
Band width for red light of wavelength 6600 Å is 0.42 mm. If red Light is replaced by blue light of wavelength 4400 Å, then the change m bandwidth will be ____________.
In Young's experiment for the interference of light, the separation between the silts is d and the distance of the screen from the slits is D. If D is increased by 0.6% and d is decreased by 0.2%, then for the light of a given wavelength, which one of the following is true?
"The fringe width ____________."
In a Young's double-slit experiment, the intensity at a point where the path difference is `lambda/3` (`lambda` being the wavelength of the light used) is I. If I0 denotes the maximum intensity, then `"I"/"I"_0` is equal to ______.
In a Young's experiment, two coherent sources are placed 0.60 mm apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be ____________.
In Young's double-slit experiment, an interference pattern is obtained on a screen by a light of wavelength 4000 Å, coming from the coherent sources S1 and S2 At certain point P on the screen, third dark fringe is formed. Then the path difference S1P - S2P in microns is ______.
In interference experiment, intensity at a point is `(1/4)^"th"` of the maximum intensity. The angular position of this point is at (sin30° = cos60° = 0.5, `lambda` = wavelength of light, d = slit width) ____________.
If the two slits in Young's double slit experiment have width ratio 9 : 1, the ratio of maximum to minimum intensity in the interference pattern is ______.
In a biprism experiment, the slit separation is 1 mm. Using monochromatic light of wavelength 5000 Å, an interference pattern is obtained on the screen. Where should the screen be moved? so that the change in fringe width is 12.5 x 105 m?
If two light waves reaching a point produce destructive interference, then the condition of phase difference is ______
The graph shows the variation of fringe width (β) versus distance of the screen from the plane of the slits (D) in Young's double-slit experiment Keeping other parameters the same. The wavelength of light used can be calculated as d = distance between the slits ______
In Young's double slit experiment, for wavelength λ1 the nth bright fringe is obtained at a point P on the screen. Keeping the same setting, source of light is replaced by wavelength λ2 and now (n + 1)th bright fringe is obtained at the same point P on the screen. The value of n is ______.
Two waves with same amplitude and frequency superpose at a point. The ratio of resultant intensities when they arrive in phase to that when they arrive 90° out of phase is ______.
`[cos pi/2=0]`
What is meant by Constructive interference?
The path difference between two interference light waves meeting at a point on the screen is `(87/2)lambda`. The band obtained at that point is ______.
With a neat labelled ray diagram explain the use of Fresnel's biprism to obtain two coherent sources.
