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प्रश्न
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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उत्तर
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.
संबंधित प्रश्न
State any one difference between interference of light and diffraction of light
Four light waves are represented by
(i) \[y = a_1 \sin \omega t\]
(ii) \[y = a_2 \sin \left( \omega t + \epsilon \right)\]
(iii) \[y = a_1 \sin 2\omega t\]
(iv) \[y = a_2 \sin 2\left( \omega t + \epsilon \right).\]
Interference fringes may be observed due to superposition of
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
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. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen?
The intensity at the central maximum (O) in a Young’s double slit experimental set-up shown in the figure is IO. If the distance OP equals one-third of the fringe width of the pattern, show that the intensity at point P, would equal `(I_0)/4`.
What are the two methods for obtaining coherent sources in the laboratory?
What is interference?
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Obtain the relation between phase difference and path difference.
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Obtain the equation for resultant intensity due to interference of light.
Explain Young’s double-slit experimental setup and obtain the equation for path difference.
Two independent monochromatic sources cannot act as coherent sources, why?
The light waves from two independent monochromatic light sources are given by, y1 = 2 sin ωt and y2 = 3 cos ωt. Then the correct statement is ____________.
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 ______.
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Two coherent light sources of intensity ratio 'n' are employed in an interference experiment. The ratio of the intensities of the maxima and minima in the interference pattern is (I1 > I2).
In biprism experiment, the 4th dark band is formed opposite to one of the slits. The wavelength of light used 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 ______
Waves from two coherent sources of light having an intensity ratio I1 : I2 equal to 'x' interfere. Then in the interference pattern obtained on the screen, the value of (Imax - Imin)/(Imax + Imin) is ______
Young's double slit experiment is performed in water, instead of air, then fringe width ______.
In Young's double-slit experiment, the distance between the slits is 3 mm and the slits are 2 m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength 480 nm and 600 run respectively. The separation on the screen between the 5th order bright fringes on the two interference patterns is ______
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Interference fringes are produced on a screen by using two light sources of intensities I and 9I. The phase difference between the beams is `pi/2` at point P and π at point Q on the screen. The difference between the resultant intensities at point P and Q is ______.
Two coherent sources P and Q produce interference at point A on the screen where there is a dark band which is formed between 4th bright band and 5th bright band. Wavelength of light used is 6000 Å. The path difference between PA and QA is ______.
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 ______.
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