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प्रश्न
In young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence, deduce the expression for the fringe width.
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उत्तर
Young’s double slit experiment demonstrated the phenomenon of interference of light. Consider two fine slits S1 and S2 at a small distance d apart. Let the slits be illuminated by a monochromatic source of light of wavelength λ. Let GG′ be a screen kept at a distance D from the slits. The two waves emanating from slits S1 and S2 superimpose on each other resulting in the formation of an interference pattern on the screen placed parallel to the slits.
Let O be the centre of the distance between the slits. The intensity of light at a point on the screen will depend on the path difference between the two waves reaching that point. Consider an arbitrary point P at a distance x from O on the screen.
Path difference between two waves at P = S2P − S1P
The intensity at the point P is maximum or minimum as the path difference is an integral multiple of wavelength or an odd integral multiple of half wavelength
For the point P to correspond to maxima, we must have
S2P − S1P = n, n = 0, 1, 2, 3...
From the figure given above
`(S_2P)^2-(S_1P)^2=D^2+(x+d/2)^2-D^2+(x-d/2)^2`
On solving we get:
(S2P)2-(S1P)2=2xd
`S_2P-S_1P=(2xd)/(S_2P+S_1P)`
As d<<D, then S2P + S2P = 2D (∵ S1P = S2P ≡ D when d<<D)
`:.S_2P-S_1P=(2xd)/(2D)=(xd)/D`
Path difference, `S_2P-S_1P=(xd)/D`
Hence, when constructive interfernce occur, bright region is formed.
For maxima or bright fringe, path difference = `xd/D=nlambda`
i.e `x=(nlambdaD)/d`
where n=0,± 1, ±2,........
During destructive interference, dark fringes are formed:
Path difference, `(xd)/D=(n+1/2)lambda`
`x=(n+1/2)(lambdaD)/d`
The dark fringe and the bright fringe are equally spaced and the distance between consecutive bright and dark fringe is given by:
β = xn+1-xn
`beta=((n+1)lambdaD)/d-(nlambdaD)/d`
`beta=(lambdaD)/d`
Hence the fringe width is given by `beta = (lambdaD)/d`
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संबंधित प्रश्न
In Young's double slit experiment, plot a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits keeping other parameters same. What information can one obtain from the slope of the curve?
Write three characteristic features to distinguish between the interference fringes in Young's double slit experiment and the diffraction pattern obtained due to a narrow single slit.
In Young's double slit experiment, derive the condition for
(i) constructive interference and
(ii) destructive interference at a point on the screen.
Two transparent slabs having equal thickness but different refractive indices µ1 and µ2are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point P0 which is equidistant from the slits?
A double slit S1 − S2 is illuminated by a coherent light of wavelength \[\lambda.\] The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen ∑ is placed behind the double slit at a distance D2 from it (see the following figure). The screen ∑ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.
Consider the arrangement shown in the figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When \[z = \frac{D\lambda}{2d},\] the intensity measured at P is I. Find the intensity when z is equal to
(a) \[\frac{D\lambda}{d}\]
(b) \[\frac{3D\lambda}{2d}\] and
(c) \[\frac{2D\lambda}{d}\]
In Young’s double-slit experiment, show that:
`beta = (lambda "D")/"d"` where the terms have their usual meaning.
The force required to double the length of a steel wire of area 1 cm2, if its Young's modulus Y= 2 × 1011/m2 is:
Two beams of light having intensities I and 41 interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π/2 and π/3 at points A and B respectively. The difference between the resultant intensities at the two points is xl. The value of x will be ______.
- Assertion (A): In Young's double slit experiment all fringes are of equal width.
- Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).
