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Question
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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Solution
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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