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प्रश्न
Using the geometry of the double slit experiment, derive the expression for fringe width of interference bands.
उत्तर
Let S1 and S2 be the two sources of light separated by distance d. A screen is placed at a distance of D from the two light sources. Let 'A be the wavelength of light.
Draw S1 S'1 and S2 S'2 perpendicular to the screen. Let O' be the midpoint of S'1 S'2.
∴ `O"'"S"'"_1 = O"'"S"'"_1 = d/2`
Let 'P' be the point at distance y from point 'O' on screen.
Finding whether point P is bright or dark depends upon the path difference. The path difference is S2P − S1P.
In ΔS1S'1P,
(S1P)2 = (S1S'1)2 + (S'1P)2
`(S_1P)^2 = D^2 + (y - d/2)^2`
`(S_1P)^2 = D^2 + y^2 - yd + d^2/4` ...(i)
In ΔS2S'2P,
(S2P)2 = (S2S'2)2 + (S'2P)2
`(S_2P)^2 = D^2 + (y + d/2)^2`
`(S_2P)^2 = D^2 + y^2 + yd + d^2/4` ...(ii)
Subtracting equation (i) from (ii),
(S2P)2 − (S1P)2
= `(D^2 + y^2 + yd + d^2/4) - (D^2 + y^2 - yd + d^2/4)`
∴ (S2P + S1P)(S2P − S1P) = 2yd
We know (S2P − S1P) = Path difference
∴ Path difference = `(2yd)/((S_P + S_1P))`
Now, D is very large as compared to y and
i.e. D >> y and D >> d.
∴ S2P ≈ S1P ≈ D
∴ Path difference = `(2yd)/(D + D) = (2yd)/(2D)`
∴ Path difference = `(yd)/D` ...(iii)
Case-I:
The point P will be bright if the path difference = nλ
`therefore (yd)/D = n lambda`
`y = (D n lambda)/d`
Equation for position of nth bright band is
`therefore y_n = (D n lambda)/d` ...(iv)
Case-II:
The point P will be dark if path difference = `(2n - 1)lambda/2`
`therefore (yd)/D = (2n - 1)lambda/2`
`therefore y = (D(2n - 1)lambda)/(2 d)`
The equation for position of the nth dark band is
`therefore y_n = (D(2n - 1)lambda)/(2d)` ...(v)
Fringe width (W): The distance between two successive bright or dark bands in an interference pattern is called fringe width or bandwidth.
∴ Fringe width (W) = yn + 1 − yn
From equation (iv),
Equation of nth bright band
`y_n = (D n lambda)/d`
The equation for (n + 1)th bright band
`y_(n + 1) = (D(n + 1)lambda)/d`
∴ Fringe width (W) = yn + 1 − yn
= `(D(n + 1)lambda)/d - (D n lambda)/d`
= `(D lambda)/d [(n + 1) - n]`
= `(D lambda)/d(1)`
∴ Fringe width (W) = `(lambda D)/d`
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