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Question
Describe geometry of the Young’s double slit experiment with the help of a ray diagram. What is fringe width? Obtain an expression of it. Write the conditions for constructive as well as destructive interference.
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Solution
i. Let S1 and S2 be the two coherent monochromatic sources that are separated by short distance d. They emit light waves of wavelength λ.
ii. Let D = horizontal distance between screen and source.
iii. Draw S1M and S2N ⊥ AB
OP = perpendicular bisector of the slit.
Since S1P = S2P, the path difference between waves reaching P from S1 and S2 is zero, therefore there is a bright point at P.
iv. Consider a point Q on the screen which is at a distance y from the central point P on the screen. Light waves from S1 and S2 reach at Q simultaneously by covering paths S1Q and S2Q, where they superimpose. 
Derivation:
In Δ S1MQ, (S1Q)2 = (S1M)2 + (MQ)2
`("S"_1"Q")^2 = "D"^2 + ["y" - "d"/2]^2` ....(1)
In `Δ "S"_2"NQ", ("S"_2"Q")^2 = ("S"_2"N")^2 + ("NQ")^2`
∴ `("S"_2"Q")^2 = "D"^2 + ["y" + "d"/2]^2` ....(2)
Subtract equation (1) from (2),
`("S"_2"Q")^2 - ("S"_1"Q")^2 = ["D"^2 + ("y" + "d"/2)^2] - ["D"^2 + ("y" - "d"/2)^2]`
= `"D"^2 + ("y" + "d"/2)^2 - "D"^2 - ("y" - "d"/2)^2`
= `("y" + "d"/2)^2 - ("y" - "d"/2)^2`
= `("y"^2 + "d"^2/4 + "yd") - ("y"^2 + "d"^2/4 - "yd")`
= `"y"^2 + "d"^2/4 + "yd" - "y"^2 - "d"^2/4 + "yd"("S"_2"Q")^2 - ("S"_1"Q")^2`
= 2yd
∴ `("S"_2"Q" + "S"_1"Q")("S"_2"Q" - "S"_1"Q")` = 2yd
∴ `"S"_2"Q" - "S"_1"Q" = (2"yd")/("S"_2"Q" + "S"_1"Q")` ....(3)
If y << D and d << D then, S1Q ≈ S2Q ≈ D
S2Q + S1Q = 2D
∴ Equation (3) becomes,
`"S"_2"Q" - "S"_1"Q" = (2"yd")/(2"D")`
∴ `"S"_2"Q" - "S"_1"Q" = "yd"/"D"`
∴ Δl = `"yd"/"D"` ....(4)
Equation (4) gives the path difference of two interfering light waves. Point Q will be bright if,
Δl = nλ = 2n`lambda/2`
where n = 0, 1, 2,….
∴ `("y"_"n""d")/"D" = "n"λ = 2"n"lambda/2` ........….[From equation (4)]
∴ `"y"_"n" = "n"(lambda"D")/"d"` ….(5)
Equation (5) represents the distance of the nth bright fringe from the central bright fringe.
Point Q will be a dark point if,
`Deltal = (2"n" - 1)lambda/2`
where n = 1, 2, 3,……
∴ `("y"_"n"^'"d")/"D" = (2"n" - 1)lambda/2`
∴ `"y"_"n"^' = (2"n" - 1)(lambda"D")/(2"d") = ("n" - 1/2) (lambda"D")/"d"` .........(6)
Equation (6) represents the distance of nth dark fringe from the central maximum.
Fringe width:
The distance between any two successive dark or any two successive bright fringes is equal. This is called the fringe width and is given by, Fringe width = W = Δy = `"y"_("n" + 1) - "y"_"n" = "y"_"n + 1"^' - "y"_"n"^'`
W = `λ"D"/"d"`
Thus, both dark and bright fringes are equidistant and have equal widths.
Conditions for constructive and destructive interference:
The phase difference between the two waves reaching P, from `"S"_1` and `"S"_2` is given by,
`Deltaphi = "y""d"/"D"((2pi)/lambda)` ..........`(∵ Deltal = "yd"/"D")`
The condition for constructive interference in terms of phase difference is given by,
ΔΦ = n2π, where, n = 0, ±1 ±2
∴ The condition for destructive interference in terms of phase difference is given by
`Deltaphi = ("n" - 1/2)2pi,` where, n = ±1, ±2
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