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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.
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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