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. 

Answer 1

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 n^th 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 n^th 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