A sigle wave
Superposition of waves
© The scientific sentence. 2010

Young's doubleslit Interference
1. Interference fringes
Simpleslit experiment explains diffraction phenomenun, whereas
doubleslit explains dinterference phenomenun.
Two light rays pass through two slits, separated by a distance d and reach a screen at a
distance, L from the slits.
As L >> d, we can consider that the lignes from the two slits to
the point P are parallel; so r_{2}  r_{1} = l = d sinθ = d x/L.
When the two rays pass through the two slits, the condition for constructive interference
is that = nλ, where n is an integral number and λ is the wavelength of the
ray (wave).
Constructive interference: d sinθ = nλ
Destructive interference: d sinθ = (n + 1/2)λ
On the screen, the bright bands correspond to the points of constructive interference,
and the dark bands correspond to the points of destructive interference. Both constite
waht we call interference fringes.
From x = L sinθ = L d sinθ/d , we have:
x_{bright} = n L λ/d for the bright bands, and
x_{dark} = (n + 1/2) L λ/d for the dark bands.
The bright bands are spaced by the distance:
Δx = x_{nextbright}  x_{bright} =
[(n + 1) L λ  n L λ]/d = L λ/d
2. Intensity distribution
For each point "P" of the screen, the intensity I of light is calculated using the
principle of superposition. It is expressed as:
I = S = (1/2) ε_{o} E_{12}^{2}, where E_{12}
is the vectorial sum of the two electric field E_{1} and E_{2} of the rays
comming from the slit S_{1} and S_{2}.
If E_{1} = E_{o} sin(ωt), and
E_{2} = E_{o} sin(ωt +φ), are the electric fiels
of each slit, then
E_{12} = E_{o} [sin(ωt) + sin(ωt +φ)].
sin(a + b) = sin a cos b + cos a sin b
sin (a  b) = sin a cos b  cos a sin b
By addition, we have sin (a + b) + sin (a  b) = 2 sin s cos b
with a + b = ωt + φ and a  b = ω t, that is a = ωt + φ/2 and b = φ/2
we obtain sin(ωt) + sin(ωt +φ) = 2 sin(ωt + φ/2) cos (φ/2)
E_{12} = 2 E_{o} cos (φ/2) sin(ωt+ φ/2), that oscillates with the amplitude
2 E_{o} cos (φ/2),
E_{12}^{2} = 4 E_{o}^{2} cos^{2} (φ/2) sin^{2}(ωt + φ/2)
If I_{o} is the intensity of light for a single slit, that is
I_{o} = constant x E_{o}^{2}, we have also
I_{12} = constant x E_{o}^{2}
= I_{o} x 4 cos^{2} (φ/2)
I_{12} = 4 I_{o} cos^{2} (φ/2)
With
φ/2 π = d sin θ/λ, we have:
I_{12} = 4 I_{o} cos^{2} (πd sin θ/λ)
= 4 I_{o} cos^{2} [(πd/λL) x]
3. Diffraction grating
For a grating with N slits, the total distribution of the intensity
on the screen corresponds to the sum of the individual electric
field for each slit "i", that is:
I = (1/2) c ε_{o} E^{2}, with E = ∑ E_{i}
If E_{i} = E_{o} sin(ωt + iφ), then:
E = E_{o} ∑ sin(ωt + iφ), with i varies from 0 to
N 1. Using the comples notation, (j^{2} =  1), we have:
E = E_{o} Img [∑ exp{j(ωt + iφ)}]
But ∑ exp{j(ωt + iφ)} = exp{j(ωt} ∑ exp{j iφ)}; and
∑ exp {j iφ} = 1 + exp {jφ} + exp {jφ}^{1} + exp {jφ}^{2} + ... +
exp {jφ}^{N  1}
We know that : 1 + x + x^{2} + x^{3} + ... + x^{n  1} = (x^{n}  1)/(x  1)
Thus:
∑ exp {j iφ} = ( exp {jφ}^{N}  1)/(exp {jφ}  1) =
(exp {jNφ}  1)/(exp {jφ}  1)
We know that : exp{jx} + exp { jx} = 2j sin x
Then:
∑ exp {j iφ} = (exp {jNφ/2}  exp { jNφ/2)/(exp {jφ/2}  exp { jφ/2}) x exp {jNφ/2}/exp {jφ/2} =
(2j sin(Nπ/2))/(2j sin(φ/2)) x exp {j(N  1)φ/2} = (sin(Nπ/2))/(sin(φ/2)) x exp {j(N  1)φ/2}
Then:
∑ exp{j(ωt + iφ)} = exp{j(ωt} (sin(Nπ/2))/(sin(φ/2)) x exp {j(N  1)φ/2}
And:
Img [∑ exp{j(ωt + iφ)}] = Img [(sin(Nπ/2))/(sin(φ/2)) exp j{ωt + (N  1)φ/2}]
Finally:
E = E_{o} sin[ωt + (N1)φ/2](sin(Nφ/2))/(sin(φ/2))
And:
I = I_{o} sin^{2}(Nφ/2))/(sin^{2}(φ/2))
Known as Single slit Fraunhofer diffraction
Where:
I_{o} = (1/2)cε_{o} E_{o}^{2} sin^{2}[ωt+(N1)φ/2]
4. Particular cases
4.1. No slits
No slit: N = 0, we get: I = 0: no effects on the screen.
4.2. One slit
Single slit: N = 1, we get: I = I_{o}, with I_{o} = (1/2)cε_{o} E_{o}^{2} sin^{2}[ωt],
One slit gives diffraction effect.
We have seen Single slit diffraction
that
I = I_{c} sin^{2} β / β^{2}, with
I_{c} = (1/2)cε_{o} E_{c}^{2} sin^{2}(ωt),
the intensity of incident rays at the center od the slit, β = (π a/λ) sin θ
Hence:
(1/2)cε_{o} E_{o}^{2} sin^{2}[ωt] =
(1/2)cε_{o} E_{c}^{2} sin^{2}(ωt) sin^{2} β / β^{2}
That is:
E_{o}^{2} = E_{c}^{2} sin^{2} β/β^{2}
Or:
I_{o} = I_{c} sin^{2} β/β^{2}
Then:
I = I_{o} = (1/2)cε_{o} E_{c}^{2} sin^{2} β/β^{2} sin^{2}[ωt]
4.3. Double slit
Double slit: N = 2, we get: I = 4 I_{o} cos^{2}(φ/2)
Where:
I_{o} = (1/2)cε_{o} E_{o}^{2} sin^{2}[ωt+φ/2]
Comparing with the result above, we get the same result:
I_{12} = 4 I_{o} cos^{2} (φ/2). Thus:
We have then:
I_{12} = 4 I_{c} sin^{2} β/β^{2} cos^{2} (φ/2)
4.4. Large number of slits
From the relationship: φ/2 π = d sin θ/λ, we differentiate to have:
Δφ/2 π = d cos θ Δθ/λ. With Δφ = 2π /N , we have:
1/N = d cos θ Δθ/λ. That is:
Δθ = λ/Nd cos θ
When N tends towards ∞ Δθ , and then Δ&phi tends towards zero.
When the number of slits is very large, N = ∞, Δφ becomes null. Δφ is
equal to the difference between the first "zero" that occurs at φ = 2π/N and 0. In this case,
we have a comb of equal intensities.
©: The scientificsentence.net. 2007.


