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Multi Source Inteference

Module by: Paul Padley

Summary: We derive the expressions for inteference from a number of sources

N-Source Interference

Lets look back at two source interference again, this time using exponential notation: We will consider both sources to be in phase with the same amplitude. Also, just like before we define E 0 E 0 to be at the point where the interference is occurring, and approximate that it is the same for both waves. We have E 1 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) E 1 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) E 2 ⃗ = E 0 ⃗ e i ( k r 2 − ω t ) E 2 ⃗ = E 0 ⃗ e i ( k r 2 − ω t ) or E 2 ⃗ = E 0 ⃗ e i ( k ( r 1 + Δ r ) − ω t ) E 2 ⃗ = E 0 ⃗ e i ( k ( r 1 + Δ r ) − ω t ) where Δ r = d sin θ Δ r = d sin θ

Figure 1
Figure 1 (Twopoint_small.png)
So I could write E 2 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) e i k Δ r E 2 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) e i k Δ r or if δ = k Δ r = k d sin θ δ = k Δ r = k d sin θ E 2 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) e i δ E 2 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) e i δ E 2 ⃗ = E 1 ⃗ ( e i δ ) E 2 ⃗ = E 1 ⃗ ( e i δ )

So when we add the E fields together we get E ⃗ = E 1 ⃗ + E 2 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) ( 1 + e i δ ) E ⃗ = E 1 ⃗ + E 2 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) ( 1 + e i δ )

Now if I consider the case of not 2 but N sources, all separated by a distance d, then this simply extends E 1 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) E 1 ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) E 2 ⃗ = E 0 ⃗ e i ( k r 2 − ω t ) E 2 ⃗ = E 0 ⃗ e i ( k r 2 − ω t ) E 3 ⃗ = E 0 ⃗ e i ( k r 3 − ω t ) E 3 ⃗ = E 0 ⃗ e i ( k r 3 − ω t ) . . . . . . E N ⃗ = E 0 ⃗ e i ( k r N − ω t ) E N ⃗ = E 0 ⃗ e i ( k r N − ω t ) then r 2 = r 1 + d sin θ r 2 = r 1 + d sin θ

r 3 = r 1 + 2 d sin θ r 3 = r 1 + 2 d sin θ

and so on,

or r 2 = r 1 + Δ r r 2 = r 1 + Δ r

r 3 = r 1 + 2 Δ r r 3 = r 1 + 2 Δ r

So now when we add the E fields up we get E ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) ( 1 + e i δ + e 2 i δ + e 3 i δ + … + e ( N − 1 ) i δ ) E ⃗ = E 0 ⃗ e i ( k r 1 − ω t ) ( 1 + e i δ + e 2 i δ + e 3 i δ + … + e ( N − 1 ) i δ ) or rewriting

Now following is a general property of geometric series: ∑ n = 0 N − 1 x n = 1 − x N 1 − x ∑ n = 0 N − 1 x n = 1 − x N 1 − x

So now we get E ⃗ = E 0 ⃗ e − i ω t e i k r 1 1 − e i δ N 1 − e i δ E ⃗ = E 0 ⃗ e − i ω t e i k r 1 1 − e i δ N 1 − e i δ or E ⃗ = E 0 ⃗ e − i ω t e i k r 1 e i δ N − 1 e i δ − 1 E ⃗ = E 0 ⃗ e − i ω t e i k r 1 e i δ N − 1 e i δ − 1 or E ⃗ = E 0 ⃗ e − i ω t e i k r 1 e i δ N / 2 ( e i δ N / 2 − e − i δ N / 2 ) e i δ / 2 ( e i δ / 2 − e − i δ / 2 ) E ⃗ = E 0 ⃗ e − i ω t e i k r 1 e i δ N / 2 ( e i δ N / 2 − e − i δ N / 2 ) e i δ / 2 ( e i δ / 2 − e − i δ / 2 ) or E ⃗ = E 0 ⃗ e − i ω t e i ( k r 1 + ( N − 1 ) δ / 2 ) sin ( δ N / 2 ) sin ( δ / 2 ) E ⃗ = E 0 ⃗ e − i ω t e i ( k r 1 + ( N − 1 ) δ / 2 ) sin ( δ N / 2 ) sin ( δ / 2 ) Now we can define k R ≡ k r 1 + ( N − 1 ) δ / 2 ) k R ≡ k r 1 + ( N − 1 ) δ / 2 ) , which makes sense, this rephrases the equation in terms of the distance from the middle of the array of sources. So the equation becomes E ⃗ = E 0 ⃗ e i ( k R − ω t ) sin ( δ N / 2 ) sin ( δ / 2 ) E ⃗ = E 0 ⃗ e i ( k R − ω t ) sin ( δ N / 2 ) sin ( δ / 2 )

Figure 2
Figure 2 (ThreePointMovie.gif)
Interference pattern from three point sources

To find the irradiance, lets simplify things by taking the real part of this E ⃗ = E 0 ⃗ cos ( k R − ω t ) sin ( δ N / 2 ) sin ( δ / 2 ) E ⃗ = E 0 ⃗ cos ( k R − ω t ) sin ( δ N / 2 ) sin ( δ / 2 )

Then I = ε c < E 2 >= ε c E 0 2 2 sin 2 ( δ N / 2 ) sin 2 ( δ / 2 ) I = ε c < E 2 >= ε c E 0 2 2 sin 2 ( δ N / 2 ) sin 2 ( δ / 2 ) or I = I 0 sin 2 ( δ N / 2 ) sin 2 ( δ / 2 ) I = I 0 sin 2 ( δ N / 2 ) sin 2 ( δ / 2 )

Figure 3
Figure 3 (MultisourceInterference__3.png)
Plot of sin 2 ( δ 50 ) sin 2 ( δ / 2 ) sin 2 ( δ 50 ) sin 2 ( δ / 2 )

Figure 4
Figure 4 (MultisourceInterference__4.png)
Plot of sin 2 ( δ 10 ) sin 2 ( δ / 2 ) sin 2 ( δ 10 ) sin 2 ( δ / 2 )

Figure 5
Figure 5 (MultisourceInterference__5.png)
Plot of sin 2 ( δ 10 ) sin 2 ( δ / 2 ) sin 2 ( δ 10 ) sin 2 ( δ / 2 )

Look at the plots, which show what the function sin 2 ( δ N / 2 ) sin 2 ( δ / 2 ) sin 2 ( δ N / 2 ) sin 2 ( δ / 2 ) looks like for N = 100 N = 100 and N = 20 N = 20 . The height of the first principle maximum is equal to N 2 N 2 . This is because as

θ → 0 θ → 0 then δ → 0 δ → 0 (recall δ = k d sin θ δ = k d sin θ ) Then lim δ → 0 sin 2 ( δ N / 2 ) sin 2 ( δ / 2 ) → ( δ N / 2 ) 2 ( δ / 2 ) 2 = N 2 lim δ → 0 sin 2 ( δ N / 2 ) sin 2 ( δ / 2 ) → ( δ N / 2 ) 2 ( δ / 2 ) 2 = N 2 or at θ = 0 θ = 0 I = N 2 I 0