D14FAGRRES3DS
Determination of the wavelength difference for two peaks,
resolved according to the Rayleigh criterion
We call the distance between the maxima b. Radius of apertures a, distance
between the apertures d, coordinate on the observation screen R,
wavelength λ, and distance from aperture to screen X.
1. Determination of Rayleigh distantance.
a ≡ .05
X ≡ 4000
R := 0 , .1 .. 50
 J1 2⋅ π ⋅ a⋅ R  
 
X⋅ λ  
g1( R) := 

  2 ⋅ π ⋅ a⋅ R  
X⋅ λ  
 
2
 J1 2 ⋅ π ⋅ a⋅ R − b  
 
X⋅ λ  
gg1( R ) := 

  2⋅ π ⋅ a⋅ R − b  
X⋅ λ  
 
0.3
0.2
g1( R)
gg1( R )
0.1
0
0
10
20
30
40
50
R
Distance b is assumed to be: see on top of graph
2. 3D Graph of pattern of two round apertures at distance b.
i := 0 .. N
j := 0 .. N
xi := ( −30) + 2.0001⋅ i
2
RR( x , y) := ( x) + ( y)
y j := −30 + 2.0001⋅ j
2
N ≡ 40
λ ≡ .0005
X := 4000
2
 J1 2 ⋅ π ⋅ a⋅ RR( x , y)  
 
X⋅ λ  
g2( x , y) := 

  2 ⋅ π ⋅ a⋅ RR( x, y)  
X⋅ λ  
 
2
 J1 2 ⋅ π ⋅ a⋅ RR( x, y − b )  
 
X⋅ λ

gg2( x , y) := 

  2 ⋅ π ⋅ a⋅ RR( x , y − b )  
X⋅ λ
 
 
M i , j := g2( xi , y j) + gg2( xi , y j)
b ≡ 24.5
M
3. Calculation of wavelength difference corresponding to b
The diffraction angle is calculated from
b
X
= ∆θ
The grating is made of round apertures of diameter a and spaced at
distance d.
From the grating formula we have for the wavelength difference
∆λ=d∆θ or ∆λ = (d/X)b.
For
d := .1
∆λ := d ⋅
b
X
−4
∆λ = 6.125 × 10
2