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What is the characteristic bump height of periodic grating below which diffraction effect cease to exist (let assume a threshold of peaks to valleys intensity of 20% as the minimum detectable by human eye). Is it significantly less than lambda?

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What makes You think that a threschold exists? Diffraction including those by gratings is a strictly linear phenomenon. –  Georg Jul 17 '11 at 18:55
    
If grating height is below wavelength then it won't disturb wavefront, right? To put it bluntly we can't distinguish structures below wavelength, can we? –  Tegiri Nenashi Jul 18 '11 at 17:22
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:=) Yes we can! –  Georg Jul 18 '11 at 17:44
    
Why diffraction is linear? One would think that as soon as the height of grating bump is large we get saturation. In other words, diffraction produced by grating with 1 mm height bumps should be identical to the one with 2 mm. So, my question is what happens when the height is vanishingly small? –  Tegiri Nenashi Jul 19 '11 at 17:01
    
Another hint: there are gratings with zero height! –  Georg Jul 19 '11 at 18:07
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1 Answer

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Assume sinusoidal grate -- sin(x) is the simplest possible periodic function. Direct axis x horizontally, and y vertically, and lets calculate diffraction at the point (x,y). According to Fraunhofer diffraction we must first calculate distance from the observation point (x,y) to the source of secondary wave (x_1, h*sin(d*x_1))) which is r=sqrt((y-h*sin(d*x_1))^2+(x-x_1)^2). Here the h is the height of the ripple, and d its frequency. This r is plugged in into the amplitude integral Integrate(i*exp(-2*pi*i*r/lambda)/(r*lambda),x1=-R..R). Mathematica expression:

Integrate[i*exp(-2*pi*i*sqrt((y-h*sin(x1))^2+(x-x1)^2)/lambda)/(sqrt((y-h*sin(x1))^2+(x-x1)^2)*lambda),{x1,-d,d}]

Maple:

Int(i*exp(-2*pi*i*sqrt((y-h*sin(d*x1))^2+(x-x1)^2)/lambda)/(sqrt((y-h*sin(d*x1))^2+(x-x1)^2)*lambda),x1=-R..R);

Neither succeeded solving it analytically, so one have to regress to numerics. Therefore, I assigned the following numeric values:

The radius of the flat mirror:

R=20 mm

Wavelength:

lambda=0.0004 mm (=400 nm)

Defect density/frequency:

d=10 (10 ripples/mm -- with higher values Wolfram alpha times out)

Defect height

h=0.0001 mm (1/1 wave)

Location

y=100 mm (10 cm above the mirror).

With all the assignments the only free variable remaining is x, so one should be able to plot intensity graph. Unfortunately, Wolfram Alpha refuse to understand what Plot[Integral[]] is and suggests some stock market graphs instead. I had to calculate pointwise. Here is an expression which calculates intensity at the axis (x=0):

Integral[i*exp(-2*pi*i*sqrt((100-0.0001*sin(10*x1))^2+(0.0-x1)^2)/0.0004)/(sqrt((100-0.0001*sin(10*x1))^2+(0.0-x1)^2)*0.0004),x1=-20..20]

Which walpfa evaluates to 0.62+3.81i

The characteristic angle of periodic grating is lambda*d so the diffraction pattern linear dimension is lambda*d*y which in our case is conveniently 1. Therefore, we could see diffraction pattern by probing only three points: x=0.0, x=0.5, and x=1.0. Here are the intensities:

x=0.0 -> 0.62+3.81*i
x=0.5 -> 7.76+3.0*i
x=1.0 -> 3.51+1.8*i

In other words, the diffraction pattern is quite noticeable when defect size is comparable to lambda. To doublecheck, what if we shrink grating height 10 fold? There:

x=0.0 -> 3.4+3.5*i
x=0.5 -> 4.1+3.1*i
x=1.0 -> 3.88+3.33*i 
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