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Overlaying optical field with Zernike aberration

  1. Feb 10, 2016 #1
    Hi there,
    optical aberrations can be expressed by Zernike polynomials Vmn(ρ,θ).
    Now, for my simulations i am using software that takes 4 inputs for creating aberrations onto an optical flat: m,n,R,A
    m,n are the Zernike orders which is perfectly clear.
    For R,A the manual says:
    R: the radius
    A: the amplitude of aberration in radians reached at R
    I am still unclear how to interpret these parameters. I was looking through the source code and could make up the following:
    ρ = sqrt( x*x + y*y )/ R*R) which i suppose does some kind of spatial scaling of the aberration
    A is the Amplitude of the Zernike polynomial A*Vmn(ρ,θ), which i suppose does some kind of height scaling

    Finally the optical field is composed as such:
    RE{Field} = RE{Field_old}*cos(A*Vmn(ρ,θ)) - IM{Field_old}*sin(A*Vmn(ρ,θ))
    IM{Field} = RE{Field_old}*sin(A*Vmn(ρ,θ)) + IM{Field_old}*cos(A*Vmn(ρ,θ))

    Can someone explain to me how the parameters are supposed to work? Since when is amplitude expressed in radians?
    I couldn't find the aforementioned formulas in any standard textbook yet (i see that the last 2 are just complex numbers multiplied, which represent the optical fields). Where are these formulas taken from?
     
    Last edited: Feb 10, 2016
  2. jcsd
  3. Feb 10, 2016 #2

    Andy Resnick

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    Zoiks... you're definitely out in the weeds here, based on your notation I recommend checking out the relevant sections in Born and Wolf. Ok- first, some background. Zernike polynomials are one way to parameterize the aberration function, which itself is a way to describe the departure of the actual wavefront from an ideal (spherical) wavefront. Zernike polynomials are defined on a circle: Vmn(ρ sinθ,ρ cosθ)=Rmn(ρ) einθ. This is the connection with your RE{field} and IM{field} expressions (and also derived in Born and Wolf).

    Because the polynomials are defined on the unit circle, you can scale by the actual radius (as you mention). Why is the amplitude in radians? Because of the way the aberration function Φ is defined: the optical field at a point P, U(P), is given as a Fresnel diffraction integral, and the aberration function gets placed in the integrand e ik(Φ+s).

    Does this help? The topic is very math-intensive, too much for my limited LaTex abilities here.
     
  4. Feb 10, 2016 #3

    sophiecentaur

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    I never heard of Zernike polynomials but the dimensions must be right in any formula.
    If A is inside a trig function then you have to be talking in terms of angles and radians are the way to specify an angle. (The argument of a trig function has to be dimensionless)
     
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