# How to get to this Surface slope formula?

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1. Jul 28, 2015

### Ornit

This is no homework; but I cannot work it out so I assume one of my assumption is off... no matter how much I looked at it I can't get to the textbook answer.
I tried searching (Google is my friend); I got to Prof. Burge's wonderful lectures slides (Arizona.edu) but he follows a different path so it didn't help.

The image below is taken from Optical System Design by Fischer & Tadic-Galeb. It can be found in the link below. The author describes a simple way of calculating surface slope but I cannot follow his steps and would appreciate help in deriving his simple formula for Maximum Slope from page 346:
http://kundoku.free.fr/O/007/Optical System Design/Optical System Design.pdf

Assuming A is in waves, I tried working out the maximum slope as the magnitude of the disturbance (A*lambda, in length units) divided by the typical length of half a bump (D/(2*n)). This gives a slope of 2*n*lambda*A/D.
My answer is 4/pi off the textbook. What might have went wrong...?

Thanks
Ornit

2. Jul 30, 2015

### soarce

The maximum slope stands for "the maximum value of the derivative" of the wave front.

3. Jul 31, 2015

### Ornit

Thanks soarce, but I still can't understand the textbook version.
Derivative with respect to what? I assumed with respect to unit length, this is why I tried dividing by the typical length of half a bump.
I completely don't understand how did pi got into the numerator of the textbook equation. What am I missing?
Thanks
Ornit

4. Aug 4, 2015

### soarce

What is your definition of "the slope"? What do you understand by "the maximum slope of the wavefront"?

Following the picture you posted I reffered to the slope of the magnitude of the wavefront with respect to the spatial dimension. Then the derivative can be taken with respect to the spatial dimension (represented by vertical axis in your picture).

5. Aug 4, 2015

### Ornit

Thanks soarce,

Thanks for your help. You actually made me find the reason for the pi... At first I approximated the wavefront by saw-tooth, missing the (clearly noted) sine wave mention, hence no π in my initial derivation. But I still can't prove the book is right.

I think the author meant that "maximum slope" would be the max derivative of the curvy wavefront with respect to distance along the Y axis in the picture (I call it x below). If w(x) is the wavefront function, I interpret maximum slope = max(w'(x)). I scaled the PV of the sine to be A*λ.

When done with a sine wave, the π comes from the normalization to one diameter, and d(sin(u)/dx=cos(u)*du/dx.

After checking and rechecking I still end up with max(w'(x))=πAλn/D instead of the πAλn/2D .
Can't find a justification for the factor of 2. .

Ornit

6. Aug 4, 2015

### soarce

The peak-to-valley amplitude is twice the amplitude of the sine?

7. Aug 4, 2015

### Ornit

True, but it doesn't help.
Let me write it down fully and properly (well, I hope not - if you find a mistake it will actually solve the problem ):

w(x)=0.5*A*λ*(sin(π+2*π*x*n/D)+1)
(0.5 and +1 are for the wavefront to swing from 0 to +1 PV as you mentioned; π is just for phase shift to have the look like the 1.5 bumps illustration; sin(2*π*x*n/D) is what I think to be the critical tern for the missing factor of 2).

Therefore w'(x)=0.5*A*λ*cos(π+2*π*x*n/D)*(2*π*n/D)

max(w'(x))=0.5*A*λ*1*2*π*n/D=A*λ*π*n/D

8. Aug 4, 2015

### soarce

How did you scale PV amplitude to be A*lambda (or A*lambda/2) ?

9. Aug 4, 2015

### Ornit

Looking at the picture. I interpret A as PV (peak to valley) in waves therefore multiplied the sine expression by A*λ. The 0.5 factor comes from the (sin(y)+1): this expression will vary between 0 and 2, therefore I multiplied by 0.5. Have I misunderstood your question?

10. Aug 4, 2015

### soarce

I was just curious. The amplitude of the wavefront is given in terms of electric field intensity and don't understand how the field amplitude is related to lambda. I was thinking that you normalized in some way the energy of the wave and then lambda appears in the amplitude of the field. The factor 1/2 it's ok.

Anyway, who is lambda ? It is some wavelength ? I was reading the page you indicated to but I didn't see any reference to lambda.

Last edited: Aug 4, 2015
11. Aug 4, 2015

### Ornit

Lambda is indeed the wavelength. The industry uses units of "waves" to measure Transmitted Wavefront Distortion (the deviation from a perfect wavefront, either plane or sphere). It is quite common to talk in units of waves (or fringes), therefore it seemed natural to me to assume those are the units (and at least it got lambda into the equation...).
Hard to believe the author had missed a factor of two. It haunts me...

12. Aug 5, 2015

### soarce

Maybe there are some different scalings to report the wavefront deviation. I have seen on internet that people speak of lambda/2, lambda/4 and lambda/8 irregularities of the surfaces.

LE: Have a look at Chapter 4, the wavefront deviations are given in terms of differences in optical paths and these differences are scaled to lambda. From here they speak of lambda, lambda/2 or lambda/4 criteria.

Last edited: Aug 5, 2015
13. Aug 5, 2015

### Ornit

Yes, people usually speak in term of fraction of lambda (aka "wave") to describe surface figure or TWD (Transmitted Wavefront Distortion). My interpretation is that this is the "A" defined in the picture. For example, if the peak-to-valley in the picture is lambda/4, A=1/4 and my scaling is A*lambda.