# Asymptotic Formula

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1. Jun 18, 2017

### JBD

In the following equation,

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} )cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$

if $$0<N_F(a) ≪ 1$$ and $$(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$$, how do you arrive at
$$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2} +\frac{1}{\frac{x^2}{η^2} − a^2} sin^2 (πN_F(a)\frac{x}{a}))$$

Please see what I have done so far and check if I have errors in it.

Relevant Equations

$$\alpha (x; a) = \sqrt{N_F(a)\eta } (1 - \frac{x}{a\eta})$$Eq. 19

where $$\eta = 1 + L/D$$, $$N_F(a) = \frac{2a^2}{\lambda L}$$ and (additional definition) $$\gamma = \eta - 1$$

Starting with this:
Eq. 20
$$P(x; a)=\frac{1}{2\lambda(L+D)} ([C(α(x; a)) + C(α(x; −a))]^2 + [S(α(x; a)) + S(α(x; −a))]^2)$$

Eq. 24 (these two)
$$C[α(x; +a)] + C[α(x; −a)] ≃ \frac{1}{πα(x; a)} sin (\frac{πα(x; a)^2}{2} ) + \frac{1}{πα(x; -a)} sin (\frac{πα(x; -a)^2}{2} )$$
and
$$S[α(x; +a)] + S[α(x; −a)] ≃ \frac{-1}{πα(x; a)} cos (\frac{πα(x; a)^2}{2} ) - \frac{1}{πα(x; -a)} cos (\frac{πα(x; -a)^2}{2} )$$

Here is what I have done so far:
$$(C[α(x; +a)] + C[α(x; −a)])^2 + (S[α(x; +a)] + S[α(x; −a)])^2 = \frac{1}{π^2α^2(x; a)} + \frac{1}{π^2α^2(x; -a)} + \frac{2}{π^2α(x; +a)α(x; −a)} [sin (\frac{πα(x; a)^2}{2})sin (\frac{πα(x; -a)^2}{2}) + cos (\frac{πα(x; a)^2}{2} )cos (\frac{πα(x; -a)^2}{2})]$$

Using equation 19:

$$=\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} )cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]$$

And then in equation 20, the outer factor:
$$\frac{1}{2\lambda (L+D)} = \frac{\gamma}{2\lambda L \eta}$$

So the new equation for P is:
$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} )cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$

But this is where I am not sure what to do anymore with $$N_F (a) ≪ 1$$ and if $$(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$$ to arrive at equation 25.

I'm trying to verify equations 25 and 27 in the paper linked below. I got confused on how to apply $$N_F(a) ≪ 1$$ to get the final result. I was able to do "Applying the Fresnel function asymptotic forms (24) to (20) and using the definition (19)" but then I got stuck here

"we deduce that if $$N_F (a) ≪ 1$$ and if $$(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$$ , we
get the following asymptotic formula:"

Eq. 25
$$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2} +\frac{1}{\frac{x^2}{η^2} − a^2} sin^2 (πN_F(a)\frac{x}{a}))$$

I could not reproduce this result (eq 25 in the paper) as well as eq 27.

Page 14 Equations 25 and 27
There are occasional typographical errors in the paper. (from what I have verified so far, pages 1-13)

[1]: https://i.stack.imgur.com/Qc5Ie.png

2. Jun 18, 2017

### Buzz Bloom

Hi JBD:

The only thing that occurs to me is to (1) replace sin(u) with u since u<<1, and (2) replace cos(u) with 1, since cos(u) ~= 1-u2.

Hope this helps.

Regards,
Buzz

3. Jun 19, 2017

### JBD

Thanks for helping. I did it differently and arrived at the final form but I had some assumptions that I don't know if allowed or not. Like for example, $$N_F(a) =\frac{2a^2}{\lambda L}$$ but since $$N_F(a) ≪1$$, I replaced it with $$\frac{a^2}{\lambda L}$$. Is this reasonable? If not, I would have arrived at a similar form but short of the factor 2. I also had to do it one more time in a fraction with large number in denominator. The numerator was one and I changed it to 2.

4. Jun 19, 2017

### Ray Vickson

$P(x;a)$ has two parameters or arguments. When you go to the asymptotic form you are letting one of the arguments get large (or maybe small); would that be large (small?) $x$ or large (small?) $a$?

Last edited: Jun 19, 2017
5. Jun 19, 2017

### Buzz Bloom

I underlined "it" in the quote because I am not sure what the antecedent of this pronoun is. I am guessing you intend it to be what is equal to NF(a). I also do not understand what would make this assumptions reasonable.

Could you post the details of your work that leads to the correct answer except for a factor of 2?

Regards,
Buzz

6. Jun 20, 2017

### JBD

$$x ≫ a$$ large x and small a

Last edited: Jun 20, 2017
7. Jun 20, 2017

### JBD

I used cos (A-B) = cos A cos B + sin A sin B (I feel bad for not noticing this. I was too focused on applying the N_f(a) ≪1 and the other condition.)

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} - \frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2(1+\frac{x^2}{a^2\eta^2})}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (2πN_F(a)\frac{x}{a})]]$$

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [\frac{1 + \frac{x^2}{a^2\eta^2}}{1 - \frac{x^2}{a^2\eta^2}}+cos (2πN_F(a)\frac{x}{a})]]$$

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{ \frac{x^2}{a^2\eta^2}}{ \frac{x^2}{a^2\eta^2}-1}+\frac{1 }{\frac{x^2}{a^2\eta^2}-1}-cos (2πN_F(a)\frac{x}{a})]]$$

$$\frac{ \frac{x^2}{a^2\eta^2}}{ \frac{x^2}{a^2\eta^2}-1} ≃ 1$$since$$\frac{x}{a\eta}$$ is a large number

$$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +1-cos (2πN_F(a)\frac{x}{a})]]$$

I used 2 sin^2 A = 1 - cos 2A

$$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})]]$$

$$N_F(a) = \frac{2a^2}{\lambda L}$$
so,
$$P(x; a)≃ \frac{\gamma}{ \eta} \frac{1}{π^22a^2\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})]$$
$$P(x; a)≃ \frac{\gamma}{π^2 \eta^2} (\frac{1}{2a^2( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} + 2sin^2 (πN_F(a)\frac{x}{a})])$$
$$P(x; a)≃ \frac{\gamma}{π^2 \eta^2} (\frac{1}{2( \frac{x^2}{\eta^2}-a^2)} [\frac{a^2 }{\frac{x^2}{\eta^2}-a^2} + 2sin^2 (πN_F(a)\frac{x}{a})])$$
$$P(x; a)≃ \frac{\gamma}{π^2 \eta^2} ( \frac{a^2 }{2(\frac{x^2}{\eta^2}-a^2)^2} + \frac{1}{( \frac{x^2}{\eta^2}-a^2) }sin^2 (πN_F(a)\frac{x}{a}))$$

A factor of 2 missing in front and an extra 2 in the denominator.

8. Jun 20, 2017

### Buzz Bloom

Hi JDB:

I now see how you derived the answer you got, and where it is different from Eq 25. What still puzzles me is where the assumption
NF(a) = 2a2/λL​
comes from.

BTW, I now also see why my original suggestion is not helpful,

Regards,
Buzz

9. Jun 20, 2017

### JBD

Fresnel Number is defined as $$N_f(a) = a^2 / \lambda L$$
where a is the size of slit, lambda is wavelength and L is distance from slit to screen
In the paper, the slit was size 2a and distance from slit to screen is L + D, but D = L so
$$N_f(a) = 4a^2 / \lambda 2L = 2a^2/ \lambda L$$

I don't know if eq 25 has typos (because I encountered some of them ) but I think it does not have typos. But then again, where did I get my math wrong? Thanks.

10. Jun 20, 2017

### Buzz Bloom

Hi JDB:

My guess is that the text you were using has typos or just plain errors. If the problem was part of some work project, I think you should just accept that you got the right answer. If the problem was related to study for a course, I suggest you review your work with the professor - he may feel inclined to give you some extra credit for finding an error in the text.

Regards,
Buzz