Fresnel diffraction from square: on axis intensity

ergospherical
Science Advisor
Homework Helper
Education Advisor
Insights Author
Messages
1,097
Reaction score
1,384
Homework Statement
Determine the factor by which the on-axis intensity from a square aperture of side length ##a## is reduced, compared to an unobstructed pattern.
Relevant Equations
Fresnel integral.
I'd appreciate if someone could check whether my work is correct. The ##x##-##y## symmetry of the aperture separates the Fresnel integral:\begin{align*}
a_p \propto \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{ikx^2}{2R} \right) dx \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{iky^2}{2R} \right) dy \equiv \left[ \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{ikx^2}{2R} \right) dx \right]^2
\end{align*}I substitute ##\frac{\pi \xi^2}{2} = \frac{kx^2}{2R}##, i.e. ##\xi = x\sqrt{\frac{k}{\pi R}}##. I also denote ##w = \frac{a}{2} \sqrt{\frac{k}{\pi R}}## to simplify the new limit,\begin{align*}
a_p \propto \frac{\pi R}{k} \left[ \int_{-w}^{w} \mathrm{exp}\left(\frac{i\pi \xi^2}{2} \right) d\xi \right]^2 = \frac{4\pi R}{k} \left[ \int_{0}^{w} \mathrm{exp}\left(\frac{i\pi \xi^2}{2} \right) d\xi \right]^2
\end{align*}and separate into sines and cosines,\begin{align*}
a_p \propto \frac{4\pi R}{k} \left[ \int_{0}^{w} \cos{\left(\frac{\pi \xi^2}{2} \right)} d\xi + i\int_{0}^{w} \sin{\left(\frac{\pi \xi^2}{2} \right)} d\xi \right]^2 = \frac{4\pi R}{k} \left[ C\left( w \right) + iS\left( w \right) \right]^2
\end{align*}The unobstructed pattern would instead correspond to ##w \rightarrow \infty##, in which case ##[C(\infty) + iS(\infty)]^2 = [1 + i]^2 = 2i##. The amplitude of the pattern with the square is therefore reduced by a factor\begin{align*}
\alpha = \frac{\left[ C\left( w \right) + iS\left( w \right) \right]^2}{2i}
\end{align*}and the intensity by ##|\alpha|^2##,\begin{align*}
|\alpha|^2 = \frac{|C\left( w \right) + iS\left( w \right) |^4}{4}
\end{align*}Does it look correct?
 
Physics news on Phys.org
The only mistake that I see is where you wrote

ergospherical said:
##[C(\infty) + iS(\infty)]^2 = [1 + i]^2 = 2i##.

##C(\infty)## and ##S(\infty) ## are not equal to 1. They equal 1/2.
 
  • Like
Likes PhDeezNutz and ergospherical
TSny said:
##C(\infty)## and ##S(\infty) ## are not equal to 1. They equal 1/2.
Ah great, thanks, I’d mis-remembered the Cornu spiral.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top