2D Fraunhofer-diffraction with infinitely long slits

[PhysicsRock]

Homework Statement

Calculate the diffraction pattern for a 2D Fraunhofer-diffraction, where monochromatic and coherent light of wave length $\lambda$ (wave number $\vec{k}=2\pi / \lambda \hat{n}$) falls onto a wall with two slits, located in the regions where $9d \leq |x| \leq 10d$. Both slits span infinitely in the $y$-direction.

Relevant Equations

$\displaystyle E(k_x,k_y) \propto \int_{-\infty}^\infty \int_{-\infty}^\infty a(x,y) e^{-i(k_x x + k_y y)} dx dy$
$I(k_x,k_y) \propto E^2(k_x,k_y)$

My issue here is the fact that the slits are supposed to infinite in the $y$-direction. With what's given in the assignment, I'd define the apparatus function $a(x,y)$ as

$$
a(x,y) = \begin{cases} 1 & , \, ( 9d \leq |x| \leq 10d ) \wedge (y \in \mathbb{R}) \\ 0 & , \, \text{else} \end{cases}
$$

Plugging this into the Fourier transform and only considering the $y$-part of it yields

$$
F(k_x,k_y) \propto \int_{-\infty}^\infty e^{-i k_y y} dy.
$$

One recognizes this as the integral representation of the delta-distribution, with a conventional factor of $2\pi$. That would mean that

$$
F(k_x,k_y) \propto 2 \pi \delta(k_y).
$$

I'm unsure whether this is what is to be expected or not. The interpretation would be that there is a single sharp peak when $k_y$ is not 0, and if it is, the $x$-part takes over and results in an oscillation, as I would expect.

 

[tech99]

As an antenna engineer, I can see the equivalence here. There is a pattern generated by the length of the slit, and we need to be far from the slit to be in its radiation far field, or Fraunhofer Region. Otherwise the pattern will vary with distance. I am not aware, however, that we get zeros anywhere on-axis, because the two halves of the slit in the y plane are symmetrical and in-phase.

 

[Steve4Physics]

Homework Statement: Calculate the diffraction pattern for a 2D Fraunhofer-diffraction, where monochromatic and coherent light of wave length $\lambda$ (wave number $\vec{k}=2\pi / \lambda \hat{n}$) falls onto a wall with two slits, located in the regions where $9d \leq |x| \leq 10d$. Both slits span infinitely in the $y$-direction.
Relevant Equations: $\displaystyle E(k_x,k_y) \propto \int_{-\infty}^\infty \int_{-\infty}^\infty a(x,y) e^{-i(k_x x + k_y y)} dx dy$

A few random-ish (and rusty) thoughts...

It is not entirely clear how $k_x$ and $k_y$ are defined but presumably they include wavelength as a factor (else your equations contains no dependence on wavelength).

If the interference pattern is projected onto a screen then $k_x$ and $k_y$ are approximately proportional to the $x, y$ coordinates of a point the screen. We notice that the position of $k_y= y=0$ is entirely arbitrary. So the appearance of $\delta(k_y)$ is confusing. An alternative approach might be to try to get the solution for slits of length $L$ and take the limit $L \rightarrow \infty$. With a bit of luck the limit should be a simple constant so you are left with a function of $k_x$ only.

The aperture function $a(x,y)$ is necessarily independent of $y$ so it might be preferable to write it simply as $a(x)$.

From symmetry considerations, the interference pattern must be independent of $y$ and hence independent of $k_y$. We intuitively expect the pattern to be same as a conventional two-slit pattern but extending from $y=-\infty$ to $y=+\infty$. The final pattern must be a function of $k_x$ only.

 

[PhysicsRock]

A few random-ish (and rusty) thoughts...

It is not entirely clear how $k_x$ and $k_y$ are defined but presumably they include wavelength as a factor (else your equations contains no dependence on wavelength).

If the interference pattern is projected onto a screen then $k_x$ and $k_y$ are approximately proportional to the $x, y$ coordinates of a point the screen. We notice that the position of $k_y= y=0$ is entirely arbitrary. So the appearance of $\delta(k_y)$ is confusing. An alternative approach might be to try to get the solution for slits of length $L$ and take the limit $L \rightarrow \infty$. With a bit of luck the limit should be a simple constant so you are left with a function of $k_x$ only.

The aperture function $a(x,y)$ is necessarily independent of $y$ so it might be preferable to write it simply as $a(x)$.

From symmetry considerations, the interference pattern must be independent of $y$ and hence independent of $k_y$. We intuitively expect the pattern to be same as a conventional two-slit pattern but extending from $y=-\infty$ to $y=+\infty$. The final pattern must be a function of $k_x$ only.

Thank you. I'll give it a shot and see where it leads me.

 

[ergospherical]

- two slits of width $d$ centered on $x_1 = -9.5d$ and $x_2 = 9.5d$
- denote a single slit by a rectangle bump of width $d$ and unit height
(i.e. $s(x) = 1$ if $|x| < 0.5d$ and $s(x) = 0$ if $|x| > 0.5d$)

- aperture function is convolution (denoted $\star$) of $s(x)$ with delta functions at positions of the slits, i.e. \begin{align*}
a(x) = s(x) \star \delta(x-9.5d) + s(x) \star \delta(x+9.5d)
\end{align*} (- think of a convolution with a delta function as stamping a copy of the other function on the position of the delta function)

- the diffraction pattern $\propto \tilde{a}(k)$, the fourier transform of $a(x)$:

- what is the fourier transform of a convolution?
- what is the fourier transform of a rectangle bump $s(x)$?
- what is the fourier transform of a delta function $\delta(x-\square)$?

 

[PhysicsRock]

- two slits of width $d$ centered on $x_1 = -9.5d$ and $x_2 = 9.5d$
- denote a single slit by a rectangle bump of width $d$ and unit height
(i.e. $s(x) = 1$ if $|x| < 0.5d$ and $s(x) = 0$ if $|x| > 0.5d$)

- aperture function is convolution (denoted $\star$) of $s(x)$ with delta functions at positions of the slits, i.e. \begin{align*}
a(x) = s(x) \star \delta(x-9.5d) + s(x) \star \delta(x+9.5d)
\end{align*} (- think of a convolution with a delta function as stamping a copy of the other function on the position of the delta function)

- the diffraction pattern $\propto \tilde{a}(k)$, the fourier transform of $a(x)$:

- what is the fourier transform of a convolution?
- what is the fourier transform of a rectangle bump $s(x)$?
- what is the fourier transform of a delta function $\delta(x-\square)$?

As far as I know, the Fourier transform of a convolution is the product of the Fourier transforms of each individual function.
The Fourier transform of a "bump" should be the difference of two exponentials.
Finally, the Fourier transform of the delta function, say $\delta(x-a)$ should just be $e^{-ika}$, correct?

 

[ergospherical]

if you work it through, the fourier transform of a rectangular bump is a sinc function (try it). you multiply that by the fourier transform of $\delta(x-9.5d) + \delta(x+9.5d)$, which is $e^{ik \cdot 9.5d} + e^{-ik \cdot 9.5d} = 2 \cos{(9.5kd)}$. so the double slit diffraction pattern is a sinc function multiplied by a cos function. have a play around with it.

 

[PhysicsRock]

if you work it through, the fourier transform of a rectangular bump is a sinc function (try it). you multiply that by the fourier transform of $\delta(x-9.5d) + \delta(x+9.5d)$, which is $e^{ik \cdot 9.5d} + e^{-ik \cdot 9.5d} = 2 \cos{(9.5kd)}$. so the double slit diffraction pattern is a sinc function multiplied by a cos function. have a play around with it.

Thank you for your help. I really appreciate it.

 

[Steve4Physics]

Can step back a bit?

In Post #1 could $k_x$ and $k_y$ actually be related to direction cosines (rather than corresponding to positions on a screen)? So $k_x$ would be related to direction in the $xz$ plane and similarly for $k_y$ in the $yz$ plane.

If that were the case then the “single sharp peak” referred to in Post #1 would be consistent with a distribution at a single angle in the $yz$ plane (rather than some angle-dependent distribution which we know we have in the $xz$ plane). The Post #1 solution would then be correct! (If so, @ergospherical's approach will of course give the same answer.)

 

[PhysicsRock]

Can step back a bit?

In Post #1 could $k_x$ and $k_y$ actually be related to direction cosines (rather than corresponding to positions on a screen)? So $k_x$ would be related to direction in the $xz$ plane and similarly for $k_y$ in the $yz$ plane.

If that were the case then the “single sharp peak” referred to in Post #1 would be consistent with a distribution at a single angle in the $yz$ plane (rather than some angle-dependent distribution which we know we have in the $xz$ plane). The Post #1 solution would then be correct! (If so, @ergospherical's approach will of course give the same answer.)

In fact, $k_x$ and $k_y$ are given as $k_x = k \cdot \sin(\alpha)$ and $k_y = k \cdot \sin(\beta)$. The angles are best specified by a picture. I'll attach it to this comment.