2D Fraunhofer-diffraction with infinitely long slits

• PhysicsRock
In summary, the author's issue is that the slit in the ##y##-direction is supposed to be infinite, but with what is given in the assignment, the function ##a(x,y)## is defined as$$a(x,y) = \begin{cases} 1 & , \, ( 9d \leq |x| \leq 10d ) \wedge (y \in \mathbb{R}) \\ 0 & , \, \text{else} \end{cases}$$This yields$$F(k_x,k_y) \propto 2 \pi \delta(k_y).$$
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.

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.

PhysicsRock said:
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.

Steve4Physics said:
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.

- 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)##?

DaveE
ergospherical said:
- 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?

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.

vanhees71
ergospherical said:
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.

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
Steve4Physics said:
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.

Attachments

• Screenshot (35).png
13.9 KB · Views: 55

What is 2D Fraunhofer diffraction with infinitely long slits?

2D Fraunhofer diffraction with infinitely long slits refers to the diffraction pattern observed when a coherent light source passes through a series of slits that are infinitely long in one dimension. This setup simplifies the analysis to a two-dimensional problem, where the diffraction pattern can be studied in the plane perpendicular to the slits.

How is the diffraction pattern formed in 2D Fraunhofer diffraction with infinitely long slits?

The diffraction pattern is formed due to the interference of light waves emanating from different points along the slits. Because the slits are infinitely long, the pattern is essentially a series of bright and dark fringes, where the intensity of light varies due to constructive and destructive interference.

What mathematical tools are used to analyze 2D Fraunhofer diffraction with infinitely long slits?

Fourier transform techniques are primarily used to analyze 2D Fraunhofer diffraction patterns. The far-field diffraction pattern is the Fourier transform of the aperture function, which describes the transmission properties of the slits. The resulting intensity distribution can be calculated using these mathematical tools.

What are the key parameters affecting the diffraction pattern in 2D Fraunhofer diffraction with infinitely long slits?

The key parameters include the wavelength of the incident light, the width of the slits, and the spacing between the slits. Changes in any of these parameters will alter the resulting diffraction pattern, such as the spacing and intensity of the fringes.

How does the number of slits affect the diffraction pattern in 2D Fraunhofer diffraction with infinitely long slits?

Increasing the number of slits generally leads to more defined and sharper diffraction patterns. With more slits, the interference effects become more pronounced, resulting in narrower and more intense bright fringes, and more defined dark fringes. This is often referred to as the multi-slit or grating effect.

Replies
4
Views
830
Replies
1
Views
1K
Replies
2
Views
2K
Replies
8
Views
2K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
2
Views
10K
Replies
4
Views
4K
Replies
4
Views
4K