# A Thin slits and accuracy of Kirchhoff's diffraction formula

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1. Jul 1, 2017

### bknock

Imagine a single slit with plane light waves incident on it with a screen (ideally far enough from the slits to simplify the math). According to Kirchhoff's diffraction formula, when a very wide slit is doubled, average intensity (averaged over all diffraction angles) doubles, and so does E_peak (the electric field at the central peak). This all agrees with energy conservation. However, when a very narrow slit (that is, much smaller than the wavelength of light) is halved, E_peak halves, and average intensity go to 1/4, which does not seem to agree with energy conservation! The average intensity should halve if the light let through the slit is halved. Calculated intensity does this because the diffraction pattern begins to lie "off of the screen" (that is, the diffraction pattern is still going at theta=pi/2).

Obviously, edge effects of the slit are becoming important for very thin slits. My question is this: is the prediction of Kirchhoff's diffraction formula correct (that is, are edge effects reflecting or absorbing extra energy to maintain energy conservation) or is the formula incorrect (that is, are edge effects affecting the screen's diffraction pattern)? Possibly the answer to this depends on the specifics of the slit. I am not an optics person, so I hope someone can help me! Basically, how much energy is expected to go through a very thin slit?

2. Jul 1, 2017

### Mgcini Keith Phuthi

You have to be careful here, the intensity is proportional to the square of the electric field i.e. $$I \alpha \epsilon E^2$$. Hence you expect the energy/Intensity to be a 1/4 of what you would get when E-field is 1/2 of its original value.

3. Jul 1, 2017

### Charles Link

For narrow slits, if the slit width $b$ is decreased, the size of the diffraction pattern including the width of the central maximum increases, so that there is complete energy conservation. (Of course, this in regards to energy that passes through the slit. A diffraction pattern made with a slit width that is narrowed by a factor of 2 will contain one half of the energy.) If the slit width drops by a factor of 2, the pattern is twice as wide. Doubling the slit width gives a central peak with 4x the on-axis intensity, but the pattern is one half as wide. $\\$ And to quantify the pattern width, the zeros in intensity of the pattern are found by the formula $m \lambda=b \sin(\theta)$ for $m=$ non-zero integer. The central peak has limits of $\theta$ such that $m=+1$ and $m=-1$ in this formula. This allows you to compute the range of angles $\Delta \theta$ covered by the central peak of the diffraction intensity pattern. Notice as $b$ gets larger, $\Delta \theta$ gets smaller.

Last edited: Jul 1, 2017
4. Jul 1, 2017

### bknock

Let me be more specific. This problem is much more subtle than freshman-level physics, so sorry for not providing the required details.

From the Huygens-Fresnel principle, we get (after doing an integral over phases assuming a distant screen) that intensity is proportional to...
a2 * sinc2[ π*a*sin(θ)/λ ]
where
a = slit width
θ = diffraction angle from -π/2 to π/2
λ = wavelength

Using the full Kirchhoff's diffraction formula gives that the intensity is proportional to...
a2 * sinc2[ π*a*sin(θ)/λ ] * (1+cos(θ))2 (eq 1)

The following have a discussion of how the above formulas can be derived:
https://en.wikipedia.org/wiki/Huygens–Fresnel_principle
https://en.wikipedia.org/wiki/Kirchhoff's_diffraction_formula

If you integrate the above intensity (equation 1) with respect to θ from -π/2 to π/2, you do not get energy conservation (as explained in my first post) when the slit is so small that the diffraction pattern begins to extend "off of the screen". To be specific, a << λ is what creates issues. Clearly, some edge effects from the slit become important in this limit, and I think the derivation of Kirchhoff's diffraction formula may have overlooked these edge effects. With this said, I am wondering if the amount of energy through the slit is correctly given by integrating the above formula (equation 1).

I numerically integrated equation 1 with respect to θ from -π/2 to π/2, and, in the a << λ limit, the result it NOT proportional to a as expected from energy conservation (but instead proportional to a2 in this limit) because sinc≈1 in this limit. The integral is proportional to a when a is large enough.

I am not an optics person, so I do not have expertise in this particular area, so any help would be appreciated!

Last edited: Jul 1, 2017
5. Jul 1, 2017

### Charles Link

There are instances in diffraction theory where it is necessary to introduce an obliquity factor such as $cos(\theta)$ to get certain integrals to converge. I will need to study this example further, but this may be one of those cases. $\\$ Editing: Upon further study of the problem, the on-axis irradiance (watts/m^2) at distance $s$ is $E_o= E_i (\frac{A^2}{s^2 \lambda ^2})$ where $A$ is the slit area (length x width) and $E_i$ is the incident irradiance (watts/m^2) (Note: $E$ is not electric field amplitude for this formula). This part is quite straightforward. The question is if the diffraction pattern is precisely as given by the Fraunhofer formula for very small slit width $b$ when the pattern becomes quite wide. It appears there may be some deviation from that $I=I_o \frac{sin^2(x)}{x^2}$ formula, but I can't make a completely conclusive statement on the matter. I would need to research it further. $\\$ Editing further: In order to conserve energy, the effective solid angle $\Omega_{effective}$ covered by the diffraction pattern needs to be proportional to $\frac{\lambda^2 }{A}$, but as the slit width $b$ gets very small, it presents difficulties, because this solid angle is necessarily bounded by $2 \pi$ steradians. In the case where slit width $b<< \lambda$, it appears there may be some anomalies. Your observation is interesting. I think it might be beyond the scope of what is normally covered in most of the textbook material on the subject.

Last edited: Jul 2, 2017
6. Jul 2, 2017

### bknock

Thanks for confirming that there is an anomaly here! If optics theory in textbooks cannot help, my hope is that some experimentalist could just find such a very thin slit and measure the intensity of what should be an almost independent-of-theta diffraction pattern (better yet, map out the obliquity factor as a function of theta!).

The obliquity factor you mentioned is the difference between the intensity formula you provided and my Equation 1. The field amplitude acquires an extra factor of (1+cosθ).

I am creating a phasor simulator for multiple slits that produces diffraction patterns after animating phasor addition at various diffraction angles. The lengths of the phasors from each slit depend on the single-slit diffraction pattern (including obliquity factor). Figuring out this very-thin-slit limit is the final thing I need before finishing, and I would happily give credit to anybody who can solve this for me! If whoever does this wants your real name (as opposed to your handle in these forums) mentioned in my acknowledgments, I can give you a link to contact me.

Last edited: Jul 2, 2017
7. Jul 2, 2017

### Charles Link

Experimentally, this would be a very difficult thing to show, either with visible, infrared, or even radio waves. $\\$ (Aside item: Experimentally I have used a narrow slit on a spectrometer to create a diffraction pattern that is perhaps 5 degrees wide (to fill the collimating mirror in order to fill the diffraction grating) for an incident HeNe laser (632.8 nm). I have never tried to get the diffraction pattern to fill a complete hemisphere, but the resulting signal would be incredibly weak and very difficult to measure). $\\$ (Continuing with first paragraph) The reason is that for very small apertures, the edges of the material play an important role. You really can not block an electromagnetic wave without creating some kind of excitation. When the aperture is larger, these edge effects are minimal, but with a very small aperture, the diffracted signal is much smaller, and also has a much wider pattern. The edge effects will likely be of some significance, and maybe even be as large as the signal itself for a very small aperture where $b<< \lambda$. In any case, you made an interesting observation, pointing out a feature inherent in the mathematics that is passed over by any and all of the textbook treatments that I have seen.

Last edited: Jul 2, 2017
8. Jul 7, 2017

### sophiecentaur

If you are dealing with identical slits and a far field situation (azimuth angle regarded as the same for all slits) you can treat the problem as 'variable separable'. This means you can find the field in any direction by 'multiplying' the form factor by the array factor. ( individual slit diffraction pattern by slit array interference pattern)

9. Jul 7, 2017

### Charles Link

The diffraction theory single slit equation, (as well as the multi-slit interference pattern), is well-established, and the conservation of energy principles apply quite accurately for both cases, including when applied in combination. In any case, the OP found, on his own, a very-limiting extreme case for an extremely narrow slit where the formula for the single slit diffraction must require a correction, and the energy conservation principles do not hold with mathematical precision. The finding has little impact on the theory as it stands, but the OP made a rather keen observation.

10. Jul 8, 2017

### sophiecentaur

I am having trouble reading the Maths expressions on my smart phone but if the slits are characterized fully with the depth as well as the width then can't the slit pattern be calculated? Very thin slits - or the equivalent- are used in microwave slot arrays and that's easier to achieve at RF, of course. I haven't sussed out where energy conservation can be violated. Energy that doesn't get through will be reflected due to the mismatch. Obvs harder to calculate if the slit material is not metallic.

11. Jul 8, 2017

### Charles Link

The energy is obviously going to be accounted for, one way or another. The problem is that the on-axis irradiance in diffraction theory in the far-field is given by $E_{on-axis} =\frac{A^2}{\lambda^2 r^2} E_i$ where $E_i$ is the incident irradiance. In the case of an extremely narrow aperture, the pattern theoretically covers the $2 \pi$ steradian hemisphere uniformly. This gives a total power $P=\frac{2 \pi A^2}{\lambda^2}E_i$ instead of the incident power $P_i=E_i A$. Normally, the effective solid angle of the diffraction pattern is $\Omega=\frac{\lambda^2}{A}$ so that the incident power is the same as that of the far-field diffraction pattern. As $A \rightarrow 0$, $\frac{\lambda^2}{A}$ needs to get large to balance the $A^2$ getting small, but is limited to $2 \pi$ by geometrical constraints. $\\$ In the demonstration of energy conservation for the diffraction pattern, the equation $\frac{\lambda}{b}=\int\limits_{-\infty}^{+\infty} \frac{sin^2(\pi b x/\lambda)}{(\pi b x/\lambda)^2} \, dx$ is used, but the $x$ in this equation is limited to $+/- \pi$, so that the integral really doesn't extend to infinity, particularly for small $b$. $\\$ Both of these are consistent with the observation made by the OP, that there must be an anomaly present for very small apertures that is not accounted for in the diffraction theory that assumes an aperture or slit width $b$ that is not exceedingly small. $\\$ I'm going to need to do a search of the literature when I get some time. This was new to me as the OP presented it, but I expect to find published papers on this anomaly in the literature. I believe we have analyzed it correctly.

Last edited: Jul 8, 2017
12. Jul 8, 2017

### sophiecentaur

I have returned to my desktop computer and I can see the Maths now.
I used the word mismatch in my previous post and I think that may be a way to describe the 'anomaly'. The diffraction pattern caused by a slit should surely consist of a reflected wave pattern which the simple -π/2 to π/2 limit doesn't take into account. The reflection is due to the edges of the slot and narrower the slit, the greater is this component.
It would be no surprise (to me) to measure detectable diffraction fringes from the back of the slit. The contrast would, of course, be very low for a single slit but for a whole diffraction grating, the ratio of slit to space ratio would be significant and would you not expect to get reflected diffraction maxima from such a grating?
Bottom line is that one method includes this and the other ignores it [Edit: by assuming that E is always forward].

Last edited: Jul 8, 2017
13. Jul 8, 2017

### Charles Link

A search of the literature turned up something in ETHZ/Photonics=a paper titled "Diffraction by a Small Circular Aperture" contributed by Alberto M. Marino and Giovanni Piredda. In a section labeled II. Bethe's solution, " The conclusion is that the power radiated by a small aperture is much less than what the Kirchhoff's theory predicts". The paper contains some very detailed analyses, but it seems to confirm the observation made by the OP @bknock . These analyses get rather mathematical, and the problem has apparently been addressed in detail by Bethe. $\\$ @sophiecentaur Although it helps somewhat to consider that there may be some consequences that result from backscatter, I think the final result is not likely to be explained simply by the inclusion of backscatter. Somehow, it has to do with the very nature of the electromagnetic propagation through the small aperture, but I think the details are beyond what most of us need to know to have a good understanding of diffraction theory. For the most part, the extra detail that the OP pointed out is only covered in some very detailed treatments of diffraction theory.

14. Jul 8, 2017

### sophiecentaur

Virtually all my EM experience is at RF and it's always interesting to see how Optics workers look at problems differently.
That nagging Energy Conservation principle makes it very difficult unless a reflected wave is included in the calculation but the details of the slit would be relevant. Does the theory assume an infinitely thin plate? Thinking in terms of a microwave slot antenna, the length would affect the measured impedance of the antenna, which would affect the Radiated Power so you couldn't just use the transmitter power to tell you the equivalent of E in your calculations. The size of the slit for a 180° beam width is small enough for the field across the slit to be non uniform so the diffraction pattern would be affected.
I can't agree with you there. Your term "backscatter" applies to a very important factor when any wave passes through a change of impedance. There will be waves inside the body of the narrow slit and there will be significant reflections on the way in and on the way out. In the absence of some other effect then we know there will always be the equivalent of mismatch loss, to account for where the missing energy has gone.

15. Jul 8, 2017

### Charles Link

Some of the energy is lost in absorption of the surface. Somehow, it is apparently physically impossible for the EM wave to go abruptly from some uniform level of electromagnetic field inside the aperture to a value of zero at the boundary of the aperture. That would appear to be what is requiring some very extensive analyses of the E&M wave contained in Bethe's solution of this problem, with the result that there are edge effects and apparently some additional absorption at the edges. $\\$ For this problem, the Kirchhoff diffraction theory assumes a uniform electromagnetic field across the aperture and zero electromagnetic field outside of this region. The Kirchhoff diffraction theory uses Huygens sources across the aperture that radiate simply in the forward hemisphere. $\\$ The Kirchhoff diffraction theory works remarkably well for most diffraction problems including those involving focusing optics, and predicts quite accurately, with complete energy conservation, the diffraction limited spot size of the focused image from a paraboidal mirror. In the case of the focusing mirror, the aperture size $b$ is essentially the diameter of the focusing mirror, so that $b>> \lambda$. $\\$ The OP's observation covers the cases where $b \leq \lambda$ and where $b<< \lambda$. For these cases, it is apparent that Kirchhoff's theory simply doesn't supply the complete answer. The cases where Kirchhoff's results don't give complete energy conservation have limited impact on the overall use of the Kirchhoff theory, but are still cases of interest. $\\$ The Kirchhoff diffraction theory also sees extensive use in diffraction grating spectrometers. Many treatments of the case of the diffraction grating derive the formula for the intensity pattern for $N$ slits or lines, $I(\theta)=I_o \frac{sin^2(N \phi/2)}{(\phi/2)^2}$ where $\phi=\frac{ 2 \pi \, d \, \sin(\theta)}{\lambda}$, without showing the energy conservation result. The energy conservation result can be readily shown with the assumptions that $sin(\theta) \approx \theta$ and that $d>> \lambda$ so that the $\theta$ limits in the integral can be set to $+/- \infty$ with little modification. (There is also a single slit diffraction pattern factor that gets included in this formula, and similar mathematics applies to show energy conservation). $\\$ In an r-f treatment of the problem, the aperture could be considered to be a region of impedance mismatch. Besides the (back-) reflected wave and transmitted wave, there could also be some absorption by the region around the aperture, and this absorption would need to be included in any energy conservation calculations.

Last edited: Jul 8, 2017
16. Jul 8, 2017

### sophiecentaur

Agreed. There have been quite a number of these 'what happens when x approaches zero (or infinity). They always throw up holes in the normally used theories.
Yes - skin depth and conductivity are important for precision.
The geometrical approach (Kirchoff) is fine for a measurement process because it's more or less a given that an instrument will be calibrated. (Sensor response would always need to be taken out of the measurements, too. if an absolute reading were needed. Haha - that's an Engineer talking.

17. Jul 9, 2017

### tech99

[QUOTE="Basically, how much energy is expected to go through a very thin slit?[/QUOTE]
If the slit is regarded as a very narrow slot antenna (which it may or may not be, because the slit can be non conducting), it can only radiate the polarisation at right angles to its direction.
The great length of the slit - thousands of wavelengths in optics - means that radiation will be concentrated on the optical axis by array action. The pattern in the plane normal to the slot will be circular.There is no minimum width for a slot antenna - a razor cut would radiate low frequencies provided the slot in more than half a wavelength long. The slot radiates due to the currents in the surrounding sheet.
There seems to be an unnecessary gulf here between antenna engineering and optics.

18. Jul 10, 2017

### bknock

Thanks Charles. Good job finding "Diffraction by a Small Circular Aperture" by Marino and Piredda! I completely agree with what you (and the paper!) are saying about the unrealistic edge assumptions used by Kirchhoff's theory. I suppose that the boundary conditions at edges are EVERYTHING at these wavelengths (for example, a Faraday cage formed by a conducting mesh essentially turns the slits/holes into waveguides or, if the conducting plate is very thin, into an antenna).

From the simplicity of Huygens-Fresnel principle, I forgot that optics was at its core hardcore electromagnetism solving Maxwell's equations!

Charles, if you contact me via here, I'll give you acknowledgments! Anyone can click the link to see my phasor-addition code!

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