# Slit sizes needed for light diffraction

• genxium
In summary: The interference pattern from double slit of width ##a## and separation ##d## is given by$$f(y) = f_{diff}(y) f_{int}(y)$$where ##f_{diff}(y)## and ##f_{int}(y)## are the field pattern from a single slit diffraction pattern with the width of ##a## and the field pattern from the two slits ###s# with separation ##d##.
genxium
I'm reading this tutorial on instructables for implementing the classical double-slit experiment. I've also read this thread for information which contains a very nice answer from @BruceW but still not resolves my confusion about the "sizes".

First of all I'd like to make a distinction between "longitudinal slit width" and "transverse slit width(s)". I believe the @BruceW answer for the referred thread is convincing for that "longitudinal slit width should be comparable to wavelength of light". Hence I'm asking mostly about "transverse slit width(s)".

In the tutorial on instructables, it's said that one can make double-slit by

Cut two parallel slits in the middle of the foil, about 1 mm (less than 1/16") apart. The tip of a utility knife pressed through will do the job well.

Which seems ambiguous in "utility knife". I have a box-cutter utility knife at hand but its blade tip seems nothing comparable to 400nm ~ 700nm as I can clearly see. On the transverse plane of double-slit experiment, do I have to cut both transverse slit widths to nearly wavelength of light? If yes how did Young make it back at his time? Or else how long should they be cut? Is this relevant to light-wavelength, beam-transverse-width(s) or polarization?

It is not difficult to calculate what will work, but it is even easier to try. Make two slits next to each other. Then hold them in front of your eye and look through them. For example, look at an LED. You should see the double-slit diffraction pattern.

genxium said:
On the transverse plane of double-slit experiment, do I have to cut both transverse slit widths to nearly wavelength of light?

You shouldn't need to cut it that small. Just try to get the slits as small as you can while keeping the edges as clean and straight as possible. The key to the double slit experiment is that path length differences between the two slits and the detection plane lead to alternating constructive and destructive interference and causes fringes to form. The slits don't need to be comparable to the wavelength of the light passing through.

genxium said:
First of all I'd like to make a distinction between "longitudinal slit width" and "transverse slit width(s)". I believe the @BruceW answer for the referred thread is convincing for that "longitudinal slit width should be comparable to wavelength of light". Hence I'm asking mostly about "transverse slit width(s)".

Note that Bruce is talking about diffraction of light from a single slit, not from two slits. If you shine a flashlight through a doorway, the opening is so big and the beam is so wide compared to the wavelength of visible light that very little diffraction happens. You don't see your flashlight beam turn into a wide cone after passing through the doorway. However, if the doorway was a circle 700 nm across, there would be noticeable diffraction of the light after passing through. Instead of a beam you'd get a wide cone.

yeah... I guess there will still be an interference pattern even if the slit width is not comparable to several hundred nanometers. There will still be diffraction from the edges of the slits, but there will also be light passing through the middle of the slit without being diffracted. So I think that's what they mean that the interference pattern will be less noticeable if the slits are not cut very thin.

Thanks for all the replies. I actually succeeded in trying out the experiment by cardboards and utility knife several days ago.

This is how the raw laser source (it's in fact "very" red but my phone's camera ruins the color) looks likeI couldn't take photos of the interference pattern(dots) because I held the cardboard and laser source both by hands :)

So roughly speaking, only "narrow enough" slits work, but it's kind of difficult for me to measure how narrow it should be. Is there some entry level reference I could read? I tried googling and reading some of the materials I found but most of them are not easy to understand for me, e.g. lots of quantum physics equations without reference.

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Hmm. I can't see the latter two pictures for some reason.

genxium said:
So roughly speaking, only "narrow enough" slits work, but it's kind of difficult for me to measure how narrow it should be. Is there some entry level reference I could read?
Even if you know how narrow the slit should be in order to be able to see the fringes, it's not like you have a precise control of how small you can carve out the slits, or do you have access to a light microscope at least?
Anyway, the interference pattern from double slit of width ##a## and separation ##d## is given by
$$f(y) = f_{diff}(y) f_{int}(y)$$
where ##f_{diff}(y)## and ##f_{int}(y)## are the field pattern from a single slit diffraction pattern with the width of ##a## and the field pattern from a double line source separated by ##d##, respectively. In particular, ##f_{diff}(y)## is a squared of sinc function and hence has a very strong central lobe but dies out very quickly at the immediate neighboring lobes, so you will probably only see fringes which are contained in the central lobe. The first zero of ##f_{diff}(y)## is given by
$$y = \frac{L\lambda}{a}$$
and suppose we expect to see ##2m+1## fringes inside the central lobe, using that the ##m##-th zero in ##f_{int}(y)## is given by
$$y = \frac{(m+1/2)L\lambda}{d}$$
We have the condition for ##d## and ##a##
$$\frac{d}{a} = m+\frac{1}{2}$$
If we want to see 15 fringes (m=7) and using ##d## as suggested in your link ##d=1\hspace{1mm}mm##, you will have to make the slit width to be ##a=0.13\hspace{1mm}mm##. I think this value is achievable with standard utility knife provided you cut it carefully.

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gracy and genxium
genxium said:
, e.g. lots of quantum physics equations without reference.
If you look at a textbook (or on-line equivalent) you are very unlikely to find QM equations in a discussion of diffraction. Did you really mean that statement?
Producing good Young's Slits images is really a matter of trial and error. In a world where Laser Pointers are cheap as chips, all the hard work has been done for you in a very affordable package.
Just put the values of wavelength (red light) and the spacing of the fringes you want into the formula in the above post and that will tell you the separation you need for the two slits. How wide should the slits be? A simple answer is that they should appear to be significantly thinner than the separation of the slits. That should mean that an individual slit diff pattern will be much wider than the pattern from the pair. One important point is that the slits need to be as near identical as possible so that the cancellation is near perfect in the null.
Alternatively:
Double slit pattern calculator
Single slit diffraction calculator
There you are: on a plate, my friend.

PS "I couldn't take photos of the interference pattern(dots) because I held the cardboard and laser source both by hands :)" Use sticky tape to hold the things in place and then you will have both hands available to take a photo!

sophiecentaur said:
Producing good Young's Slits images is really a matter of trial and error. In a world where Laser Pointers are cheap as chips, all the hard work has been done for you in a very affordable package.

Exactly. It's much easier to exercise nowadays

@sophiecentaur, about the "quantum physics reference" problem I'm afraid that my previous reply was not stating the situation correctly. I'm intentionally trying to learn how to calculate "how light beam behaves while traveling thru a single-slit" if the wavelength of light, and <transverse_width, transverse_height, longitudinal_depth> of the single slit are given. Thus I came across some information that involves QM -- but it's not good for me to say that "lots of quantum physics equations without reference" bcz it's actually not a lot and as you said, I could be reading improper information.

As @blue_leaf77 pointed out that single-slit-diffraction plays a role here and the answer is very satisfying. However, I learned this method to calculate the single-slit-diffraction intensity distribution before and had some questions about it, but it seems better to create another thread to ask when my questions are ready

P.S. it's shameful but true that I didn't have sticky tapes or glue at home back then

genxium said:
single-slit-diffraction plays a role here and the answer is very satisfying
It is 'allowed' to multiply the two patterns together so narrow slits will provide wide enough beams to let the double slit pattern come through recognisably. If the two slits are too wide then the levels of the double slit pattern will drop off more quickly.
Have fun with your craft knife. Back in the day of 'real' photography, people used to use photo negatives of a large, carefully drawn black on white image. That would give excellent slits on fine grain film.

Hi genxium! It's nice to hear you are trying to do the double slit experiment. I wrote a blog post on doing the double slit experiment at home, but regretfully PhysicsForums does not have blogs anymore. But... here's what I wrote regarding construction of slits in my blog post:
DennisN blog said:
CONSTRUCTION OF SLITS
There are various ways to construct slits; I have tried these, and I preferred options 2 and 3.
1. Razor blades (be careful) and aluminium foil. Paint the aluminium foil black (low gloss) with e.g. spray paint. Tape two razor blades together and cut two parallel slits in the foil. Make a number of versions, since it's easy to tear apart the foil.
2. A hair (40-120 µm). One barrier is equivalent to two slits with infinite width, so you can get an interference pattern by using just a single thin barrier like a hair. If needed, paint the hair black with a marker pen or spray paint. Tape up the hair on some frame (e.g. old dia slides).
3. A thin copper wire. Extract one of the thin copper wires inside an electrical wire, cut it to a suitable length and paint it black.
4. A lead for mechanical pencils can also work, but thinner barriers are better.
5. A sewing needle. Paint it black. Notes: Concerning needles and leads, you can also try taping a number of them together. This will result in two or more slits with finite width. Don't tape them too tight together though, there should be some space between them.
6. A CD. Reflection from a laser via a CD can also show interference patterns, but they can be difficult to interpret, so I do not recommend it.
7. Prefabricated slits. You could also buy a slit setup from some science shop.
I will try to send you a copy of my blog post in a while (check your conversations). The blog post contains various info and hints, and some links to relevant instructional videos. Here is a nice one from MIT: Thomas Young's Double Slit Experiment.

When you have done the standard double slit experiment you could also try experimenting with a single pinhole and a double pinhole system (which also is described in the video clip above), those patterns are also very nice to see for yourself. Good luck!

EDIT: I copied the old blog post into a message to you (including some pictures), check your conversations...

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Drakkith
Try using 2 pieces of .5 graphite set by a standard plastic stand then shine a light or wave of of water, just basic ideas no complicated ideas.

## 1. What is the relationship between slit size and light diffraction?

The size of the slit used in a diffraction experiment is directly related to the amount of diffraction that occurs. A smaller slit will produce a wider diffraction pattern, while a larger slit will produce a narrower pattern.

## 2. How do I determine the appropriate slit size for my experiment?

The appropriate slit size for a diffraction experiment depends on several factors, including the wavelength of light used, the distance between the slit and the screen, and the desired width of the diffraction pattern. Generally, the slit should be smaller than the wavelength of light to produce a clear and defined diffraction pattern.

## 3. Can I use any size slit for light diffraction?

No, the size of the slit must be carefully chosen to produce the desired diffraction pattern. Using a slit that is too large can result in a diffraction pattern that is too narrow and difficult to observe, while a slit that is too small can produce a diffraction pattern that is too wide and lacks detail.

## 4. How does the distance between the slit and the screen affect diffraction?

The distance between the slit and the screen is an important factor in diffraction experiments. As the distance increases, the width of the diffraction pattern also increases. Therefore, a larger distance between the slit and the screen requires a smaller slit size to produce a clear and defined diffraction pattern.

## 5. Are there any other factors that can affect the size of the diffraction pattern?

Yes, other factors such as the angle of incidence, the refractive index of the material, and the shape of the slit can also affect the size of the diffraction pattern. It is important to carefully consider all of these factors when determining the appropriate slit size for a diffraction experiment.

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