# Question of size in double slit experiment [duplicate]

I have a few questions about double slit experiment.

1. In order to get an interference pattern with electromagnetic waves, what is the relationship between widths and separation of slits with the wavelength of the wave?

2. What about if we want to get an interference pattern for electrons? What is the wavelength of electrons when they behave as waves?

3. What about bigger objects? I've read that the interference pattern has been reproduced with big molecules. Is there any theoretical or practical limit on the size of particles for which an interference pattern is possible? This is my main question.

## marked as duplicate by Pieter, stafusa, Jon Custer, sammy gerbil, glSMay 10 '18 at 7:43

1. The interference pattern for a double slit experiment is

$$I(x)=I_0 \cos^2 \left ( \frac{\pi d x}{\lambda D} \right )$$

where $I_0$ is the maximum intensity (the intensity of the central maximum), $d$ is the slit separation, $\lambda$ is the wavelength of the incoming waves (regardless of whether they are EM waves, waves in a lake, etc.), and $D$ is the separation between the plane of the two slits and the observing screen. Note that the above formula is valid only under the assumptions that $d\gg\lambda$ and $D\gg d$ (i.e. we are operating in the far field regime and the width separation is much greater than the wavelength of the light).

If you also wish to take into account the modulation by the diffraction pattern, this becomes

$$I(x)=I_0 \left ( \frac{\sin \left ( \frac{\pi ax}{\lambda D} \right )}{\frac{\pi ax}{\lambda D}} \right )^2 \cos^2 \left ( \frac{\pi d x}{\lambda D} \right )$$ where $a$ is the width of each slight. I've highlighted in bold the quantities you were interested in.

1. You can obtain the De Broglie wavelength of electrons from the De Broglie equation:

$$\lambda_{DB} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$

assuming the speed of the electron is much less than $c$, otherwise the relativistic expression for the momentum $p$ must be used.

For example, if an electron ($m = 9.11 \times 10^{-31}$ kg) is accelerated through a p.d. of $1$ kV, then it will have an energy of $E=1$ keV $= 1.60 \times 10^{-16}$ J and so its De Broglie wavelength will be $\lambda_{DB} = 3.88 \times 10^{-11}$ m.

On a side note, this corresponds to a $32$ keV photon (X-rays). Comparing the energies needed to achieve a certain wavelength can actually give you some pretty deep insight. If you have a lattice whose intermolecular spacing is such that, in order to obtain a diffraction pattern, you need the incident light to be of $\lambda \approx 3.9 \times 10^{-11}$ m, then you now know it is easier to do so (from an energetic point of view, at least) by accelerating an electron rather than exciting a photon, since the latter requires more energy than the former. There are many such applications in structural analysis and determination, where resolution on the atomic scale must be achieved. (You may also wish to compute the energy a neutron must have to have a De Broglie wavelength similar to that of an electron or to the wavelength of a photon -- things are even more interesting there).

1. I would suggest you visited this question, as mentioned in the comments. But I thought I might make a quick note.

All matter also displays a wave-like nature, even though it can be quite well hidden: for example, the De Broglie wavelength of a walking human being is on the order of $10^{-36}$ m.

Since all matter is also wave-like, everything exhibits diffraction effects when it goes through an aperture. A book I read had a very interesting example of a man (or woman) walking through a door -- unfortunately I cannot recall the book's title but I ran some calculations and here are my results.

When a $60$ kg human walks through a door, he is diffracted in such a way that, in order to be deflected by $1$ cm from his/her intended path, he/she would need to walk for roughly $10^7$ times the distance to the edge of the observable universe. The diffraction effect is minimal, but it is still very much there! In fact, if for some reason the value of Planck's constant $\hbar$ were much higher, then the effect would be a lot more noticeable. Although I'm sure there would be plenty of other interesting consequences too.

• Amazing answer, thanks. My guess was that particles bigger than a certain size will be too big to pass through the corresponding slits, and this is the practical limit. Your equations seem to confirm this. What is that size limit? – Transcendent Apr 13 '18 at 10:09
• I had a look on the equations again, and I realized by choosing larger D, we can have wider slits. Let's assume D<10 m. – Transcendent Apr 13 '18 at 10:14
• Do we get interference patterns by simply shooting atoms to slits, or should atoms be modified/be in some special state first? – Asmani Apr 13 '18 at 10:16