# How far apart can the slits be in a double-slit experiment using direct sunlight?

In a normal double slit experiment, I'm told that sunlight doesn't produce a visible interference pattern because there is no stable phase relationship between the two slits.

However, sunlight bouncing off a CD does produce interference-based rainbows, so sunlight can interfere if the slits are close enough.

How close do the slits need to be to see interference when direct sunlight falls on them?

• What do you call "the normal double slit experiment"? You can get bright, spatially coherent light from sunlight with a simple lens that is being focused on a first slit. This will give you as flat a wavefront as you want by placing the two slits far enough away from the first slit. If you don't want the color fringes, insert a simple narrowband color filter and they are gone. If you are doing the experiment without the first slit, it's like having many entrance slits overlaid on each other, which destroys the interference. That's not "normal" but poor experimental technique. – CuriousOne May 13 '16 at 7:28

The visibility of interference fringes from a double slit depends on the how correlated the fields from the source at the two slits are. The typical slit-distance for which the fringes are manifestly visible is called the transverse coherence length $d_{\rm coh}$ of the source. For a spatially extended source, this quantity primarily depends on two parameters, namely the size $S$ of the source and the distance $R$ of the double-slit from the source. The quantitative relation is, (please see "Optical Coherence and Quantum Optics" by L. Mandel and E. Wolf for the derivation)

$d_{\rm coh}\approx\frac{R\bar{\lambda}}{S}$

where $\bar{\lambda}$ is the mean wavelength of the source. The inverse dependence on $S$ can be intuitively understood from the fact that a spatially extended source is a collection of many uncorrelated point sources, each of which produces its own double-slit pattern. The pattern due to the full source is the incoherent sum of these intensity fringes. Since the individual point sources were uncorrelated, these fringes add up in an uncorrelated manner and have the effect of washing out. As a result, the larger the source, the lesser the coherence length and consequently more difficult for interference to manifest. Upon plugging in the numbers in the above formula, you will find that $d_{\rm coh}$ for the sun for a double-slit placed on earth is of the order of tens of microns. Thus, the distance between the two slits must be less than this transverse coherence length of sunlight, i.e less than few tens of microns to observe interference. Regarding your example, since the surface topography of a compact disc varies on those scales, the interference effects are manifestly visible to the naked eye.

P.S: To address the point raised by one of the comments, when you focus the sunlight using a lens and do a double slit experiment, the slit distance can indeed be larger. This is due to the fact that you have now prepared a secondary source (tiny point source) from the primary source (sun), and the interference effects will now be governed by the coherence of the secondary source. The transverse coherence length according to the above formula (taking the limit $S\to0$) is indeed large.

• Ok, how far apart for slits until they would not make a pattern? Might as well spell it out clearly. – Jon Custer May 13 '16 at 13:10
• Sorry. I have made the edit now. – Girish May 13 '16 at 13:14
• @Girish, that's amazing. Are you saying that a focusing lens is all you need to provide coherent sunlight to a double-slit? I was under the impression that, even with the lens, you still had to have a small aperture. That's what Young did if I'm not mistaken...he created a 'point source' at the window by means of an opaque covering with a pin-hole, then used a focusing lens to direct the light over a table surface with his double-slit. – David Reishi May 13 '16 at 18:51
• @David: Well you would need an opaque screen with an aperture exactly aligned with the focal spot of the lens. This is in order to prevent the extraneous sunlight (the part not falling on the lens) from participating in the double slit experiment, and obscure your interference pattern. But apart from that, a lens is all you need and that is indeed quite interesting. – Girish May 14 '16 at 3:43
• @Girish, let's say we have a window covering with a pinhole. That pinhole serves to isolate relatively spatially coherent light if I'm not mistaken, right? So what role does the lens then play in terms of the spatial coherence? Does it make the light even more coherent? I thought the lens's role in this case was to simply take the spatially coherent light emerging from the pinhole and "get it to where you need it in one piece," so to speak. – David Reishi May 14 '16 at 7:16