What happens when the limb of the Moon passes in front of a distant star? What should we expect to see? Obviously, the star changes from bright (before the limb comes close) to invisible (after it is completely hidden by the limb), but if we could measure its intensity very carefully, what details might we expect to see?
I will attempt to give a very basic introduction to the topic in this document, together with some simple examples which may serve to provide a flavor of the sort of phenomena one might expect in real life. Those who are interested in the details should probably consult a university textbook on optics. I will mention several complicating factors, but treat none in great detail.
I'd like to thank Roger Easton, of RIT's Center for Imaging Sciences, for his advice and guidance on this problem. Thanks also to Brian Loader for catching an error in the horizontal scale of graphs in the section on partially resolved disks.
You may be familiar with the phenomenon of diffraction: the interference of waves from a source when they encounter an obstruction. We use the example of light passing through a slit and falling on a distant screen to illustrate the effect in our introductory physics classes:
In the case of monochromatic light passing through a narrow slit onto a distant screen, we find a simple pattern of light -- fringes of constant spacing -- on the distant screen. There's even a simple formula relating the wavelength of the light and the width of the slit to the angular size of the fringes. How convenient ....
When the Moon passes in front of a star, it also produces diffraction. But the situation is a bit different: this time, light passes through a "slit" which has only a single edge:
The result, it turns out, will again be a set of fringes falling onto the distant surface of the Earth:
But this time, the fringes will not have a constant spacing, and the mathematical formula required to calculate their location and amplitude will not be so easy; moreover, as the Moon's limb moves through space, the pattern of fringes will glide across the ground.
The term "Fresnel diffraction" is often used to describe situations like the lunar occultation: a distant source of light sends waves past an obstruction of some kind and onto a distant screen. Now, there are several natural lengths in this problem:
How do these lengths combine to interact? It turns out that the crucial combination is the square root of the product of L and λ,
which I will denote by the lower-case letter b. This gives the characteristic spacing between fringes on the Earth's surface, and so can be compared with x, the distance on the surface away from the geometric shadow's edge.
For our particular case of a lunar occultation observed in the visible, b is typically
b = sqrt [ L * λ ] = sqrt [ 384 x 10^(6) m * 500 x 10^(-9) m ] = 14 m
If we look at very large regions of the Earth, on scales x >> b, then the situation is simple: far from the geometric shadow's edge, for positive x the Earth's surface is uniformly bright; far within the geometric shadow, for negative x, the Earth's surface is completely dark. But in between, near the geometric shadow's edge on scales of meters or tens of meters, the brightness of the Earth's surface will vary in a complicated manner.
The goal, then, is to calculate the intensity of light falling on the Earth's surface at a distance x from the shadow's edge. It turns out that we can do so using the following equation:
Here I_{0} is some normalization factor which scales the entire pattern. I will fiddle with it throughout the examples herein to yield roughly the same overall intensity for all situations, since it is the relative variations in the signal with position that concern us.
We can split this complex integral into two real integrals if we wish:
Note that one must add up the pieces of each integral BEFORE squaring the result.
In the examples which follow, I performed numerical integration on a grid of stepsize 0.1 meter, considerably smaller than the characteristic length scale of 14 meters. I then tabulated the values of I(x) every 0.2 meters.
So, what sort of pattern do we get? The simplest case involves light of a single wavelength -- I'll choose λ = 500 nm. If we were to float in a balloon high above our telescopes, we would see a pattern of light and dark fringes on the ground like this:
It is easier to see the quantitative behavior on a slice through the pattern:
Note that the intensity reaches a value of exactly one-quarter of its nominal value at the location of the geometric shadow's edge.
If we zoom in on the first few fringes, we can see that the distance between each fringe decreases as we move away from the shadow's edge.
We have so far considered the light from a star to be monochromatic: all of a single wavelength. However, real starlight is a mix of different wavelengths. How will that affect the observations of an occultation?
In order to illustrate the effects of polychromatic light on diffraction patterns, I will consider four cases:
In each of the polychromatic cases, I will approximate a true continuum of light by adding together contributions from 11 wavelengths uniformly spanning the passband. So, for example, the "narrow" case will be simulated as the sum of light of 500, 505, 510, 515, ..., 545, 550 nm. I give each contribution equal weight. In real life, stellar spectra would vary significantly over the entire visible range, giving some wavelengths more weight than others. My simulations here therefore provide a sort of "worst case" set of examples. I suspect that a typical CCD-based video camera, observing a typical star through a typical atmosphere, yields a passband which lies somewhere between my "intermediate" and "wide" examples.
So, what happens when we include light of more than a single wavelength? In a sense, we must add together the diffraction patterns of each wavelength; I will perform the addition within the complex integral (though the results of adding the results afterwards look very similar). Each wavelength of light will have its own characteristic linear fringe size on the ground.
Thus, the diffraction fringes created by each wavelength will be slightly different in size. When we add the patterns together, it is reasonable to expect some cancellation of the diffraction fringes; that cancellation ought to become more more significant as the passband grows wider.
Let's find out if that is true. Below is the fringe intensity for the monochromatic case:
Next, look at the pattern for the "narrow" case:
The "intermediate" passband:
Finally, the "wide" passband:
Here are all four models shown at once for quick comparison:
Note that in all the cases, the first fringe is significantly brighter than the nominal intensity. As the passband grows, however, subsequent fringes very rapidly decline into uniformity.
Just for fun, we can look at the fringes as they would appear on the ground to an observer floating high above; from the very sharp, monochromatic case:
to the narrow band:
to the intermediate band:
and finally the wide band:
Note added Mar 21, 2018: It appears that I made a small error in the calculations in this section. Although the text states that the star in this example is 10 milliarcsec in diameter, it seems that my calculations used a disk only 5 milliarcsec in diameter; probably a confusion of radius and diameter. Sigh. Thanks to Tarek H. for his detective work!
Although stars are so far away that they appear as point sources to the naked eye, some are close enough and large enough that their angular size produces detectable effects. Let us see what happens if the light source occulted by the Moon's limb is not a point source, but instead a very small disk.
I will consider here only the simplest model of a star:
In real life, the apparent angular size of a star certainly does change with wavelength (generally growing with wavelength); moreover, the limbs of stars usually appear fainter than the central portions of their disks (an effect called "limb-darkening"). It is not too difficult to add these complications to the mathematical computations described below.
There are (at least) two ways that we can modify our calculations to incorporate the non-zero angular size of a light source. They ought to (and do) produce the same results, but one method turns out to be a lot quicker in this simple case.
We then create a second, smaller image, which I'll refer to as "the 2-D kernel". This should be the projection of the angular distribution of the light source onto the Earth's surface.
A uniform disk of angular size θ will turn into a filled circle of linear diameter D on the Earth's surface, where
The 2-D kernel is simply a picture of this circle on the ground. For example, given the standard distance L = 384,000 km, a star of angular diameter θ = 1 milliarcsec produces a circle of linear diameter D = 1.86 m. The 2-D kernel for a star of angular diameter 10 mas ("mas"