Re: [Phys-L] two-slit interference for non-laser sources

[John Denker <jsd@av8n.com>]

On 10/04/2016 06:17 PM, Carl Mungan wrote:
> I know this question has been discussed on web forums before, as one can
> find by googling. But I'm not very satisfied with the complexity of some of
> those discussions.

The discussions I found were incomprehensible.

[snip]

> Assuming you buy my explanation (do you?) here are two follow-on issues:

I snipped what went before, because IMHO what follows should be considered
primary, not secondary or tangential.  It is simple and well connected to
the fundamental physics:
 
> How does the diameter of the sun affect things?> Well, the angular diameter
> of the sun is about 1/2 degree. That will mean each fringe is actually a
> set of fringes spread over an angle of about 1/2 degree. 

Right.

> So to avoid
> blurring out the fringes, they need to be farther apart than that. That is
> why I chose the value of d above that I did: it means fringes spaced by
> about (500 nm)/d in rad or about 3 degrees.

The key idea here is to consider the diffraction pattern to be a 
/superposition/ of different patterns.  Essentially, we are imagining
each point on the surface of the sun to be a separate source, creating
a separate diffraction pattern.

To make this unforgettable, consider what would happen during an eclipse,
where the source is crescent shaped:
  http://www.michaelfrye.com/wp-content/uploads//2012/05/Claudia_Eclipse1.jpg
  http://www.michaelfrye.com/2012/05/30/winging-it-through-the-eclipse/


A high-power (small-d) diffractor is hard to come by, but if you have
one, it covers a multitude of sins.

  Conversely, some lasers (as mentioned in the Subject: line) produce a
  source with a very high brightness (i.e. large power per unit phase space)
  which is tantamount to a source subtending a very small angle.  With
  such a source, you can make a crude diffractor by drawing with a blunt
  crayon and still get a decent diffraction pattern.

  A third option is to use a pinhole source, but that sacrifices intensity.
 
> Okay, so the sun works. How about using an old-fashioned light bulb? Put it
> say r = 1 m away. Then for the same slits we have d^2/r = 1e-10 m, still
> plenty good enough. (Use a bulb with a clear envelope and a reasonably
> small filament to keep its angular size small.)

That's all correct as far as it goes, but there is more to the story.

In particular, the foregoing is how you know when-and-if there will be a
sharp central peak in the diffraction pattern, i.e. a bright peak right
where the geometrical ray model suggests a deep shadow.  However it
doesn't tell the whole story about the higher-order peaks.

The spread of the higher orders depends on wavelength.  Since the sun
puts out many wavelengths, once again we must consider a superposition
of diffraction patterns.  At the extreme, one could imagine the sun to
put out white light, with all possible wavelengths, resulting in a
hopelessly smeared-out diffraction pattern.  We are to some extent
rescued by the biophysics of the human eye, which responds to wavelengths
over about one octave only.

  Conversely, a broadband photometric measurement might disagree as to
  whether a clean diffraction pattern was produced.

  You could improve things by sticking in a sharp colored filter.

The superposition idea gets used over and over in this argument.  It
works because almost everything that happens is linear.  The only
relevant nonlinearity occurs within your eye, which responds to
intensity, which is a nonlinear function of wave amplitude.  This
point is probably lost on most students the first time, but it makes
a good lesson unto itself.  When faced with a complicated system like
this, it pays to think about what's linear and what's not.

  Not every problem is a nail, and not every tool is a hammer, but
  if you're holding a hammer and you see a nail, a nifty line of
  attack suggests itself.

==============

The foregoing argument can be re-expressed in fancier language, namely
in terms of coherence lengths ... but you still need every bit of that
argument in order to figure out the coherence lengths!  (Either that or
you have to stick in some hard-to-explain assumptions about the coherence
lengths.)

Note that the two halves of the argument correspond to two different
coherence lengths:
 -- One expresses the side-to-side coherence:  If you have two slits,
  each illuminated by a different flashlight, there will be no diffraction
  pattern.
 -- Another expresses the coherence in the longitudinal direction, or
  (equivalently) in the time direction.  If you use a femtosecond pulse
  laser, where the pulse is only one cycle long (as opposed to a high-Q
  almost-monochromatic wave train) there will be no diffraction pattern.

A laser tends to create fairly large coherence lengths because of the
Q of the laser cavity.  Then, if there is mode-locking, the laser will
be even more monochromatic than you might have expected.  (And if you
really want to understand why a laser is different from filtered sunlight,
there is even more to the story.)