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Re: Thomas Young's experiment



On Saturday, Nov 1, 2003, Ludwik Kowalski wrote:

On Saturday, Nov 1, 2003, Bernard Cleyet wrote:

try replacing the card with a strip whose
depth is the same as its width.

I am not performing an experiment. Why would the
length of the card slip (which divides a beam from
a pinhole into two separate parts) have to play a
role? I suppose the slip card was black in order to
minimize stray light. What is wrong with thinking that
the card should act as if it were a wire?

Unfortunately, I was not able to figure out what
fraction of the beam area was actually covered by
the card. There were no illustrations in Young's
paper. To act as a single slit each uncovered area
was probably very narrow. Thus the slip card
shadow was probably at least 90% of the area of
the diverging beam.

In a single-wire experiment the situation is usually
opposite; the area of the wire is a small part of the
beam area. I can not prove it but I suspect that in
going from one extreme to another one must be able
to observe a transition from "peaks of equal widths"
to a typical diffraction pattern -- a "central strong
peak about twice as wide as other (weaker) peaks."
That may be an experiment worth performing.

It is not either one (equidistant fringes due to two
coherent sources) or another (a dominant central
peak due to diffraction from a single wire). The two
phenomena coexist, as I just realized by performing
an experiment at school.

Our lab has no windows and I used a 1 mW He-Ne
laser beam. The beam diameter, on the white screen
2.5 meters away, was about 5 mm,; the diameter
near the laser output (where my wires were) was
about 1 mm.

Observation #1: (wire diameter 1.7 mm). The
beam is totally dark; it is blocked by the wire.

Observation #2: (wire diameter is 0.9 mm) I saw
the central diffraction peak and seven equidistant
(interference) peaks inside of it. The distance dx
between the interference peaks was about 1.8 mm.
I suppose that Young saw some of the seven peaks
but only the central peak was white. The size of my
wire was about the same as the size of his slip card.
Most of the sun beam was blocked by his card and
what was passing (on each side) was equivalent
to two slits or two pinholes.

A also saw fringes under other diffraction peaks,
as shown in our textbooks. They would not be
visible in white light. Can I get the wavelengt of
my laser light from these data (known to be
633 nm)? Assuming the distance detween the
equivalent slits is d=0.95 mm, and knowing
that D=2500 mm I have lambda=dx*d/D=646 nm.
Not bad, considering crudness of my estimation
of d (and poor accuracy of measuring dx).

It is a very easy experiment, use it to celebrate
the 200 anniversary of Young's paper. I suppose
that the experiment can also be performed with
a common laser pointer. My copper wire was
from a telephone cable.

Observation #3 (which can be presented as
a puzzle):
The wire was replaced by a hair. Before
introducing the hair into the laser beam I saw
a circle on my screen; its diameter was 5 mm.
The circle was still present (as it should) when
the hair was inserted. With the hair in I could
see three diffraction peaks: the dominant one
was 63 mm wide while the other two were much
less intense and about 30 mm wide. So far all is
normal; the distance 63 mm corresponds to
the hair diameter of 0.05 mm.

But I also saw equidistant peaks inside the
diffraction peaks (but only outside the 5 mm circle).
The distances between them were about 1.5 mm.
Can the equidistant peaks be attributed to Young's
diffraction? This question is legitimate because my
"equivalent slits" were essentially semi-circles of
widths 0.5 mm each. And the distance between
their "centers of intensity" were probably close to
0.25 mm. With these numbers I get dx=1 mm. Why
is not closer to 1.5 mm (as measured)?
Ludwik Kowalski.