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Re: [ATM] No fringes
You don't have to be at the radius of curvature at all, IMO. It sounds like you have the two faces matching well enough. You simply illuminate the setup with your bright, monochromatic light source, and view it all from relatively close up. I don't know whether a green LED will be good enough or not. I would remove the shellac specks, however.
I used ordinary fluorescent lights and some greenish theatrical lighing gels (about $5 each) that had the correct spectral transmission charateristics to filter out everything except for the green mercury lines.
A diffuser would definitely help.
Christopher Dalla Piazza <dalchri@hotmail.com> wrote: I can't get fringes to appear on the interference test.
I am using a green LED which I assume is monochromatic enough. There are a
few little specks of shellac which I left on my convex lens on purpose to
act as the shim and to also prevent glass on glass contact.
I have the concave reference R=185" facing me. In front of that, I place
the convex, matching surface against it. I know that the surfaces must be
getting close together because I have to overcome that vacuum force when I
try to separate them.
Next I just hold the LED next to my eye just like in the spit test and find
the effective radius of curvature when the light fills the lens. This is
much close than 185" since the convex R=43" alters the focus, perhaps about
60" away from the test stand.
Light fills the lens but no fringes. I tried increasing the wedge to the
point that I could see a double image of the LED inside focus with no
change.
Do I need to create a pinhole or a slit for the LED? Texereau just shows a
Neon lamp and a prism at the radius of curvature. Is the orientation of the
air wedge important relative to my eye-source axis? For example, if I place
the source to the right of my eye, does the thick part of the air wedge need
to be on the right and not the top or bottom?
Am I completely misinterpreting the illustration? I wanted to avoid making
a diffuser but is that what needs to be done instead of being at the radius
of curvature?
Thanks again!
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