Re: Building 30" Drum Scope

On May 30, 10:38 am, Dave Typinski <möb...@xxxxxxxxxxxxx> wrote:
Hayden <haydenflem...@xxxxxxxxx> wrote:

On May 30, 3:38 am, Dave Typinski <möb...@xxxxxxxxxxxxx> wrote:
Hayden <haydenflem...@xxxxxxxxx> wrote:

See figure 3a above with the description "Holding up
a card in front of the subject, because every point on
the card receives light rays from numerous points on
the subject". If the aperture is so big.. like 2 foot,
won't numerous points on the subject especially if
it's near the aperture

Aha!  Yes, that's a good point--but we weren't talking about imaging
objects close to the aperture.  Stars are as far away from the
aperture as you can get, right?  ;-)

Pinhole cameras are used on close objects like cars,
buildings, etc. So if you have a 2 foot diameter pinhole
camera focusing at a car 5 meters away,

Um... I may know nothing about optics, but /that/ ain't gonna happen.
You'd be tring to image details smaller than the "pinhole."  How could
that possibly work?  

What? If the pinhole is 2 feet across. The details in
the target you would see is govered by
theta = 1.22 lambda/diameter or 0.23 arcsec. So
you can see 0.23 arcsec of the object which is
smaller than the size of the pinhole and the focal
plane would be 130 miles away, no?

Hmm.. are you saying a pinhole can only image
details the same size as a pinhole?? So if you have
2 foot aperture.. you can only see details in the
target that is 2 foot across so if the target is only
2 feet in size, you would see just a single disc? Hmm...

About the car 5 meters away, well, thinking it over,
focusing 5 meter away from aperture with a 130
miles focal length for a 2 feet aperture is not
possible unless the distance from target is more than
130 miles focal. It's like you can't focus object with
an 8" SCT that is closer than focal length of 2032mm.

Back to the pinhole able to image only detail the
size of pinhole. So if you use the 2 foot diameter
lensless aperture on the moon and stars. What
details would you see? I think it is still governed
by the formula theta = 1.22 lambda/diameter or
the resolving power of the pinhole so if diameter
is 1mm, resolving power is 138.4 arcsec.

how long
would the focal point have to be such that the
diffraction disc formed won't come from numerous
points of the subject? Can you calculate it, for
practice?

Nope, because I know not how.

My SWAG (scientific wild ass guess) is that a pinhole camera provides
a minimum angular resolution based on wavelength, the limit being
determined by diffration effects.  Whether the pinhole is 2mm or 2
feet, the image plane is still going to "see" features with a solid
angle no smaller than the angle subtended by the diffraction pattern
of a point source.

But isn't it that the angle subtended is governed by
theta = 1.22 lambda/aperture? So if the aperture is 2 feet,
it can resolve 0.23 arcsec, if 2 mm, 69.20 arcsec. No?
Are you saying both 2 feet and 2mm aperture can
resolve the same detail? But note diffraction pattern
are formed by the aperture and not inherent in the
object or target.


Or something like that.

illuminate the same diffraction
disc in the focal plane located 130 miles away??
But then thinking it over, I think the trigo tan theta=y/x
can make it work where the diffraction disc is not
overlapped by numerous points on the subject if
it is far enough. Is this what you have in mind?
Anyway, how do you imagine the diffraction disc
would look like 130 miles away. It would be more
than 2 foot in diameter!

Yep.

Not more than 2 feet.. it seems less.

Why?  For a 0.01mm slit, the distance between the first two minima of
the diffraction pattern imaged a meter from the slit is certainly
larger than 0.01mm.  Should work the same way for round holes and
pinhole cameras, right?

That's because the target is not at infinity. If it is nearer, then
the focal length would be further away and bigger.


If target is an infinity, how big exactly would the
diffraction disc be 130 miles away with the 2
foot diameter aperture?  I wonder if it's r= 1.22 lambda f/D =
1.22 (0.000550) x 209,214,000mm / 600mm= 234mmx2=
468mm or 18" in size. 2 foot is 24", The diffraction disc
seems to be smaller than the diameter. Agree with this?

Yep.  So maybe it's smaller after all!
--

One should add blur size introduced by geometric optics
which is govered by size = a (p/p+q). If p is an infinity,
size is close to the aperture. See:

http://www.huecandela.com/hue-x/pin-pdf/Prober-%20Wellman.pdf

Hay

Dave- Hide quoted text -

- Show quoted text -

.

Relevant Pages

  • Re: Building 30" Drum Scope
    ... a card in front of the subject, ... If the aperture is so big.. ... You'd be tring to image details smaller than the "pinhole." ... can make it work where the diffraction disc is not ...
    (sci.astro.amateur)
  • Re: Building 30" Drum Scope
    ... aperture is a plane wave (no phase variation across ... how do you imagine the diffraction disc ... would look like 130 miles away. ...
    (sci.astro.amateur)
  • Re: Building 30" Drum Scope
    ... since it depends on the aperture size only. ... have a 30 inches aperture scope, but without any lens. ... It's a camera obscura or pinhole camera. ... diameter at one end with the film at the other. ...
    (sci.astro.amateur)
  • Re: Building 30" Drum Scope
    ... aperture is a plane wave (no phase variation across ... can make it work where the diffraction disc is not ... how do you imagine the diffraction disc ... would look like 130 miles away. ...
    (sci.astro.amateur)
  • Re: Building 30" Drum Scope
    ... since it depends on the aperture size only. ... have a 30 inches aperture scope, but without any lens. ... It's a camera obscura or pinhole camera. ... distance between the pinhole and the film and lambda is the wavelength ...
    (sci.astro.amateur)