| Topic: |
Science > Physics |
| User: |
"PD" |
| Date: |
18 Apr 2006 12:12:36 AM |
| Object: |
Cute problem |
Suppose you are sufficiently downstream from a double slit experiment
that the separation between the maxima is large (meaning, larger than
the size of your iris). If you look toward the two slits with one eye,
what will you see? Will you see two lit slits?
PD
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| User: "Old Man" |
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| Title: Re: Cute problem |
18 Apr 2006 10:10:03 PM |
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"PD" <TheDraperFamily@gmail.com> wrote in message
news:1145337156.029370.176580@g10g2000cwb.googlegroups.com...
Suppose you are sufficiently downstream from a double slit experiment
that the separation between the maxima is large (meaning, larger than
the size of your iris). If you look toward the two slits with one eye,
what will you see? Will you see two lit slits?
No. That's what
delta_theta = lambda / d
is all about.
[Old Man]
PD
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| User: "Ken S. Tucker" |
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| Title: Re: Cute problem |
19 Apr 2006 01:40:41 AM |
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Old Man wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1145337156.029370.176580@g10g2000cwb.googlegroups.com...
Suppose you are sufficiently downstream from a double slit experiment
that the separation between the maxima is large (meaning, larger than
the size of your iris). If you look toward the two slits with one eye,
what will you see? Will you see two lit slits?
No. That's what
delta_theta = lambda / d
is all about.
[Old Man]
Yup Old Man is right, (looks like OP is a troll), it's just
aperature width vs resolution, standard stuff in amateur
astronomy. Is it more complicated???
Ken
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| User: "PD" |
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| Title: Re: Cute problem |
19 Apr 2006 07:16:27 AM |
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Ken S. Tucker wrote:
Old Man wrote:
"PD" <TheDraperFamily@gmail.com> wrote in message
news:1145337156.029370.176580@g10g2000cwb.googlegroups.com...
Suppose you are sufficiently downstream from a double slit experiment
that the separation between the maxima is large (meaning, larger than
the size of your iris). If you look toward the two slits with one eye,
what will you see? Will you see two lit slits?
No. That's what
delta_theta = lambda / d
is all about.
[Old Man]
Yup Old Man is right, (looks like OP is a troll), it's just
aperature width vs resolution, standard stuff in amateur
astronomy. Is it more complicated???
Ken
It certainly wasn't intended as a troll. It was tossed out not as a
Great Unanswered Question, but as a small exercise with an interesting,
practical result that is worth going through the analysis of. Not
everything posted here has to be a Great Unanswered Question, right?
PD
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| User: "Timo Nieminen" |
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| Title: Re: Cute problem |
18 Apr 2006 12:34:35 AM |
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On Tue, 17 Apr 2006, PD wrote:
Suppose you are sufficiently downstream from a double slit experiment
that the separation between the maxima is large (meaning, larger than
the size of your iris). If you look toward the two slits with one eye,
what will you see? Will you see two lit slits?
My guess is that you'd see one slit with a width that would give you a
single slit central maximum approximately the width of your pupil.
I thought about the N-slit grating pattern. The overall modulation tells
you the width of individual slits; how can you get this information unless
at least the central peak (of the overall modulation) falls on the pupil?
A larger spread than that, and the individual slits must be narrower than
the eye can resolve. Similarly, the distance between the slits is
responsible for the spacing of the lines. How can you get this information
unless at least 2 lines fall on the pupil? A wider distance than that at
the eye and the spacing must be closer than the eye can resolve. Finally,
the width of each line tells you the overall width of the grating. As long
as at least the central maximum of a line falls on the pupil, you should
be able to resolve the grating-as-a-whole.
For the question above, except perhaps for when one line is centred on the
pupil, and the adjacent order just miss it, it'll be the case that the
pupil is smaller than the central maximum of any individual line, so the
apparent width of the whole grating will result from the resolution limit
of the eye, and no individual feature of the grating will be visible.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "PD" |
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| Title: Re: Cute problem |
18 Apr 2006 03:02:01 PM |
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Timo Nieminen wrote:
On Tue, 17 Apr 2006, PD wrote:
Suppose you are sufficiently downstream from a double slit experiment
that the separation between the maxima is large (meaning, larger than
the size of your iris). If you look toward the two slits with one eye,
what will you see? Will you see two lit slits?
My guess is that you'd see one slit with a width that would give you a
single slit central maximum approximately the width of your pupil.
You guess well!
d=slit separation
D=distance from slits to eye
L=wavelength
p=pupil diameter of eye
The fringe separation is LD/d and want to insist that this is larger
than the pupil diameter by a sizable amount, so that
LD/pd > 1
The angular resolving power of the eye is about L/p, and the angular
separation of the slits as seen by the eye (if resolvable) is d/D. So
to be able to resolve them, we must have d/D > L/p, or LD/pd < 1
Thus if the eye is placed far enough out that the fringes are separated
by more than the pupil diameter, it will be impossible for the eye to
resolve the two slits, and the eye will see a single broad and fuzzy
slit at best. This is an easy thing for even amateurs to test.
PD
I thought about the N-slit grating pattern. The overall modulation tells
you the width of individual slits; how can you get this information unless
at least the central peak (of the overall modulation) falls on the pupil?
A larger spread than that, and the individual slits must be narrower than
the eye can resolve. Similarly, the distance between the slits is
responsible for the spacing of the lines. How can you get this information
unless at least 2 lines fall on the pupil? A wider distance than that at
the eye and the spacing must be closer than the eye can resolve. Finally,
the width of each line tells you the overall width of the grating. As long
as at least the central maximum of a line falls on the pupil, you should
be able to resolve the grating-as-a-whole.
For the question above, except perhaps for when one line is centred on the
pupil, and the adjacent order just miss it, it'll be the case that the
pupil is smaller than the central maximum of any individual line, so the
apparent width of the whole grating will result from the resolution limit
of the eye, and no individual feature of the grating will be visible.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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