Subject:
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Re: Math/Optics Problem
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Tue, 14 Aug 2001 18:33:23 GMT
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Viewed:
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97 times
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In lugnet.off-topic.geek, David Eaton writes:
> So. Let's assume I have a photograph of Bob standing in front of his house.
> Knowing Bob the way I do, I know that Bob is 6 feet tall. And asking him
> later, I know that his house is 30 feet tall. Now I want to find out how
> far
> in front of his house Bob was when the photograph was taken. In the
> picture,
> Bob is, say, 2 inches tall, and the house is, say, 4 inches tall. As far as
> I know, I need one more piece of information to solve the puzzle: the
> "fisheye" constant of the photograph. (I don't actually care about the
> answer-- I want to know how to solve the problem)
Alrighty. Since I have been studying this stuff for the past month and
am supposed to pass a test about it, let's see what I can do.
> #1 Will the fisheye constant be consistant for the same camera?
Depends. Does it "focus"? The "fisheye" constant you're referring to
is, in technical terms, called the focus. It is constant for any individual
lens and depends on the curvature radii of the two surfaces, as well as the
coefficient of the material you're using (n = speed of light in the
first media/speed in 2nd). The equation is:
1/f = (1 - n)*(1/R1 - 1/R2) with R1 and R2 as the radii.
There is no set equation for the "fisheye constant" of the combination,
and it depends on the distance between them. The "focus" of a camera can be
adjusted by changing that distance.
If the two lenses are adjacent, the equation for the comprehensive focus is:
1/f = 1/f1 + 1/f2
> #2 If not, will it at least be constant for the same zoom level?
No, just for the same "focus". If you "focus" the camera it will be
constant.
> #3 are there any other "missing pieces" to the formula?
hmm, well, you've got the magnification of the images. M = -q/p = H'/H ,
where q=distance of the image to the lens and p=distance of the object
to the lens, H is the length of the object, H' = length of the image. The
neg. sign is not relevant here, since the image you get on the camera *is*
inverted.
(Now, all this holds true for a *single* lens. I am assuming that the
distance between the two lenses is so small relative to the rest of this
that it is irrelevant. I may be making a horrible mistake by doing this,
though. Normally, you'd have to figure out the image made by the *first*
lens, then use it as the *object* of the second lens, then figure out the
image made by the second lens. But, *ignoring* that, you get:)
You know that
M of Bob = 2 inches / 72 inches = 1/36 = q(bob)/p(bob)
M of house = 4 inches / 360 inches = 1/90 = q(house)/p(house)
q is the same for both - it is the distance between the lens of the camera
to the film.
p(house) = p(bob) + d
(d is the distance between bob and the house - what you want to find)
Putting those together:
90*q = p(house) = d + p(bob)
36*q = p(bob)
(90 - 36)*q = d + p(bob) - p(bob)
54*q = d
As you can see, you are missing q.
A cute little pic for ya:
http://www.geocities.com/shiri_writing/optics.jpg
> #4 what is the formula for figuring out the distance between the two
> objects?
Well - there is none. There's a formula for figuring out the distances
between the object, focus, and image:
1/f = 1/p + 1/q
And then the magnification, above:
M = -q/p = H'/H
Using these two equations, you can usually find just about anything.
> #5 can I use that same formula to find out how far away from the
> *camera*
> the objects are?
That *is* the formula above.
HTH-
-Shiri
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Message has 1 Reply:
 | | Re: Math/Optics Problem
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| (...) Really? I woulda thought it had more to do with the zoom rather than the focus. But then again, I'm really not sure how the lenses in cameras work. I guess my assumption (based on nothing at all I think) is: {A}--------B--------...E)--F--[G] (...) (23 years ago, 14-Aug-01, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Math/Optics Problem
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| So. Let's assume I have a photograph of Bob standing in front of his house. Knowing Bob the way I do, I know that Bob is 6 feet tall. And asking him later, I know that his house is 30 feet tall. Now I want to find out how far in front of his house (...) (23 years ago, 13-Aug-01, to lugnet.off-topic.geek)
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