T3D Re: Focal length

[john bercovitz <bercov@xxxxxxxxxxx>]

Relinquish all hope (of brevity) all ye who enter here.  Actually,
I'm going to be extremely brief considering the size of the subject 
at hand.  I'll skip all the interesting details.

Let me go over the basics of focal length and focussing and see how
much that clears up.  Then we can go back and pick up anything that's
missing.

Important points of a lens:

|<---- x --->|<-f->|      |<-f->|<-x'->|

O            +     +  ()  +     +      o

|<---------s------>|<- h->|<----s'---->|


>From left to right,
O is the object
+ is the front focus - where light from infinity, running 
                       backward through the lens, would focus
+ is the primary principal point
() is the lens itself
+ is the secondary principal point
+ is the back focus - where light from infinity, running 
                      normally through the lens, would focus
o is the image plane

h is the hiatus which is the space between the principal points but 
that's not too important unless you are trying to find the exact 
distance from O to o.  The distance from the front focus to the 
primary principal point and the distance from the secondary principal
point to the back focus are equal and are the focal length, f, of the 
lens.  (Now no nitpicking please; remember I'm trying to leave out 
the interesting little details. 8-)

There are two formulas which will tell you where the object lies and 
where the image lies if the image is in focus.  They are actually the 
same formula because one can be derived from the other but one will
generally be more useful than the other in a given situation so use
whichever works out better.

Gaussian formula:
1/s + 1/s' = 1/f

Newtonian formula:
x*x' = f*f

In your case you are going to tack 100 mm to the back of a 45 mm lens.
So let's just say the lens is focussed for infinity and you add in the
100 mm.  Then the Newtonian formula is more appropriate for finding 
where the in-focus object will lie relative to the lens.

x*x' = f*f

x * 100 = 45 * 45

x = 20.25 mm

So the in-focus object will lie 20.25 mm in front of the front focus
of the lens.  You can find the location of the front focus by pointing
the back of the lens at some bright distant object and then moving a 
white card to and from the front of the lens to see where that distant 
object focusses.  Probably it will focus very close to the front of 
the lens in your case.

When actually taking a picture, measure forward another 20.25 mm from 
the front focus and place your object there.

As an aside, you aren't going to have very much depth of field.  You
may find this setup most suitable for taking pictures of fairly flat 
objects, coins, for instance.

s' is the distance from which you should view the image made by the 
camera.  s' is what we have been calling "the operating focal length".
It is equal to f + x'.  s' isn't exactly correct but it is close.
It is exact if the lens is symmetrical.  Not too many are, but then
most lenses aren't horribly asymmetrical either.  If you enlarge the 
image in an enlarger or by projection, you just multiply s' by the 
magnification you used and you have your new viewing distance.  

> By "operating focal length" you mean the actual focal length when 
> focused at a particular point, as opposed to the focal length printed 
> on the barrel?

This is correct but of course the focal length doesn't really and truly 
change.  The distance from the secondary principal point to the film is 
what actually changes and that also changes the perspective.  For a more 
exact accounting, see the perspective paper I will send you to put on 
the web site.

> I forgot about the "taking" length equals the "viewing" length part.  
> I'll have to look at the formula to see how I might jigger it to work 
> when the taking and viewing focal lengths are not the same.

Can't really be done (without a computer 8-).  Changing perspective 
point changes stretch/squash which is the scale in the depth direction 
but it does not change the scale of the other two directions, up and 
down.  It is changing stereobase which rescales all three dimensions
exactly in proportion to each other.  Changing perspective in stereo is 
like changing perspective in mono:  If you take a closeup portrait with
your 15 mm wide angle, you can be sure that viewing the image from 
"normal" distances will give you the impression that the subject has a
prodigious proboscis.

Let's say you are going to make an image which is 24 mm high with that
"145 mm" lens.  Let's say the image projected onto the screen is three
feet high.  Then the ortho seat is (145/24)*3' = 18' from the screen.  
So it's not that tough to view the images made by this lens from the 
correct distance.

John B


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End of TECH-3D Digest 361
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