Re[2]: ATM Appropriate barlow ?s

[Dwight Elvey <elvey@hal.com>]

David Simons <hmsimons@hotmail.com> wrote:
> 
> I would also like to see some kind of Barlow formula that describes the 
> negative lens focal length, factored with the distance to the eyepiece = 
> amplification.
> 
> Dave
> ===========


Hi David
 Here is a post that I sent to Jim that should answer your
questions. If it doesn't, ask me about what you might have
troubles with.
 Also, if you are using the page from the ATM that shows
the focal length of negative lenses, there is an error.
The drawing shows both u and v being measured from the
positive lens. The measurements should both be referenced
from the position of the negative lens. I think using the
star drift method can produce far more accurate result,
my self.
see: http://www.atmpage.com/focal.html  but note error in drawing!
Dwight


lifedata@vol.com wrote:
> Dwight Elvey <elvey@hal.com> said:
> 
> >needs to be inwards 1/2 focal Barlow length and the eyepiece out 1/2
> >focal Barlow length from where the normal eyepiece focal positions
> 
> Thanks for the additional info.  It's clear you have to know the FL
> of the Barlows in order to do this kind of thing correctly.  None of
> the Barlows I've seen in catalogs tell the FL.  May I assume they are
> on the lenses themselves?

Hi Jim
 Like all things, there is an exact way and a close enough
way. First to understand why I came up with the 1/2-1/2.
It is some algebra but not that bad.
 All lenses obay the simple lens rule:

 1/f = 1/a + 1/b

I haven't accounted for sign conventions here because I
find it easier to think what is what without it. In some
books, you'll see one of the terms as negative. That is only
because they like to keep right as positive and left as
negative idea. I like to think of distances from the lens
as positive from the side the image comes from so bare
with me.
 What the equation says is that an image formed at "a" from
the lens will project to an image at "b" for a lens of focal
length "f". This is a vary useful equation for all optics.
It doesn't do any correction for spherical aborations or
compensate for the effects of the thickness of lenses but
is still vary useful in determining where to put things.
 It is a shame that the manufactures don't publish focal
lengths of Barlows but let us just work at though they did.
We can get an approximate focal length with some simple
measurements.
 If you know the distance from the open end of the Barlow
and the lens inside, in combination with the rated X, you
can calculate the focal length and use the above formula
to determine the position of the final optics.
 I'l start with the more interesting case where the Barlow is
something more interesting than 2X. Lets say it is 2.5X so
we can see how thing work out.
 I take my Barlow and measure with a wood dowel, as a depth
guage, ( binging careful not to scratch the center of the
lens and onlt measuring at the edge ) that the lens it 4.5 inches
from the end of the Barlows tube. I know that each focal length of
the lens that I move the eyepiece away from the lens creates one
additional X. So, ( 2.5 - 1 ) * f = 4.5 or f = 3.
The -1 in the equation is because it is 1X at the lens
it self ( no magnification ).
Now we know the focal length of the lens, we can do other
things with it.
 We know that it is really a negative lens, being concave
on both sides so f = -3 in reality ( By the way, the
magnification factor follows that 1X for one focal length
for any kind of lens, including pasitive lenses. Positive
lenses are not generally used like Barlows because the lengths
of the tubes needed are quite a bit longer than the more compact
negative lens arrangment. ) We know two of the values for
determining the final position. We know the "f" and one of the
other 2, we'll call it "b". So,  1/-3 = 1/a + 1/4.5. Solving
for a = -1.8. So, what does the - mean in direction. On
the positive lens, it means the distance on the opposite
side of the lens from the other image. In this case, it
must mean that both images are on the same side of the lens.
We know that b was the image that is right at the front
of the eyepiece ( this is the one we use the eyepiece as
a magnifying glass to look at ), so "a" must be the image
that comes from the primary. We also know that this image
is on the other side of the lens from the primary because
of the - value. Draw all of this out.
 Putting this all together, we now know that the Barlows
lens will be inwards 1.8" from the original focal position and
that the eyepiece will be moved outwards by 4.5 - 1.8 or
2.7".
 This all assumes that the manufacture of the Barlow made
the tube the right length and the back edge of the lens
was truly the effect position to measure from to get
the focal length.
 Another way to determine the focal length would be to
do some drift time measurments of a star across the field
of view with a stop watch. Do this first without the Barlow
and once with the Barlow. This will give you the true
effective magnification ( not the stated ), by the ratio of
the times. Now slide the eyepiece part way out of the Barlow
by a measured amount and measure the time again. Do the
ratio of this time with the original time to get the increased
magnification factor. Now subtract the the two magnification
factors and divide this into the length that you increased
the position of the eyepiece. This is the true focal length
of the Barlow. Do this once more to make sure you
made no mistakes and write it down.
> 
> One more question.  If you were to have one Barlow, would you make it
> a 2X or something else?

 I would think a 1.5 is a better size since most eyepieces
come in steps of 2 to start with. There are of course other
purposes of using Barlows than just filling in eyepiece ranges.
Often they are used to keep some of the eye relief of the
eyepiece, such as one would do for eye glasses. In
which case, 2.5X, 3X or 3.5X might be a better value.
2X always seemed to be the least useful.
Dwight