From: "Dwight K. Elvey" 
To: atm{at}shore.net
Reply-To: "Dwight K. Elvey" 


>From: "Steve Follett" 
>
>Hi all,
>
---snip---
>
>1).  How do you calculate the effect of a focal reducer on system
>focal length?  If you start with 12M focal length and you use a 1000mm
>lens as a reducer what is the resulting focal length.  Does the relative
>position of the reducer and the eyepiece effect the result (as with a barlow
>lens) ?

Hi
 I can get you started on thinking about lenses and focal distances.
Each lens has a parameter called the focal length. This can be positive or
negative. This value is related to what is called the thin lens formula (
1/f = 1/-a + 1/b ). 'a' and 'b' are distances to the source and image
formed by the lens. I have written the equation such that the polarities
are that things towards the right of the lens are positive and things
towards the left are negative. This is so that it will be easier later to
know what polarities to use in calculations. Keeping track of what the
polarities are is the most confusing part. This is why I chose this
convention to describe this problem ( I use a different convention myself
that doesn't make as much sense ). We will also assume that 'a' is the
source image and 'b' is the resulting image.
 In a compound lens system, it is all about moving these focal points
around. One has to think in terms of where the new image forms if I take
the first image and modify it with the next lens. In a telescope, we always
want the final image to form at the focal plane. We can work from there
towards the primary or from the primary towards the focal plane ( about the
same ).
 Lets take the case of using a focal reducer that has a positive
focal length of 15 inches. Lets use this in a telescope that had a focal
length of 48 inches. We will try a few different locations for the reducer
to see how it effects the operation on the telescope. For those that like
to think in metric, change everything to cm. It also helps to draw out a
scaled picture to keep track of what is happening.
 As stated at the start, we will place the source star off to the
left so that our focal plane is to the right. Plugging infinity, for 'a',
into our equation and it shows that 'f' and 'b' are equal ( as we expect ).
 Lets see what would happen if we placed the reducer in front
of the original focal plane by 5 inches. This is 5 inches towards the
primary from the focal plane. We know where the original focal point was
from the primary, what we want to know is where the reducer moved the new
focal point to.
 This is where keeping track of polarities from the sides of the lens
keep us honest. The new source for 'a' of the reducer is on right side of
our lens ( the original focal point ). The original equation measured 'a'
as a negative number if it was on the left and a positive number on right.
Since the source is on the right, it is a positive number. So, we plug this
in as:

1/15 = 1/-(5) + 1/b

 We now pull out our calculator and solve for 'b'.

 b=3.75

 We have modified the focal length by the ratio of 3.75/5 or 0.75.
This increases the f-ratio by 1.333.

 Now, lets change the reducer by moving it to 2.5 inches in front of
the focal point instead of the 5 inches above.

1/15 = 1/-(2.5) + 1/b

then:

 b=2.14

 In this case, the focal length was changed by 2.14/2.5 or 0.856

 You can see from the above that the distance of the lens to the
original focal plane changes the effect of the lens ( this is true for
Barlows as well ). You can see that keep a clear thought on the polarities
of the measurements of old focal point image to the new focal point image
is the key to using the equation.
 I'll add here that I didn't compensate for the effects of the
thickness of the glass. The equation is called the "thin lens
formula". The glass modifies the distance where the images form. If
you really wanted to be accurate, you'd need to measure the effective
center planes for the light entering and leaving the lens, since these are
the true 'a' and 'b' distances. Still for a first pass measurement of what
your looking for, you can assume that what I had above was correct.
 I left the other questions for others to answer.
Dwight

PS:
For the calculator impaired
1/15 = 1/-(5) + 1/b

 is 1/15 - ( 1/-(5) ) = 1/b
 the -'s cancel to

  1/15 + 1/5 = 1/b

  I type the following into my calculator:

   15,  1/x, +, 5, 1/x, =, 1/x

  This displays the value b.

--- BBBS/NT v4.00 MP