[Laser] CONCLUSION TO >>> focal length question

[George Vastianos]

PLEASE READ ALL THE MESSAGE

Thanks to all of you who answered my question. From your (following) replies
i made a list with the most important (according to my understanding) key
points:

[1] The area that covers the Fresnel Lens defines the amount of light that
the
     receiver will accept. So the biger is allways better {{From Tim's Toast
reply}}
[2] The focal length defines how wide will be the angle at the position
(inside the
     receiver box) where the incoming light (that passes from the lens) will
be
     focused on. So sort focal lengths produce very wide angles at the focus
position
     and long focal lengths produce narrow angles. {{From Yves' reply}}
[3] In order to have no loss in the reception of light we must install a
sensor (photodiode
     or phototransistor at the focus position...) that will have a reception
angle (according
     to its technical specifications) equal or wider from the angle that is
produced from
     the incoming light (as described in [2]). If the sensor's reception
angle is narrow
     compared to the produced input angle then we have light loss....{{From
Yves' reply}}

FINALLY I BOUGHT FROM eBAY 4 PIECIES OF FRESNEL LENSES:
SIZE 180mm x 260mm WITH FOCAL LENGTH 300mm FOR $20 (4 PCS+SHIPING!!!!)

SO IN ORDER TO CALCULATE THE PRODUCED ANGLE AT THE FOCUS
POSITION WE MUST TAKE THE INTO ACCOUNT THE "WORST" CASE AND ASSUME
THAT THE SIZE IS 260x260:


            lens
-----     (
    |        ( \
    |        (    \  (hepotenuse side)
260/2    (       \
    |        (          \
------    (--FL---> [focus position]
    |        (          /
260/2    (      /
    |        (    /
    |        (  /
------    (

>From trigonometry we know that

hepotenuse side ^ 2 = (260/2)^2 + Focal Length(FL)^2 <=>
hepotenuse side = Sqrt(130^2+300^2)=Sqrt(16900+90000)=327mm

Also the Sinus of the half input angle at the focus position is equal to
Sin(A/2)=(260/2)/327=130/327=0.398 <=>
So A/2=23.4 degrees So the Input angle at the focus position is 2*23.4= 46.8
degrees

IN ORDER TO DONT HAVE LIGHT LOSS DURING RECEPTION THE SENSOR MUST HAVE
RECEPTION ANGLE WIDER THAN 46.8DEG.

IN MY QSO TESTS IAM PLANNING TO USE INFRARED LIGHT SO I FOUND THAT THE
PHOTODIODE  BPW34 HAS HALF RECEPTION ANGLE +-60 DEG. SO THE RECEPTION ANGLE
IS
120 DEG.

IN MY OPINION THE BPW34 WITH THE 120 DEG. 60% WIDER FROM THE INPUT ANGLE OF
46.8DEG IS GOOD CHOISE. SO THE PHOTODIODE WILL RECEIVE ALL THE INCOMING
LIGHT WITH NO LOSSES

PLEASE WRITE ME YOUR OPINION AND IF MY THOUGHTS ARE CORRECT & IF I
UNDERSTOOD CORRECTLY YOUR POINTS....

THANKS FOR YOUR TIME
73 DE SV8GXC


From: <F1AVYopto at aol.com>
> The focal length choice  depends on the photo detector you use.
> At the focal point, the light flux  from the TX must converge without
losses
> on the photosensitive area. The  convergence angle must be the same
> the chose PIN photodiode sensitivity angle. For flat and thin photodiodes
> the focal length can be short. For  small diameter photodiodes with the
> sensitive area distant from the glass window  or for the ones with
integrate
> lenses or convex clear plastic case, the focal  must be long. The 330mm
focal
> length lenses seem easiest to match with  standard photodiodes. /// 73
Yves

From: "Tim Toast" <toasty256 at yahoo.com>
> Yes, either lens would collect the same amount of light and i suspect
either one
> of those would be fine as long as your detector has a field of view wide
enough
> to 'see' the entire lens from its position. Although the longer focal
length version
> would give a narrower field of view and slightly better image quality,
both of
> which help cut down on background noise. I know from photography, a
shorter
> focal length lens does have some advantages in some cases (faster) but i
don't
> know if that really applies to optical communications as much as it does
in
> photographing dim stars. P.s. - I sketched this out on a piece of paper
and
> just measured the angles with a compass to avoid the math, so i hope i'm
not
> too far off :)  With the 390mm dia. 220mm FL lens, the detector needs at
least
> a 100 degree FOV to see the whole lens. With the 330mm FL lens, the
detector
> only needs to have about a 80 degree FOV. I'm assuming that's a square
lens
> (550mm diagnal) //// Tim Toast

COMMENTS: TIM ABOUT THE FASTER.... I BELIEVE THAT DEPENDS FROM
WHAT KIND OF SENSOR YOU ARE USING...ACCORDING TO SPEED ORDER
(FROM FASTER TO SLOWER) WE HAVE: PHOTOTRANSISTORS, PHOTODIODES
AND PHOTORESISTORS

From: "Charles Pooley" <ckpooley at sbcglobal.net>
> You might add a second lens, either + or -, to make it a telescope. Then
it will
> tend to place a nearly parallel beam of light coming from a distant point
source,
> onto the detector when the focus is adjusted. This has the advantage of
allowing
> easy placement of a pinhole blocker to confine the field of view. Also, to
acco-
> modate the placement of a tilted bandpass filter (tilted as a tuning
measure, and
> to allow a camera to get the image from light reflected from the BPF). I
plan to
> use 1/2 of a cheap pair of binoculars for first experiments, then scale up
to larger
> area later. /// Charles Pooley  KD6HKU

COMMENTS: CHARLES THE IDEA OF BINOCULARS IS USED MANY
YEARS NOW SUCCESSFULLY FROM MANY PEOPLE (ESPECIALLY
IN LASER QSOs) THE GOOD THING ABOUT THAT IS THAT U USE
THE OTHER HALF FOR TO LOCK THE TARGET BY EYE.....ABOUT
THE FILTERS AND THE USE OF A CAMERA ....THOSE THINGS ARE NOT
IN MY PLANS...MY GOAL HAS TO DO WITH THE USE OF FRESNEL LENS
ONLY. BTW GOOD LUCK WITH YOUR EXPERIMENTS. 73