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#1
22-02-2008, 04:47 PM
 Bucky1379 Registered User Join Date: Jan 2007 Location: Melbourne, Australia Posts: 27
'Scope magnification with a webcam

Hi

I have a Philips 900CN with a 1.25" adaptor (which I haven't been able to use yet) having removed the lens as per instructions. I have a question though.

I know that I calculate magnification for my scope using the Primary focal length divided by the eyepiece focal length (eg 1200 / 9 = 133.3333) but how do I calculate it with a no eyepiece and a lensless CCD chip at the focal point.

I have searched the internet a found 1 reference saying the result is about equivalent to a 4mm eyepiece but is this true and how is it calculated?

Thanks for any help understanding this.

Steve
#2
22-02-2008, 04:55 PM
 iceman (Mike) Sir Post a Lot! Join Date: Sep 2004 Location: Gosford, NSW, Australia Posts: 36,709
Hi Steve

A very similar question was asked the other day - check this out:

#3
22-02-2008, 08:18 PM
 bojan amateur Join Date: Jul 2006 Location: Mt Waverley, VIC Posts: 5,814
Hi Steve,
When using camera in prime focus, there is no such thing as "magnification".
Instead, there is "scale", expressed in "/pixel or pixels/" or pixels/deg.
What you really want to ask is "how big field of view I will have when I attach the camera?"
This will depend on focal lens of your telescope and the linear size of the camera sensor.
The formula is alpha=2*atan(d/(2*L)), where
alpha is field of view expressed in radians (if you want degrees, you have to multiply this by 180/pi)
d is linear dimension of sensor in mm
L is focal length in mm
#4
22-02-2008, 10:06 PM
 Bucky1379 Registered User Join Date: Jan 2007 Location: Melbourne, Australia Posts: 27
Thanks guys.

All the info is very understandable and makes sense. I also appreciate the formula's etc so I can work it all out for myself .

I could have just connected it all up and expected usable results because I see the photos on the web that other people get with similar equipment but I prefer to understand what I am doing and be able to estimate what will happen if I change something.

It's great being able to get answers so easily and quickly from people who have already worked this stuff out. Thank you.
#5
23-02-2008, 06:26 PM
 bojan amateur Join Date: Jul 2006 Location: Mt Waverley, VIC Posts: 5,814
You are very welcome :-)
#6
27-02-2008, 10:03 AM
 Geoff45 (Geoff) PI rules Join Date: Jul 2006 Location: Sydney Posts: 2,369
Another way to look at things is to make a picture of the object. If it is on A4 paper and you hold it a tad more than 30 cm away, the magnification shown in the picture is (approximately) 40 divided by the angular width of sky shown on the short side, so for example, if the moon just fills the short side of the paper (20cm), the mag is 80 (40/0.5).

What I mean here by "magnification" is that if you hold the pic of the moon up to the sky at 30 cm from your eye, it will span 40 degrees, which is 80 times more than the real moon.
Geoff

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