# Magnification question

### #1

Posted 13 December 2012 - 07:54 PM

/Ira

### #2

Posted 13 December 2012 - 08:45 PM

Jarad

### #3

Posted 13 December 2012 - 08:59 PM

/Ira

### #4

Posted 13 December 2012 - 09:03 PM

The reason I ask is that I was using a Mallincam at the prime focus of my C8, and someone asked me what the magnification was. I had no idea how to answer him.

/Ira

Ira:

Generally one thinks of Image Scale when using a camera. It is possible to estimate some sort of "magnification" with a video camera like the Mallicam but one would have to know the screen size, how far the observers eye was from the screen, the focal length of the telescope and the size of the image sensor.

The magnification would be the ratio of the angular sizes. Say the scope had a focal length of 2000mm and the image sensor was 7mm in the long dimension. Say the screen was 10 inches across, the image filled the screen and the observer was 20 inches from the screen.

The viewers angle would approximately be:

Angle = 57.3 deg/rad x 10 inch/20 inches = 28 degrees.

That image represents a field of view that is:

TFoV = 57.3 deg/rad x 7mm/2000mm = 0.20 deg

The magnification would be the visual angle divided by the field of view:

Magnification = 28 deg/.2 deg = 140x

You can plug in the actual numbers and get some sort of estimate, this basically says that the image of an object appears 140x larger when it's on the screen than it does naked eye. Of course if you are closer to the screen, the magnification is greater...

Jon

### #5

Posted 13 December 2012 - 09:05 PM

/Ira

### #6

Posted 13 December 2012 - 09:33 PM

### #7

Posted 13 December 2012 - 10:34 PM

Sounds complicated. I know it had to be more than 40x because I had a very wide split of the Trapezium, even with the Mallincam focal reducer in the optical train.

/Ira

It's not too complicated... My guess on the size of the Mallicam chip was pretty good, it's actually 7.4mm in length. If I had dimensions of the monitor, the viewing distance and information about any focal reducers, it would plug right in.

Jon

### #8

Posted 14 December 2012 - 07:04 PM

Magnification is How much bigger it appears to your eye, and your eye, nor the eyepiece, is not in the optical chain.

So the system and comparison is not applicable.

### #9

Posted 14 December 2012 - 08:14 PM

There is not a magnification as such, the scope simply creates an image of size X, based on the focal length of the scope and the subtended angler of the object being viewed. The size is in mm or whatever the scope focal length units are.

Magnification is How much bigger it appears to your eye, and your eye, nor the eyepiece, is not in the optical chain.

So the system and comparison is not applicable.

If you do a little out of the box thinking and realize that magnification is a ratio of angles, then I believe it is possible to determine how much larger an object appears to the eye when it is displayed on the computer screen than it is naked eye.

An example:

- Naked eye: The object is 1000 feet away and 1 foot high. This represents an angle of

AngleNakedEye = 57.3 deg/rad x (1/1000) = 0.057 degrees

- Assume that the Mallicam system with it's 7.4mm sensor, 2000mm focal length scope and 10 inch monitor (horizontal) shows the objects height to be 2.8 inches. If the observers eye is 20 inches from the screen, one can calculate the angle it subtends:

Angular size at the eye = 57.3 deg/rad x (2.8 inch / 20 inches) = 8.0 degrees.

Magnification = 8.0 degrees/0.057 degrees = 140X.

The eye sees the object on the screen as being 140 times larger than it is naked eye. This is a measure of the magnification. One does not normally think of magnification in this way, it changes with the distance from the screen.. but nonetheless, it fits within the fundamental definition of magnification, how many times larger does the magnified image appear than it appears naked eye.

The method for determining the magnification of the telescope-Mallicam-screen-eye system was described in my previous two posts.

This is not the traditional way we think of magnification but is useful nonetheless.

Jon

### #10

Posted 15 December 2012 - 01:23 AM

With prime focus photography, one image is virtual, at infinity (the sky itself), the other is a real image, formed in the focal plane. The former has an angular size (is measured in degrees), the other has a linear size (measured in mm). You cannot compare them, so therefore magnification is not defined in this case.

When you're asked this question again, you can simply reply that "magnification" is only defined when you look into an eyepiece with your eye; but for what you're doing magnification is undefined, and only the linear size of the image (measured in mm) can be calculated.

It will be hard to convince them, because for the average layperson "magnification" is the most important thing ever in a telescope. It's the first thing they think about when they see a scope: "how much does in magnify?" To them, a telescope "without magnification" is like a car without an engine. But the reality is, magnification is only defined for virtual images (we're talking astronomy here, when the object is infinitely far away, not general optics). No virtual image, then magnification makes as much sense as a square circle. End of story.

It's easy to calculate the linear size of the real image. All you need to know is the focal length of the telescope, and the angular size of the object.

IMG = linear size of image

F = focal length of telescope = 2032mm for a C8

OBJ = angular size of object

Then:

IMG = F * arctan(OBJ) = 2032 * arctan(OBJ)

Let's calculate the image sizes of a few objects.

The Moon: OBJ = 30arcmin. Then IMG = 17.7mm

http://www.wolframal...rctan(30arcmin)

Jupiter: OBJ = 30arcsec ... 50arcsec. Then IMG = 0.3mm ... 0.5mm

http://www.wolframal...rctan(30arcsec)

M31: OBJ = 3deg. Then IMG = 106mm = 10.6cm

http://www.wolframal...32*arctan(3deg)

Click on those links, edit the query and enter your own object sizes, and you'll be able to calculate the image size for anything.

If there's a focal reducer or a barlow in the optical chain, then you'll have to adjust the focal length. E.g., with a 2x barlow it will be 4064.

You can do the same calculation on any pocket calculator with trigonometry functions, or on a calculator app on a smartphone. This comes in handy when you're outside shooting pictures and you may or may not have (easy) access to the Internet.

### #11

Posted 15 December 2012 - 04:38 AM

"Magnification" is a ratio - the ratio between the angular size of the naked-eye image, and the angular size of the image seen in the scope. Both images are at infinity, and as such only possess angular size. You can talk about magnification in this context because the two images compared are similar.

With prime focus photography, one image is virtual, at infinity (the sky itself), the other is a real image, formed in the focal plane. The former has an angular size (is measured in degrees), the other has a linear size (measured in mm). You cannot compare them, so therefore magnification is not defined in this case.

It is true that with prime focus photography magnification does not make sense. Photographers think in terms of image scale, the ratio between the linear size of the object's image at the focal plane the angular size of the object, the units could be mm/degree.

But Ira's situation is not prime focus photography, it is an observer viewing an object on a monitor from a measurable distance and it is possible to determine the angular sizes of both the object seen naked eye and the image of the object as observed on the computer screen. The magnification factor can be calculated.

Also, at the small angles involved here, it is not necessary to use the Arctangent in calculating the linear size of an object. The small angle approximation for both the Tangent and the Sine, i.e. that they are equal to the angle as measured in radians, is more than sufficient. In this case, the maximum error is about 1 part in 10,000.

Jon