Have I got this wrong?

[Samop]

I have  a reflecting 'scope.

As I understand it, it's a long toilet roll with a mirror at the end, and yhe mirror concentrates the light and reflects it into another mirror that points it sideways into the eyepiece and yhat provided the magnification.

Right so far?

So, if I then attach my camera body in place of the eyepiece, where does the magnification come from?

I appreciate I'm missing something basic here...

[vlaiv]

Yes, reflecting telescope works the same as regular camera lens or telescope with front lens. Details are a bit different - it bends light via reflecting it of curved surface rather than refracting it when passing thru an piece of glass with curved surfaces - but you guessed it  - curved surfaces are the key

When you attach the camera - there is no magnification. There is "projection".

Magnification is the term we use to denote angular magnification - or linear increase in angles of incoming rays (which makes object appear larger to us). With camera sensor - we have projection. We take incoming light rays and project them onto a flat plane.

We can talk about "scale" (which can be understood as magnification in some sense) - which determines ratio between angle and distance at focal plane (on sensor surface).

To see the difference between scale and magnification - imagine you have an image printed on a piece of paper - and you look at that image from one meter away and from 10 meters away. Object on image will appear bigger (magnified) when image is placed 1 meter away in comparison to 10 meters away - although both images have the same scale.

Another example would be to print a copy but scaled down %50 and hold piece both pieces of paper at the same distance - here we will have impression that one image is twice as large as the other - although they are at the same distance - this is what scale does. Effect is almost the same - but causes are different.

In any case - scale in photography is determined by two factors - size of sensor (or sensor pixels, depending how you want to look at it) and focal length of lens element (does not matter if it is mirrored system or glass lens system or even combined system like with catadioptric telescopes / camera lenses).

To understand where magnification comes from - you need to study ray diagrams like this one:

This is example of magnification by telescope (and eyepiece) - we have incoming light rays at some angle theta zero on the left, and we have bigger angle theta e on the output (or exit).

Similarly, instead of using eyepiece - we can just look at arrow labeled focal point and its distance from dotted line (optical axis or center of the sensor) - it is directly related to how big theta zero is - larger incoming angle further away this arrow is (or point on sensor) - this is projection part and it depends on F0 or focal length of objective lens.

 

 

[Mick J]

Perhaps you can think of your toilet roll as a fixed lens, what you get is what you get image wise. What you need to get is focus and possibly how long the shutter is open, then it gets complicated but doable.   this may help.

(how vlaiv puts that together in a couple of minutes is beyond me)

Edited November 8, 2023 by Mick J

[Mr Spock]

Here's a photo taken with a Nikon D500 and 100mm f7.4 scope with a x2 Barlow. It will give you an idea of the image scale possible at 1480mm on a crop frame sensor.

Best viewed full size 

 

[Louis D]

3 hours ago, Samop said:

So, if I then attach my camera body in place of the eyepiece, where does the magnification come from?

From the curved primary mirror at the bottom of the tube.  If it was flat, there would be no increase in image scale.  For that matter, no image would be formed for the camera's sensor.

Ever messed around with a shaving/make-up mirror where one side is concave and thus magnifying?  Same idea.

[Hingday37]

I have never got this before.

[ollypenrice]

Another way to think of it is to remember what 'magnify' means. It means 'to make larger.' When you look at the moon with naked eye the moon's image on your retina is a certain size. When you use optical aid, the size of the image on your retina will be made larger than it was before, and by a definable amount, 10x, 50x, 100x etc. This is all nice and easy: we have an image on a retina and that is what we magnify.

Now point an imaging telescope at the moon. It collects and focuses light and projects an image onto the chip. Is this image 'magnified?' Magnified over what? There is no definitive baseline size equivalent to the image on the retina so we have nothing to 'magnify.' We certainly don't magnify the moon since our image is a fraction of the moon's size. 

The only magnification to be found in photography is in the macro world. Some macro lenses can project an object's image onto the chip, an image which is larger than the object itself, and this is, quite reasonably, known as magnification in macro photography. In fact, all telescopes compress the moon's image to a tiny fraction of the moon's real size. This is really negative magnification!

Olly

[Mandy D]

@Samop I think I see the problem, here. Without the eypiece there is no magnification of the image formed by the primary mirror. Look into your focuser tube with no eyepiece and the telescope pointed at the Moon and you will see an extremely bright, but very small image of the Moon. I recommend wearing sunglasses if you try this. I did it accidentally, just this morning!

To figure out what size this image is at the focal plane, we can simply place a camera there and calculate, from the camera's resolution and sensor size, the actual size of the image in mm. @Mr Spock has kindly provided a suitable image for me to do this with, but unfortunately, he has scaled the image. If I resize it using GIMP to the number of pixels across his sensor (5568) then count the pixels across the height of the Moon, knowing the length of the sensor (23.4 mm) I can calculate the height of the Moon in mm on his sensor. I make it about 3580 pixels, so 23.4 x 3580/5568 = 15 mm. I think his Barlow is actually giving a bit more than 2x magnification, but with Barlows there is some adjustment possible, so we can ignore that.

Now, if we project Mr Spock's image onto the sensor of my D800, it will cover a smaller proportion of the space available, as being a full-frame camera as opposed to the D500 being a crop sensor camera, so the image will appear smaller when we print the full image at the same size. Now, if you use a mobile phone with a very small sensor (after carefully removing it's lens to keep things equal), the opposite happens and the image will spill over the edges. So, the number of pixels we grab for a given focal length (assuming the image fits on the sensor) depends only on the focal length and pixel size, not the sensor dimensions. As it turns out, my D800 has larger pixels than the D500, so I get a lower resolution image at the same focal length.

So, now we know what size the image is that is formed by the primary mirror of the telescope, we can consider it as a photograph placed at the focal plane of the telescope, so we now look at it with a magnifying glass, which we call an eyepiece, to make it look bigger and dimmer. Hence, your magnification comes from the eyepiece. A longer focal length of telescope will produce a larger image to start with, so less magnification is required from the eyepiece to get the same view.

Finally, let's look at the actual magnification of the Moon by the telescope, rather than angular magnification. It is the latter that we are after when using a telescope visually as that will determine how big the image is on your retina. But, what is the ratio of the image formed by the telescope at prime focus (or after Barlow amplification in this case) and it is simply image size / Moon size. So, the Moon is about 3475 km, or 3 475 000 000 mm in diameter and our image is 15 mm, which gives a 'magnification' of about 4.3 x 10^-9.