<blockquote>
<p>Magnification using a CMOS sensor</p>
</blockquote>
<p>The concept of magnification is most easily defined when we are comparing apples to apples, not apples to oranges.</p>
<p>When we use a telescope (and assuming normal or corrected vision) the situation is as follows.</p>
<p>Light entering a telescope from a point in space, it is essentially parallel, and when it is correctly focused light exits the eyepiece parallel as well. In this case the term magnification really applies to angles. If we have a 900 mm telescope and 7.5 mm eyepiece giving 120x magnification, then a 30 <em>arc seconds</em> wide Jupiter will appear  1 deg apart to our eye.</p>
<p>So here &quot;apples to apples&quot; means angles to angles.</p>
<p>There is no easy  way to talk about magnification when a recording device like photographic emulsion or a CMOS sensor is placed at the focus, because in this case angles have been converted to positions.</p>
<p>Here &quot;apples to oranges&quot; means angles to positions.</p>
<p>Instead we can talk about scale.</p>
<p>In [this comment] the OP correctly obtained 1.65 arc seconds per pixel using pixel size divided by focal length which gives the angle in radians and then the multiplier</p>
<p><span class="math-container">$$\frac{180}{\pi} \frac{3600}{1000} = 206.2648$$</span></p>
<p>which converts radians to degrees, then to arc seconds, then divides by 1000 because that equation specified micrometers for pixels and millimeters for focal length.</p>
<p>If you wanted an <em>apparent magnification</em> based on viewing your image on a screen, you can use the following which is a ratio of angles</p>
<p>\frac{\text{camera pixel}}{\text{focal length}}</p>
<p><span class="math-container">$$M_{apparent} = \frac{\text{angle}}{\text{angle}} = \frac{\frac{\text{screen pixel}}{\text{viewing distance}}}{\frac{\text{camera pixel}}{\text{focal length}}} = \frac{\text{screen pixel}}{\text{camera pixel}} \times \frac{\text{focal length}}{\text{viewing distance}}$$</span></p>
<p>Which means of course that the farther we hold the screen away from us, the smaller Jupiter looks, or the larger the pixel size on our screen, the bigger it looks.</p>
<p>But of course that only works if our device displays with a 1:1 correspondence between image pixels and screen pixels. If the device scales the pixels (most do) then you have a bigger challenge.</p>
<p>That's why recorded images through telescopes rarely if ever display &quot;magnification&quot;. For microscope images, if they show a bacteria and say &quot;magnified 5000 X&quot; then they are comparing the size of the bacteria to the size of the printed image in the book.</p>

CC BY-SA 4.0
2020-12-25T02:50:26.897Z
Thanks so much for this summary. I'll be spending some time digesting all this! Would you happen to recommend an entry-level text on this subject! Again, thank you for taking the time to respond!
ed huff
CC BY-SA 4.0
2020-12-25T06:02:01.417Z
I'm glad to hear that this is helpful! I can't help with the reading recommendations, thought, sorry.
uhoh
CC BY-SA 4.0
2020-12-25T13:11:55.223Z
No problem on the text — I'm sure I'll find something via a few minutes searching. Again, thanks.
ed huff
CC BY-SA 4.0
2020-12-24T23:21:54.040Z
2020-12-24T23:21:54.040Z
uhoh
<p>I'm reworking my question regarding magnification of my 60mm, ⨍=700mm refractor. The video eyepiece is a CMOS sensor with a pixel size of 5.6μ x 5.6μ. The number of pixels is 720 X 480 with a 0.3 megapixel resolution. That said, I'm clueless regarding an algorithm from which you could decipher the magnification given these physical characteristics of the CMOS sensor. Any thoughts would be appreciated. Thank you!</p>

CC BY-SA 4.0
2020-12-24T17:24:48.737Z
From http://astronomy.tools/calculators/ccd : (   Pixel Size (µm)   /   Telescope Focal Length (mm)   )   X 206.265 = resolution (arc seconds per pixel)
Aaron F
CC BY-SA 4.0
2020-12-24T18:40:23.753Z
Thanks for pointing this table out Aaron. Here's probably a stupid question, but if I come up with a resolution of 1.65"/pixel, how would this correlate to actual magnification and field-of-view (FOV) as designated with optical eyepieces. Again, thanks!
ed huff
CC BY-SA 4.0
2020-12-29T00:57:37.637Z
What you can try is to use the FoV calculator on the same site - add your sensor to the view, then add some eyepieces to the same view. It'll list the eyepieces with their magnifications as multiples, while listing the angular resolution of your sensor (as detailed in uhoh's answer). Eyepiece FoV are circular and the sensor's is rectangular, so you'll want to find an eyepiece that has a radius which is greater than the rectangle's height and less than the rectangle's width. Long story short: about 200x for you. You'll want to buy a cheap 0.5x focal reducer unless only interested in planets :-)
Aaron F
<amateur-observing>
Magnification using a CMOS sensor
ed huff
# Magnification using a CMOS sensor

(Asked by: username_0 on 2020-12-24)

I'm reworking my question regarding magnification of my 60mm, ⨍=700mm refractor. The video eyepiece is a CMOS sensor with a pixel size of 5.6μ x 5.6μ. The number of pixels is 720 X 480 with a 0.3 megapixel resolution. That said, I'm clueless regarding an algorithm from which you could decipher the magnification given these physical characteristics of the CMOS sensor. Any thoughts would be appreciated. Thank you!

**Tags:** amateur-observing

> **Comments:**
> **username_1**: From http://astronomy.tools/calculators/ccd : (   Pixel Size (µm)   /   Telescope Focal Length (mm)   )   X 206.265 = resolution (arc seconds per pixel)  
> **username_0**: Thanks for pointing this table out username_1. Here's probably a stupid question, but if I come up with a resolution of 1.65"/pixel, how would this correlate to actual magnification and field-of-view (FOV) as designated with optical eyepieces. Again, thanks!  
> **username_1**: What you can try is to use the FoV calculator on the same site - add your sensor to the view, then add some eyepieces to the same view. It'll list the eyepieces with their magnifications as multiples, while listing the angular resolution of your sensor (as detailed in username_2's answer). Eyepiece FoV are circular and the sensor's is rectangular, so you'll want to find an eyepiece that has a radius which is greater than the rectangle's height and less than the rectangle's width. Long story short: about 200x for you. You'll want to buy a cheap 0.5x focal reducer unless only interested in planets :-)  

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## 1 answers

### Answer by username_2 on 2020-12-24. Score: 2

> Magnification using a CMOS sensor

The concept of magnification is most easily defined when we are comparing apples to apples, not apples to oranges.

When we use a telescope (and assuming normal or corrected vision) the situation is as follows.

Light entering a telescope from a point in space, it is essentially parallel, and when it is correctly focused light exits the eyepiece parallel as well. In this case the term magnification really applies to angles. If we have a 900 mm telescope and 7.5 mm eyepiece giving 120x magnification, then a 30 *arc seconds* wide Jupiter will appear 1 deg apart to our eye.

So here "apples to apples" means angles to angles.

There is no easy way to talk about magnification when a recording device like photographic emulsion or a CMOS sensor is placed at the focus, because in this case angles have been converted to positions.

Here "apples to oranges" means angles to positions.

Instead we can talk about scale.

In [this comment] the OP correctly obtained 1.65 arc seconds per pixel using pixel size divided by focal length which gives the angle in radians and then the multiplier

$$\frac{180}{\pi} \frac{3600}{1000} = 206.2648$$

which converts radians to degrees, then to arc seconds, then divides by 1000 because that equation specified micrometers for pixels and millimeters for focal length.

If you wanted an *apparent magnification* based on viewing your image on a screen, you can use the following which is a ratio of angles

\frac{\text{camera pixel}}{\text{focal length}}

$$M\_{apparent} = \frac{\text{angle}}{\text{angle}} = \frac{\frac{\text{screen pixel}}{\text{viewing distance}}}{\frac{\text{camera pixel}}{\text{focal length}}} = \frac{\text{screen pixel}}{\text{camera pixel}} \times \frac{\text{focal length}}{\text{viewing distance}}$$

Which means of course that the farther we hold the screen away from us, the smaller Jupiter looks, or the larger the pixel size on our screen, the bigger it looks.

But of course that only works if our device displays with a 1:1 correspondence between image pixels and screen pixels. If the device scales the pixels (most do) then you have a bigger challenge.

That's why recorded images through telescopes rarely if ever display "magnification". For microscope images, if they show a bacteria and say "magnified 5000 X" then they are comparing the size of the bacteria to the size of the printed image in the book.

> **Comments:**
> **username_0**: Thanks so much for this summary. I'll be spending some time digesting all this! Would you happen to recommend an entry-level text on this subject! Again, thank you for taking the time to respond!  
> **username_2**: I'm glad to hear that this is helpful! I can't help with the reading recommendations, thought, sorry.  
> **username_0**: No problem on the text — I'm sure I'll find something via a few minutes searching. Again, thanks.  

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