The Magnification Issue

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The Magnification Issue
I know this is an absurd kind of question. We don't care about "power" calculations. I'm just curious though. Conventionally, if i use a 10x eyepiece with my 40x objective we say the microscope is 400x. I have an eyepiece camera that includes a .5x reduction lens. What would be the power equivalent.

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Re: The Magnification Issue
Optical magnification would be 20x, assuming the 0.5x is actually doing a 0.5x reduction. A generic 0.5x will not actually give exactly that on every system.

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Re: The Magnification Issue
You are saying that the camera is equivalent to a 20x eyepiece?

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Re: The Magnification Issue
No, what I was saying is that 40 x 0.5 = 20. The camera has nothing to do with what the optical magnification is. This is why optical magnification is not useful for comparing images taken across different systems, while field of view is.

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Re: The Magnification Issue
Ok I wont ask it any more. Lol.

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Re: The Magnification Issue
It's not a bad question to ask, it just doesn't end up leading to the answer you really want, since we apply our own magnification to the final image when we view it on the screen. Field of view deals with all the issues of cropping and resizing by skipping it and looking at the final result.

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Re: The Magnification Issue
Ok. What I want to know is what eyepiece the camera is roughly equivalent to. I know it crops less than my old Omax camera.
Re: The Magnification Issue
You need to include the sensor sizes [physical dimensions, not pixel count] in your calculation to do that, Don.DonSchaeffer wrote: ↑Mon Jul 13, 2020 2:04 pmOk. What I want to know is what eyepiece the camera is roughly equivalent to. I know it crops less than my old Omax camera.
MichaelG.
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Re: The Magnification Issue
ok. Thanks.

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Re: The Magnification Issue
I'd suggest you skip the calculations and measure directly. Put a ruler in focus, take a look at it through your 10x eyepieces and through your camera to directly measure the field of view in each and compare.
Re: The Magnification Issue
If the specs at http://www.aibome.com/product/36_146.html as posted in your other thread are correct then you can estimate:
According to Wikipedia a 1/2.5" sensor is 5.76 mm wide. Assume the objective and 10x eyepiece are normally working with a 20 mm diameter intermediate image. The 0.5x relay should project that to a 10 mm diameter image in the plane of the sensor, of which the sensor sees only 5.76 mm. So the horizontal field of view you see with the camera should be approximately half (5.76/10) of what you would see with a 10x eyepiece.
I think it is not necessarily straightforward to say what eyepiece magnification is equivalent since, from what I understand, eyepieces above 10x typically crop the intermediate image also. If you assume that an inexpensive 20x eyepiece maintains roughly the same apparent field of view then it would be cropping by a factor of 2 relative to the 10x and give a field of view of your specimen that is similar to your camera.
According to Wikipedia a 1/2.5" sensor is 5.76 mm wide. Assume the objective and 10x eyepiece are normally working with a 20 mm diameter intermediate image. The 0.5x relay should project that to a 10 mm diameter image in the plane of the sensor, of which the sensor sees only 5.76 mm. So the horizontal field of view you see with the camera should be approximately half (5.76/10) of what you would see with a 10x eyepiece.
I think it is not necessarily straightforward to say what eyepiece magnification is equivalent since, from what I understand, eyepieces above 10x typically crop the intermediate image also. If you assume that an inexpensive 20x eyepiece maintains roughly the same apparent field of view then it would be cropping by a factor of 2 relative to the 10x and give a field of view of your specimen that is similar to your camera.

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Re: The Magnification Issue
Yes that is probably the best. I did it with my other camera (you may recall). The camera did cut off a fair bit of the field. I think this one cuts off less. I will try again.Scarodactyl wrote: ↑Mon Jul 13, 2020 7:46 pmI'd suggest you skip the calculations and measure directly. Put a ruler in focus, take a look at it through your 10x eyepieces and through your camera to directly measure the field of view in each and compare.

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Re: The Magnification Issue
Forgive me: at 80 years old my arithmetic (or algebra) is very rusty. Using my 60x objective .01 mm looks like 3.8 cm or 38 mm. What would be the magnification factor then? Is it only 380 power?
Re: The Magnification Issue
Hello, there are many interesting issues in this thread, here are a few   
One is: If .01mm in a "ruler" becomes 38mm, then the following is true, 0.01 x =38, so x equals 3,800 as the magnification. Any known dimension such as 0.01mm or even 1mm measured and compared as above can yield a final magnification.
A second issue is where does a "calculated" magnification occur? If the objective is 40x, a 0.5x intermediate lens is present, and the ocular is 10x , this yields a calculated magnification of 200x. As I understand it, that 200x magnification occurs at a distance of 10 inches from the eye point of the ocular. If the projection distance is less than 10 inches, then the fraction of 10 inches present as the projection then needs to be also multiplied. So if a distance of 5 inches projection is used, the magnification at that point would be 100x. Since there are +/ errors in all of the above, such a calculated magnification is very much an approximation.
Back to the first example above, I photograph a known dimension ruled slide or ruler with a given setup along with the images of my subjects. That known dimension can then be carefully remeasured in the final output: such as a photographic print, on your computer monitor, or even on a projection screen to let one calculate the final magnification you have produced. This is very simple and direct.
Using appropriate proportions one can use measurements like the above to accurately calculate the actual size for objects photographed under a microscope. I do not know how to produce equations in this application, otherwise I would create a fully detailed example.
One is: If .01mm in a "ruler" becomes 38mm, then the following is true, 0.01 x =38, so x equals 3,800 as the magnification. Any known dimension such as 0.01mm or even 1mm measured and compared as above can yield a final magnification.
A second issue is where does a "calculated" magnification occur? If the objective is 40x, a 0.5x intermediate lens is present, and the ocular is 10x , this yields a calculated magnification of 200x. As I understand it, that 200x magnification occurs at a distance of 10 inches from the eye point of the ocular. If the projection distance is less than 10 inches, then the fraction of 10 inches present as the projection then needs to be also multiplied. So if a distance of 5 inches projection is used, the magnification at that point would be 100x. Since there are +/ errors in all of the above, such a calculated magnification is very much an approximation.
Back to the first example above, I photograph a known dimension ruled slide or ruler with a given setup along with the images of my subjects. That known dimension can then be carefully remeasured in the final output: such as a photographic print, on your computer monitor, or even on a projection screen to let one calculate the final magnification you have produced. This is very simple and direct.
Using appropriate proportions one can use measurements like the above to accurately calculate the actual size for objects photographed under a microscope. I do not know how to produce equations in this application, otherwise I would create a fully detailed example.

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Re: The Magnification Issue
I knew it. I got the whole thing oversimplifed and calculated wrong. Forgive me. I was going by an image on my monitor and I multiplied wrong. I'm staying out of this calculation business.

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Re: The Magnification Issue
There are some misconceptions here.Lifephoto wrote: ↑Tue Jul 14, 2020 1:31 amA second issue is where does a "calculated" magnification occur? If the objective is 40x, a 0.5x intermediate lens is present, and the ocular is 10x , this yields a calculated magnification of 200x. As I understand it, that 200x magnification occurs at a distance of 10 inches from the eye point of the ocular. If the projection distance is less than 10 inches, then the fraction of 10 inches present as the projection then needs to be also multiplied. So if a distance of 5 inches projection is used, the magnification at that point would be 100x. Since there are +/ errors in all of the above, such a calculated magnification is very much an approximation.
When Don is talking about a 0.5x lens, it is part of a camera mount. No 10x eyepiece is involved there.
A viewing ocular typically projects the image to infinity (which your eye lens then refocuses onto your retina which is a whole other can of worms in making calculations). If you have a high enough eyepoint eyepiece and move your eye towards and away from the eyepiece the image does not change in magnification.
I think you're making this harder on yourself than it needs to be. I don't know what it would mean for it to look like 3.8cm, don't worry about that. All you need to do is see how much of the ruler is in the field of view:DonSchaeffer wrote: ↑Mon Jul 13, 2020 8:20 pmForgive me: at 80 years old my arithmetic (or algebra) is very rusty. Using my 60x objective .01 mm looks like 3.8 cm or 38 mm. What would be the magnification factor then? Is it only 380 power?
(It should be a LOT less than 62mm in real life!)
Then take a picture of the ruler with your camera and compare the width.
Re: The Magnification Issue
DonSchaeffer wrote: ↑Mon Jul 13, 2020 8:20 pmForgive me: at 80 years old my arithmetic (or algebra) is very rusty. Using my 60x objective .01 mm looks like 3.8 cm or 38 mm. What would be the magnification factor then? Is it only 380 power?
DonSchaeffer wrote: ↑Tue Jul 14, 2020 2:42 amI knew it. I got the whole thing oversimplifed and calculated wrong. Forgive me. I was going by an image on my monitor and I multiplied wrong. I'm staying out of this calculation business.
38 ÷ 0.01 = 3800
Note that this is the overall magnification, on your screen
... which is a concept that you disparaged in a previous discussion
Ref. viewtopic.php?f=24&t=9109
MichaelG.
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Re: The Magnification Issue
I know I'm making too much of a big deal about this. I guess it started with the notion that microscopes are advertised as 1,000x. They arrive at that number by simply multiplying objective times eyepiece. What an lovely formula that is? So my mind says, so why can't I multiply objective times my camera. Then I'm faced with, "ok what is my camera number?" It seems a simple enough notion. How do you arrive at it? It's certainly not worth the fuss I'm making. Even an estimate based on how it looks would make me happy.

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Re: The Magnification Issue
It is very straightforward. How large is your subject? Let's say it's 1mm. How large is your subject when projected on your sensor? To find out, you take a photograph of the subject. You then bring this photo into a photo editing software with some sort of measuring capabilities. I use imageJ for this.
You need to know how large your sensor is (in mm), and how many pixels there are along one of its sides.
You divide the width of your sensor (in mm) by the number of pixels along it's wide side. This gives you a number which represents the width (in mm) of each pixel.
You then measure how many pixels your subject covers on the image.
You then multiply the width of your subject (in pixels) with the pixel width you calculated earlier.
This will tell you how large the subject is when projected on the sensor.
If you divide the width of your subject (on the sensor) by it's size (in real life), you get the magnification. If it was 10mm on sensor and 1mm in real life, your optical magnification is 10x.
So magnification on sensor is very straightforward. The way eyepiece magnification is calculated is actually more abstract.
You need to know how large your sensor is (in mm), and how many pixels there are along one of its sides.
You divide the width of your sensor (in mm) by the number of pixels along it's wide side. This gives you a number which represents the width (in mm) of each pixel.
You then measure how many pixels your subject covers on the image.
You then multiply the width of your subject (in pixels) with the pixel width you calculated earlier.
This will tell you how large the subject is when projected on the sensor.
If you divide the width of your subject (on the sensor) by it's size (in real life), you get the magnification. If it was 10mm on sensor and 1mm in real life, your optical magnification is 10x.
So magnification on sensor is very straightforward. The way eyepiece magnification is calculated is actually more abstract.
Re: The Magnification Issue
.DonSchaeffer wrote: ↑Tue Jul 14, 2020 10:24 am[…]
Then I'm faced with, "ok what is my camera number?"
[…]
Even an estimate based on how it looks would make me happy.
You already have that, Don
... You did it with the check on the 60x objective
3800 ÷ 60 = 63.33333333333333 [ish]
MichaelG.
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Re: The Magnification Issue
ok I'm a happy camper. It's amazing how little I care about validity.
Re: The Magnification Issue
If I knew what you meant by that, I might know whether to be pleased or offended.
MichaelG.
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Re: The Magnification Issue
Maybe he was hoping somebody would say “42” ??
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Re: The Magnification Issue
I measured the FOV of a video and photograph by photographing/filming a grating having 500 lines/millimeter. For instance, with my 63x objective, the width of the FOV had 102 lines and the height 56 lines, which calculates to a FOV of 102/500 and 56/500, or 0.2x0.11 millimeter. If the size of an object is known, it is easy now to calculate the total magnification by just measuring the object size on the photograph relative to the FOV size (length of width) and divide by 0.2 or 0.11. And it's also easy to calulate the real size of an object if unknown,

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Re: The Magnification Issue
No reflection on you guys. Thanks for this discussion. It's my muzzy mind. I just want a talking point.