Jump to content

sgl_imaging_challenge_2021_3.thumb.jpg.30e9b298c34c80517e8b443ce153fce3.jpg

What is the telescope's prime magnification with NO eyepiece lens?


Recommended Posts

Hi everyone!
I am new to astrophotography so please pardon me if my inquiry is a novice question.
I have a Canon EOS DSLR camera. I also have a Galileo brand 700x76mm reflector telescope with 1.25" eyepiece.
I just ordered a T-mount for Canon DSLRs and a t-ring 1.25" adapter to slide into my telescope. (from B&H)
Those adapter parts will arrive later this week.
I will be viewing it in the prime configuration where I will use no eyepiece lens.
I may or may not need a Barlow lens in this configuration. (I have been told it depends upon the telescope, whether or not you need a Barlow, so I will buy that later if I end up needing it to reach the telescope's focal plane).

My question is:
How much magnification will I see from the telescope's mirrors alone with no eyepiece glass and no Barlow lens in between the telescope's focuser and my camera's sensor?
What is the magnification that the telescope's mirrors alone create with no eyepiece in the focuser?

I know that if I have to drop in a Barlow lens, that will change the magnification depending on the Barlow's power.
I was just wondering how large the moon will look compared to what I see with the naked eye.
I know that if I was using an eyepiece, I would simply divide the 700mm focal length by the eyepiece size (700mm/20mm= 35x magnification or 700mm /6mm = 117x magnification) etc...

Thanks!
icon_salut.gif

twothumbsup.gif
Banane09.gifparty05.gifastronomer1.gif

  •  
Link to post
Share on other sites
  • Replies 65
  • Created
  • Last Reply

Top Posters In This Topic

Top Posters In This Topic

Popular Posts

So as you see, telescope is not magnifying at all! It is minifying - making Moon that is 3474 km in diameter, appear to be only 6.1mm!  

I love simple answers.    

That's possibly not the best response to get continued answered from people.They are giving you the answers and information you need to learn in order to understand. My answer gave exactly the in

Posted Images

The scope just does the focusing, it's the eyepiece that does the magnification. I'm no expert in AP but I think you're really asking about image scale matching with the camera chip size. I'll leave that for an imager to answer in more detail - and better than I can explain. :)

  • Like 1
Link to post
Share on other sites

No magnification, you need an eyepiece and an eyeball for magnifiv=cation.

What youy get is an image size, defined by the tangent of the angle subtended by the object multiplied by the scope focal length.

If you put a 2x barlow in then the object size is doubled.

So moon = 0.5 degrees, scope appears to be 700mm so the image is:

S = tan(0.5)*700 = 6.1mm.

Edited by ronin
  • Like 1
  • Thanks 1
Link to post
Share on other sites
1 hour ago, ronin said:

No magnification, you need an eyepiece and an eyeball for magnifiv=cation.

What youy get is an image size, defined by the tangent of the angle subtended by the object multiplied by the scope focal length.

If you put a 2x barlow in then the object size is doubled.

So moon = 0.5 degrees, scope appears to be 700mm so the image is:

S = tan(0.5)*700 = 6.1mm.

So as you see, telescope is not magnifying at all!

It is minifying - making Moon that is 3474 km in diameter, appear to be only 6.1mm!

:D

 

  • Like 2
  • Haha 5
Link to post
Share on other sites

At prime focus there is no magnification as the image can scaled to any size, so what is the magnification then?

Im no expert, but the way you measure the image "magnification" is with arc seconds per pixel. That way, no matter how much you crop or move away from the image, the arc seconds per pixels stays the same. :) 

HTH

Link to post
Share on other sites

In all seriousness, telescope without eyepiece or camera is projection device rather than magnifying device. As pointed out by ronin, telescope forms image of object on focal plane of a certain size. That size depends on focal length of telescope and actual size and distance of object (or in other words - angular size of object).

Magnification would imply conversion of scale of single unit - so we can say that Moon is magnified x60 if angular size of Moon is 30 degrees given that we see Moon without magnification to be 30' (or 0.5 degrees). Telescope on its own does not convert scale of single unit - it does unit conversion (projection) - from angular to planar / spatial so we end up going from angular size to length.

Now eyepiece / telescope combination does magnification. Telescope / prime focus camera does not do magnification in that sense, it is also projection device - with addition of sampling rate. Sampling can be seen as number of pixels per mm or number of pixels per angular size - this is what gives illusion of "magnification" or zoom to an image. It is not really magnification or zoom, it is just sampling resolution. Same image can be both small or big, depending on monitor that we use to display it, and distance to observer - look at any image on computer screen from "normal" distance and then move 10 feet away - it will look much smaller :D  (but it is the same image, sampled at the same resolution). So resolution is not measure of magnification / zoom in normal sense.

 

  • Like 1
Link to post
Share on other sites

eye roll.....

 

OK, thanks for the "CLARIFICATION"!!  LOL!! Clear as mud now!  ha-ha!

The telescope without an eyepiece lens surely will make the image look larger than if I was standing next to the telescope with my naked eye. YES???

And it will surely make the image on my camera's sensor look larger than if I just took a picture with my camera and a basic 35mm lens for example. YES???

I am just trying to figure out (roughly) how much magnification hooking a DSLR up to a telescope in prime mode will get me, compared to say, using a 35mm lens or a 50mm lens or an 18mm lens for example.

Do I simply look at the focal length of the telescope and compare that to any normal DSLR lens' length?  So 700mm will be 20x more zoomed in than a 35mm DSLR lens?! (for example)

 

Edited by GalileoCanon
Link to post
Share on other sites

"Magnification" is all relative to a baseline size. I think you're asking "how much bigger will the picture image look at prime, versus naked eyeball, and telephoto lens at XXmm ". All I can do is give some comparative photos which might give you an idea. You'll have to consider the focal length of the telescope used; at "prime" focus, it would be like having that focal length lens on the DSLR. In my case, the telescope used for comparison has a focal length of 2000 mm.

1st picture: Nikon D3400. 18mm lens, 1/4 sec, f/3.5 at ISO3200     Same camera used in all pictures

2-Nikon D3400. 300mm lens, 1/4 sec, f/6.3 at ISO 3200

3Nikon D3400. EYEPIECE PROJECTION, Edge HD 800, 25mm EP (40x magnification) 1/160 sec at ISO 100

4-Nikon D3400. PRIME FOCUS, Edge HD 800 (2000mm F/L) 1/60 sec at ISO 400

See the relative difference in size of the Moon? Frame size is the same in all four photos. These were all taken the same night, so relative Moon size is the same. This was last week's Supermoon from the SE USA.

 

DSC_1016.JPG

DSC_1017.JPG

 

DSC_0949.JPG

DSC_0931.JPG

Edited by Luna-tic
  • Like 2
Link to post
Share on other sites
7 hours ago, GalileoCanon said:

Do I simply look at the focal length of the telescope and compare that to any normal DSLR lens' length?  So 700mm will be 20x more zoomed in than a 35mm DSLR lens?! (for example)

Yep. You got it! :)

  • Like 2
Link to post
Share on other sites
7 hours ago, GalileoCanon said:

The telescope without an eyepiece lens surely will make the image look larger than if I was standing next to the telescope with my naked eye

If you think about it, when you shave with a flat mirror you stand a certain distance away. But if you put a convex curve on the mirror you have to stand a little nearer cos the focal point is closer. If you put a concave curve on the mirror you'd have to stand further away. You're not magnifying anything, you're just changing the point of focus.

In telescopes, a lens is used to magnify or reduce the size of the focused image, at the focal point. Depending on the size of the lens used and the focal length of the scope objective (mirror or lens), you get a different size of magnified image. With a normal camera, the lens does the magnifying at a fixed focal length. When you switch the lens for a telescope you're working with a different fixed focal length - the primary focal length of the scope. Hope that helps. :)

 

Link to post
Share on other sites

Thanks Luna-Tic!

That helps a lot!!

I guess the "proper" semantics for what I was trying to ask in the first place is, how large will the image of the moon look in my pictures compared to  how large it looks at 18mm and at 55mm with my camera's kit lens. OR How large will the image look compared to how it looks on my 50mm lens.

My camera has a crop sensor.  If I'm not mistaken isn't a 50mm lens pretty close to what the naked eye sees? If I hold my camera to my right eye with a 50mm lens isn't that pretty close to what I'd see in size if I looked at the image directly with my left eye?

So to filter past all the semantics couldn't we just say that the image that my DSLR sensor will view through the telescope will be magnified about 14x using the prime setup??  (700mm / 50mm = 14x magnification with the telescope in prime setup)

 

Edited by GalileoCanon
Link to post
Share on other sites

One way to think about it I suppose is that a telescope is really a modified microscope for looking at distant objects. An eyepiece is really a microscope. The purpose of the primary lens or mirror is to collect as much light as possible from the distant object and to focus and project an image of it at (or near) the focal plane for close inspection by the eyepiece (microscope). 

The size of the image formed by the primary lens or mirror is proportional to its focal length.   The longer the focal length, the larger the image.  With my Canon 450D DSLR attached to my 1000mm Newtonian telescope the full moon almost fills the short side of a full image. The same camera attached to 700mm telescope would provide an image in which the full moon was 0.7 times smaller. Just over half the size in other words. Of course the actual size of the full moon on the OP's DSLR will depend on the size of the sensor. But in approximate terms, even if it's a full frame DSLR, I would expect the OP's telescope to give an image of the moon about half (or slightly more) the size of the full image. 

Link to post
Share on other sites
12 minutes ago, GalileoCanon said:

Thanks Luna-Tic!

That helps a lot!!

I guess the "proper" semantics for what I was trying to ask in the first place is, how large will the image of the moon look in my pictures compared to  how large it looks at 18mm and at 55mm with my camera's kit lens. OR How large will the image look compared to how it looks on my 50mm lens.

My camera has a crop sensor.  If I'm not mistaken isn't a 50mm lens pretty close to what the naked eye sees? If I hold my camera to my right eye with a 50mm lens isn't that pretty close to what I'd see in size if I looked at the image directly with my left eye?

So to filter past all the semantics couldn't we just say that the image that my DSLR sensor will view through the telescope will be magnified about 14x using the prime setup??  (700mm / 50mm = 14x magnification with the telescope in prime setup)

 

Maybe you've said but what sort of DSLR have you got? 

Link to post
Share on other sites
11 hours ago, GalileoCanon said:

I think you guys are so entrenched in the terminology and semantics of telescopes that you can't give a simple answer.

That's possibly not the best response to get continued answered from people.They are giving you the answers and information you need to learn in order to understand.

My answer gave exactly the info requested, the size of the moon on camera sensor with your scope, and the link to the tool required to calculate it for yourself for your exact camera.

 

  • Like 3
Link to post
Share on other sites

Good point Stu! Thanks guys! I REALLY appreciate the assistance. I guess this is a new language to me at this point.

It's a bit overwhelming at this point. The good news is that I'm a quick learner.

When I got a DSLR, all the language was new and overly complicated compared to point and shoots I'd used previously (aperture, ISO, shutter priority, lens zoom measured in millimeters, f-stops, f3.5 vs f22, field of view, EF-S vs EF lens mounts, crop sensors vs full frame, STM, white balance, exposure, polarizers,  etc...)

That stuff all makes sense to me now.

Ouroboros, I have a Canon EOS Rebel XS DSLR. (or 1000D)

I checked and it has a 22.2mm X 14.8mm image sensor. It is 3888 X 2592 pixels.
My telescope is a Galileo brand 700x76mm reflector.

Thanks also for that astronomy too link Stu. That really helped. Also seeing the image you posted of the moon in the camera sensor in my scope really helped clarify things. Thanks!

Edited by GalileoCanon
  • Like 1
Link to post
Share on other sites

So you are interested in two things (if I'm getting that right):

1. how larger object would be compared to certain lens?

Simple ratio of focal lengths of a given lens to 700mm of your reflector (provided we are talking about same camera used).

2. what would actual size of object be on your image?

For this you need to know angular size of the object and pixel size of your camera (that is roughly sensor size divided with resolution, so in your case it would be 5.7um - or 5.7 micrometers, or you can look it up online).

Then you need to use following formula:

http://www.wilmslowastro.com/software/formulae.htm#ARCSEC_PIXEL

To get resolution of image in arc seconds per pixel - in your case it is 1.68"/pixel

Use that value, and divide target size (in arc seconds) and you will get number of pixels that target will be on image. So for Moon which is 30 arc minutes (that gives 1800 arc seconds) will be ~1070 pixels wide on the image.

 

Link to post
Share on other sites
7 minutes ago, GalileoCanon said:

Good point Stu! Thanks guys! I REALLY appreciate the assistance. I guess this is a new language to me at this point.

It's a bit overwhelming at this point. The good news is that I'm a quick learner.

When I got a DSLR, all the language was new and overly complicated compared to a point and shoots I'd used previously (aperture, ISO, shutter priority, lens zoom measured in millimeters, f-stops, f3.5 vs f22, field of view, EF-S vs EF lens mounts, crop sensors vs full frame, STM, white balance, exposure, polarizers,  etc...)

That stuff all makes sense to me now.

Ouroboros, I have a Canon EOS Rebel XS DSLR. (or 1000D)

I checked and it has a 22.2mm X 14.8mm image sensor. In pixels it has 3888 X 2592 pixels.
My telescope is a Galileo brand 700x76mm reflector.

OK. So the same size sensor as mine. You'll get a slightly larger image size than I get with my 450D attached to my Evostar ED80 which has a focal length of 600mm.  So the full moon will look something like the lunar eclipse image I took below. M31 is about 3 times the apparent diameter of the moon and will be nicely framed if you place it on the diagonal of the frame. That should give you an idea of the image size you'll get. 

image.jpeg

  • Like 2
Link to post
Share on other sites

Cool!
Thanks vlaiv and ouroboros!
That really helps!

So vlaiv, at 1070 pixels, the moon will fill 41% of my image vertically and 28% of my image horizontally. That really clears things up in extreme detail. Thanks a lot!!

Thanks for providing your picture of the moon ouroboros. That also really clears it up.

I appreciate the info guys! Thanks!

Edited by GalileoCanon
Link to post
Share on other sites
1 minute ago, GalileoCanon said:

So vlaiv at 1070 pixels, the moon will (roughly) fill 1/2 of my image vertically and 1/3 of my image horizontally? That really clears things up in extreme detail. Thanks!!

Yes, well, probably closer would be to say 1/4 horizontally (3888 is close to 4000) and 1/2.5 vertically. But do have a look at Stu's post from above to get the feel of the size of moon on image - it clearly shows how much of image would Moon take up.

  • Like 1
Link to post
Share on other sites
12 minutes ago, GalileoCanon said:

Cool!
Thanks vlaiv and ouroboros!
That really helps!

So vlaiv, at 1070 pixels, the moon will fill 41% of my image vertically and 28% of my image horizontally. That really clears things up in extreme detail. Thanks a lot!!

Thanks for providing your picture of the moon ouroboros. That also really clears it up.

I appreciate the info guys! Thanks!

A rough rule of thumb for lunar imaging is 100mm of focal length results in a diameter of 1mm on the sensor. ie 700mm focal length gives a diameter of approximately 7mm on the sensor.

  • Like 1
Link to post
Share on other sites

Cornelius Varley.

If that's the case, with my sensor the moon would roughly fill 1/3 horizontally and 1/2 vertically.  (7/22 and 7/15)   Thanks!

 

Ouroboros,

I used your lunar picture from above, made the moon 117% larger to accommodate for my 700mm telescope instead of your 600mm. I saved it in grey scale to show the moon as it will look normally. Then I resized it to the 3888 pixel width of my sensor.

The image I attached below should be as near as possible to what I will see I guess. :) Nice!!
Thanks again guys!

moon.jpg

Edited by GalileoCanon
  • Like 1
Link to post
Share on other sites

vlaiv,

I also used your calculator site. I used my Canon 1000D camera with the measurements you provided.

I did it both with a Barlow lens added to the configuration and without a Barlow lens in the image below.

Thanks!!

 

Capture.JPG

Edited by GalileoCanon
Link to post
Share on other sites

Thanks again for everything guys. I really appreciate the calculations, links, etc..

I now have an idea how magnified the image will be (about ~ 29X roughly) and how large it will appear in my camera sensor. (1/4 to 1/2 the screen)

An update: I got my t-mount and t-ring adapter from B&H a couple days ago. I had to wait until 2:30am for the moon to rise so I could try it out.
I first got the moon into focus without the camera attached with no Barlow lens and with my basic 6mm telescope eyepiece.
I then connected the camera. (I was happy to see that my t-ring adapter DOES accept a 1.25" eyepiece if needed by the way.)
From the start I could see a blurred bright blob of a moon on my camera's screen. When I dialed the focuser closer down, it got a bit more in focus.
Then I hit the stop on the focuser and couldn't dial in any closer.
SO, as anticipated I will need to buy a Barlow lens so that I can bring the focal plane out to where my camera's sensor can reach it.
Will a 2X power Barlow be enough to reach the focal plane with my DSLR sensor or will I need a 3X? I'd like to keep the power as low as possible,
so that I don't lose any more brightness than necessary. I want my camera to be able to reach it obviously though.

 I know they make 1.5x, 2x, 3x etc... Will 1.5x be enough? Is 2X enough??

I will state in advance that I am well aware of the limitations of this power telescope and this setup without an equatorial tracking tripod/mount, etc...
My expectations are properly set, so no need to reiterate the limitations of this setup.

What power Barlow lens would you guys recommend?
Also, now that I know that the t-ring adapter WILL accept a 1.25" eyepiece if needed, can you recommend an eyepiece size that is best for viewing the moon (and maybe the Andromeda Galaxy)?
If I use an eyepiece lens will I still need the Barlow lens to reach the focal plane if I don't use the prime viewing setup?
Thanks guys!

Edited by GalileoCanon
Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
  • Recently Browsing   0 members

    No registered users viewing this page.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue. By using this site, you agree to our Terms of Use.