Will a given focal length always yield the same optical magnification, all else being equal?

[Dark Penguin]

When working with telescopes, it's easy to calculate the theoretical magnification as the focal length of the objective divided by the focal length (FL) of the eyepiece, but the diameter of the objective doesn't enter into the equation at all. You could theoretically obtain 600x using two-inch objective lens/mirror combined with the correct ocular, but the resulting image would be near to useless given its limited light-gathering capability.

With the zoom lens I have now, the shortest FL and widest aperture available are 70 and f/4.5 respectively. I think I've mentioned elsewhere how gratified I was to discover that with this setup I can actually get some nice images of the night sky, with much fainter stars being resolved than if I use my other lens at 18mm and f/3.5. For example, here's one I took of Delphinus last night with the 70mm lens, single frame only and with a waxing gibbous moon, yet! (Post-processed as usual in Photoshop to improve contrast and add diffraction spikes):



The above image has been cropped down from its original 4000x6000 pixel size.

Suppose I were to buy another 70mm lens, only a wider one that would permit an aperture as wide as, say, f/2.5. So now I should be getting a lot more light on my sensor with each shot, but would the apparent magnification stay the same? Using the new lens, if I took the exact same shot under the same conditions, and cropped it exactly as I did the above, would Delphinus have the same size in the image? Assuming that any recent Nikon or Nikon-compatible lens will fit onto my crop-format D5600, is there any limit to how much lens my camera can use?

I probably wouldn't buy another 70mm lens; I'd want something different from what I have now. But I'm definitely looking for more aperture without having to go down to extremely short FLs, though I'll be looking into those types of lenses as well, eventually.

 

[bmike]

Characteristics of Best cameras and lenses for nightscape and astro photography, Clarkvision.com

The focal length will be the same, the light gathering capabilities of the lens changes.

The physical aperture on your 18mm f/3.5 mm lens is 18/3.5 = 5.14mm with an area of
20.7499053 sq mm

The physical aperture on your 70mm f/4 lens is 70/4 = 17.5 mm - nearly 3 times as large - but it lets in far more light with an area of 240.528188 sq mm

Correct me if I did the math wrong...

 

[Astrozoid]

Suppose I were to buy another 70mm lens, only a wider one that would permit an aperture as wide as, say, f/2.5. So now I should be getting a lot more light on my sensor with each shot, but would the apparent magnification stay the same? Using the new lens, if I took the exact same shot under the same conditions, and cropped it exactly as I did the above, would Delphinus have the same size in the image? Assuming that any recent Nikon or Nikon-compatible lens will fit onto my crop-format D5600, is there any limit to how much lens my camera can use?

Yes 70mm is going to be the same magnification.

No limit on the size of the lens you can use really. If you put it on a C14 SCT, it would have like 3000mm of focal length. And if you wanted, you could put a gf 2x Barlow on it and go to 6000mm

But if you want speed with faster focal ratios, it's best to use a prime.

Jerry

 

[Dark Penguin]

http://www.clarkvision.com/articles...-and-lenses-for-nightscape-astro-photography/

The focal length will be the same, the light gathering capabilities of the lens changes.

This is what I thought. So, if I understand how all this works, if I get a larger lens of the same FL, but with more "glass", I should be able to seriously ramp up my astrophotography.

The physical aperture on your 18mm f/3.5 mm lens is 18/3.5 = 5.14mm with an area of
20.7499053 sq

Does the f number allow you to calculate the effective entrance pupil diameter?

The physical aperture on your 70mm f/4 lens is 70/4 = 17.5 mm - nearly 3 times as large - but it lets in far more light with an area of 240.528188 sq mm

Correct me if I did the math wrong...

Well, I did say f/4.5, not 4, though I assume the same principle applies. Since the maximum aperture is presumably dependent on the entrance pupil, I assume this is how you are able to calculate the light gathering capability?

 

[1llusive]

http://www.clarkvision.com/articles...-and-lenses-for-nightscape-astro-photography/

The focal length will be the same, the light gathering capabilities of the lens changes.

This is what I thought. So, if I understand how all this works, if I get a larger lens of the same FL, but with more "glass", I should be able to seriously ramp up my astrophotography.

The physical aperture on your 18mm f/3.5 mm lens is 18/3.5 = 5.14mm with an area of
20.7499053 sq

Does the f number allow you to calculate the effective entrance pupil diameter?

The physical aperture on your 70mm f/4 lens is 70/4 = 17.5 mm - nearly 3 times as large - but it lets in far more light with an area of 240.528188 sq mm

Correct me if I did the math wrong...

Well, I did say f/4.5, not 4, though I assume the same principle applies. Since the maximum aperture is presumably dependent on the entrance pupil, I assume this is how you are able to calculate the light gathering capability?

Yep.

Focal length / focal ratio = aperture diameter

Diameter / 2 = radius

Then apply πr^2 if you're curious about aperture area. Area may better show the differences in lens' light gathering ability. Yes, more diameter always equals more area, so either allow you to compare lenses. However, area increases by squares so it more easily shows how much the differences add up with relatively small changes to diameter.

 

[Tristimulus]

Am pragmatic so a practical example is a good start:

Use a 1000mm lens and the diameter of the average moon (30 arc minutes wide) will be very close to 8.7mm on the image sensor.

Note that the image of the moon with this focal lenght is 8.7mm regardless of what image sensor is put behind the lens!

Use a 400mm and the diameter of the projected image of the moon is about 7mm.

Use a 135mm and the diameter is 2.35mm.

This apply to every other case, sun, planets, deep sky objects - and even aunt Annie.

Practical info:

1000 mm focal lenght: plate scale - one degree = 17.45mm.

1000 mm focal lenght: plate scale - one arc minute = 0,29mm.

Example:

If using a 135mm lens, just calculate 17,45 x 0,135 = 2.35mm (plate scale per degree).

This simple calculation holds up well for plate scale, and also for finding approximate field of view for moderate wide angle to telephoto lenses (and telescopes).

 

[bmike]

Dark Penguin wrote:

Well, I did say f/4.5, not 4, though I assume the same principle applies. Since the maximum aperture is presumably dependent on the entrance pupil, I assume this is how you are able to calculate the light gathering capability?

The f number is a ratio. If you divide the focal length by the f number you will get the physical size of the hole that light comes through. Assume this is a circular and you can calculate the area of this circle. The bigger the circle, the bigger the hole in the bucket, the more light you let in. Basically your wide lens, even though it is ‘faster’ has a very small hole to let light in.

The catch is that the longer the focal length the shorter your exposure will be if you are not using a tracking mount.