10 replies to this topic

[FirstSight]

What is the shortest distance in front of a telescope that an object can be brought into focus?  To what extent does the answer to this depend on:

- telescope focal-length

- telescope focal-ratio

- eyepiece focal-length ?

 

Assume that sufficient aperture to satisfactorily resolve the object is not an issue.

 

This question came up last weekend when I was exhibiting two of my telescopes in the North Carolina Science Museum's Astronomy Days event, my 90mm f/6.3 WO Megrez 90 refractor, and my 12" f/4.9 reflector.  The event was indoors during daytime, so the only object available to view through the telescopes was an image of the moon hung on a balcony about 60 feet diagonally above our exhibit site.  I could get the moon image to come to focus in the 90mm f/6.3 refractor using a 13T6 Nagler, but even with my lowest-power available eyepiece (31T5 Nagler) I could not come anywhere near close to bringing the 12" f/4.9 reflector into focus - I could only do so by holding the 31T5 in hand several inches outward beyond the farthest available out-focus in the focuser.  In principle, the same problem would have occurred with my refractor, had the moon image been at a sufficiently close distance.

 

I ask this in the refractor forum because refractors are sometimes used for terrestrial viewing, where too-close proximity could potentially be a problem (at least without a sufficiently long focal-length eyepiece handy) e.g. bird-watching, whereas 12" reflectors are almost exclusively used for astronomical observing, where objects being too close to come into focus in any available eyepiece is obviously never any problem.  And so, it's more likely someone here in refractors may have studied this issue.

[MooEy]

Length of the tube, baffles and focal ratio. To focus an object that is nearer to you, you will have to rack the focuser outwards. If you use a long enough extension tube, you can probably use a telescope as a macro lens. The baffles comes into the equation as it will stop down more as the object gets closer. 

 

~MooEy~

[james7ca]

As an object moves closer in from infinity the amount of displacement of the image that is formed in space by the lens or mirror is determined by the focal length of the optical system. However, depending upon the physical design and focal length of the eyepiece you will see differences in where the eyepiece needs to be placed in order to focus on the image. The focal ratio will determine the allowable depth of focus, or the range of critical focus (i.e. how far you can be displaced from the actual point of focus and still maintain sharpness in the image).

 

For a simple lens (a so-called "thin" lens) and when the image is the same size as the object the distance between the lens and the object and the lens and the point of focus will be the same and equal to twice the focal length of the lens when focused at infinity. So, if the lens had a focal length of 500mm (at infinity) and if you placed an object 1000mm (1 meter) in front of the lens then the point of focus would be 1000mm behind the lens.

 

You can get an approximation of the distances involved by using the thin-lens formula:

 

1/ o + 1/ i = 1/ f

 

Where "o" is the object distance, "i" is the image distance (from the lens) and "f" is the focal length.

 

So, when an object is located at infinity you have 1 / o equal to 1 / infinity which is zero and thus the image distance is equal to the focal length since you have:

 

  0 + 1 / i = 1 / f , or simply 1/ i = 1 / f.

 

In the case you cited, the focal length of the refractor was 90mm x 6.3 = 567mm and the object distance was approximately 60 feet, which is about 18300mm.

 

This gives:

 

 1 / 18300mm + 1 / i = 1 / 567mm

 

and thus solving for i (the image distance):

 

 1 / i = 1 / 567mm - 1 / 18300mm

 

 1 / i = 32 / 18300mm - 1 / 18300mm  (since 1 / 567 is equal to or the same as 32 / 18300)

 

 1 / i = 31 / 18300mm

 

cross multiply: 31 x i = 18300mm

 

 i = 18300mm / 31 = 590mm

 

So, your focus point would need to be extended by 590mm - 567mm or 23mm.

 

Doing the same for the 12" f/4.9 reflector (focal length 1490mm):

 

 1 / 18300mm + 1 / i = 1 / 1490mm

 

 1 / i = 1 / 1490mm - 1 / 18300mm

 

 1 / i = 12 / 18300mm - 1 / 18300mm

 

 1 / i = 11 / 18300mm

 

Cross multiply: 11 x i = 18300mm

 

 i = 18300 / 11 = 1660mm

 

So, your focus point would need to be extended by 1660mm - 1490mm or 170mm.

 

I could have made a math mistake here, but you should get the general idea.

Edited by james7ca, 03 February 2017 - 07:12 AM.

[sg6]

Add an extension tube and objects can be closer. However the image formed will be greater and so dimmer. Stick an object at the focal length of the main objective and the image formed is at infinity - oddly this is what the eyepiece produces for you when you focus on an object.

 

Your minimum distance is defined by the length the the focuser tube will allow you to move the eyepiece outwards. As the closer the object is to the main objective then the further out the image formed by the objective is. The purpose of the focuser is simply to allow you to put the eyepiece at the image location.

 

As James says the simple rule is:

1/ o + 1/ i = 1/ f

 

And that does not have a minimum distance.

[Jon Isaacs]

To add to what James and sg6 have written, the amount of outward focuser travel required depends on the focal length of the telescope and distance to the object. 

 

If one starts with the 1/f = 1/o + 1/i then one can derive the formula 

 

 i = ( f x o)/(f-o)  where i is the image distance from the lens, o is the object distance and f is the focal length of the scope.

 

The amount of outward focuser travel (oft) required is i-f so 

 

oft = (f x 0)/(o-f) -f

 

This gives somewhat different values that James's results, I checked my numbers by plugging them back into the original equation:

 

For a focal length of 567 mm and an object at 60 feet, I get:

 

oft = 18.142 mm 

 

For a focal length of 1490 mm, I get

 

oft = 132.165 mm

 

As the object distance approaches the focal length, this blows up and is a singularity at o = f.

 

jon

[james7ca]

Jon, my numbers aren't exact because I rounded everything to two or three significant digits and then used whole numbers for the ratios. If I run the digits out and use exact decimals the extension in focus for the refractor comes out to 18.141696... mm, the same value you calculated.

[Jon Isaacs]

Jon, my numbers aren't exact because I rounded everything to two or three significant digits and then used whole numbers for the ratios. If I run the digits out and use exact decimals the extension in focus for the refractor comes out to 18.141696... mm, the same value you calculated.

 

 

James:

 

I cheated, I used MathCad..  I just typed in the equation 1/f = 1/i + 1/o, clicked on solve for variable.  I had my solution.  Then I just set f = 567mm and o = 60 feet and it did the rest, it handles the units and it's fast.  

 

I did double check everything including plugging the numbers back into the original equation and did the derivation on paper to be sure I hadn't goofed up..   I wanted to be sure I had done it correctly because I knew your engineering background.. 

 

Jon

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[orion61]

Near focus is usually set to the Practical distance of the design of the Telescope in question. An example is Questars 90mm Close up lens. The diameter is the same as their regular Astronomical telescope, but the tube is extended so near focus is a matter of feet instead of Yards or meters.

I am trying to give the answer I would guess the common person would understand.

Most Astronomical tubes are set from about a block away to infinity, while spotting scopes can focus down to a few yards in some cases.

So I would say near focus is set to what the designer intended it's targets for.

[SonnyE]

Sorry, but I don't get tangled up in the mathematics. I'm a simple guy.

I got my Orion ED80T CF mainly as a camera lens to take pictures of DSO and specifically nebulae.

But in playing in my back yard, I've had it in focus on a cross arm and insulator 50-60 feet away. And a highline tower about a mile away.

And as many Nebula as I can thousands of light years away.

 

In short, probably about the most useful lens I own. As I had hoped it would be as I was making my choices on my way into Astronomy.

Mostly, I enjoy OSC (one shot color) long exposure nebula photography. I knew that was what I wanted to do.

So there you have it, 50-60 feet, out to the Universe. (And without the headaches of reflectors)

[CounterWeight]

Just curious if you are making a test set up and/or using an 'artificial star'?

Edited by CounterWeight, 06 February 2017 - 10:05 PM.

[olivdeso]

Just curious if you are making a test set up and/or using an 'artificial star'?

it will depend on what do you want to test...

 

If you just want to align/colimate your lens, you will still need about 60 feet with most of refractor. Of course you can use a diagonal and extension tubes to get the perfect focus at short distances. But these extension tubes will also bring some stress to the focuser, probably resulting on flexure, which is to be absolutely avoided while collimating of course.

 

Now if you want to perform a full star test, you will need much more distance (abour 200x the telescope diameter while using a 9µ star) and the focus will not be the main concern anymore