# Stargazing Aperture of ISO telescope

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1. May 6, 2012

### Quagz

a bit random to this but could any of you give me an example aperture for a telescope collecting ISO far-infrared radiation. (to assist a theory about Andromeda)

2. May 6, 2012

### Staff: Mentor

Re: Astrophotography

An example aperture? What do you mean?

3. May 7, 2012

### Quagz

Re: Astrophotography

The aperture (light/radiation gathering area in meter squared) of a telescope collecting ISO Far-infrared radiation :)

4. May 7, 2012

### Staff: Mentor

Re: Astrophotography

Yes, I know what an aperture is, but I don't know what you mean by asking for an example aperture. I'm sure apertures range from a few millimeters to multi-meter designs.

5. May 9, 2012

### Quagz

Re: Astrophotography

The Aperture is different for every telescope depending on what radiation it is gathering, i kneed to know an aperture for a telescope gathering ISO Far-infrared radiation.

6. May 9, 2012

### sas3

Re: Astrophotography

No the aperture is in reference to how much radiation you want to gather not what type of radiation. Normally bigger is better, maybe you are thinking of the focal point, that does change for the different types of light.

If not then I am confused by your question also???

7. May 9, 2012

### Staff: Mentor

Re: Astrophotography

As Sas3 said, the aperture isn't usually dependent on the type of light you want to capture. A bigger aperture simply captures more light overall, of any type.

8. May 9, 2012

### Redbelly98

Staff Emeritus
I've moved this discussion to a new thread, since it's not really related to Astrophotography.

As others said, your question is worded rather strangely. ISO -- the Infrared Space Observatory -- is one specific telescope with a definite aperture, so it is odd to ask for an "example aperture" when asking about a specific telescope. It's kind of like asking "please give an example of John Smith's last name". Unless you mean something entirely different by ISO?

According to Wikipedia, the ISO has an aperture of 60 cm.

9. May 10, 2012

### collinsmark

Also, try researching/googling "resolving power" of a telescope.

The ability for a telescope to resolve two objects of an given angular distance is a function of the telescope's aperture. It is also a function of the wavelength of light being observed. This is the result of diffraction.

The detail is proportional to the aperture, and inversely proportional to the wavelength.

Putting it a different way: The bigger the aperture, the smaller the diffraction. The bigger the wavelength, the bigger the diffraction.

Resolving power can be expressed as

$$\sin \theta = 1.220 \frac{\lambda}{D}$$

Where $\theta$ is the minimum angular separation in radians, $\lambda$ is the wavelength of the light, and $D$ is the telescope's aperture.

And since $\theta$ is bound to be small for any practical telescope application, you might want to make the approximation (for small $\theta$), $\sin \theta \approx \theta$

Although the above is fine and good, it really represents the maximum resolving power. Other factors such as atmospheric "seeing" can reduce the effective resolution (for Earth based telescopes) to something worse than what is given above.

Last edited: May 10, 2012
10. May 11, 2012

### Quagz

@collinsmark Many thanks for you reply and also understanding the question unlike some :)