<h2>Single &quot;pixel&quot; dish</h2>
<p>If you have a single point feed at the focal plane (one single-mode waveguide or one <span class="math-container">$\lambda/2$</span>-sized antenna) then the antenna's reception pattern (or <a href="https://en.wikipedia.org/wiki/Radiation_pattern" rel="nofollow noreferrer">radiation pattern</a>) will be similar to it's <a href="https://en.wikipedia.org/wiki/Diffraction" rel="nofollow noreferrer">diffraction pattern</a>. For a circular aperture and ignoring perturbations due to the small secondary or its supports this pattern will be the <a href="https://en.wikipedia.org/wiki/Airy_disk" rel="nofollow noreferrer">Airy disk</a>.</p>
<p>The half-angle of the first minimum is given by the familliar equation used to express the resolution of any telescope, optical or radio:</p>
<p><span class="math-container">$$\theta = 1.22 \frac{\lambda}{d}$$</span></p>
<p>where <span class="math-container">$\lambda$</span> is the wavelength and <span class="math-container">$d$</span> is the diameter of the telescope's aperture.</p>
<p>So if you have a 20 meter dish and you are receiving a signal at 3 GHz, your wavelength is 0.1 meters and the half-width of the angular acceptance is 0.006 radians or 0.3 degrees.</p>
<h2>Arrays of telescopes</h2>
<p>If you have an array of radio telescopes and you are using interferometry to generate a synthetic aperture (like the <a href="https://en.wikipedia.org/wiki/Very_Large_Array" rel="nofollow noreferrer">VLA</a> or <a href="https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array" rel="nofollow noreferrer">ALMA</a> or the <a href="https://en.wikipedia.org/wiki/Event_Horizon_Telescope" rel="nofollow noreferrer">Event Horizons Telescope</a> (EHT)) then your field of view can be much smaller but it is defined in the software running in the <a href="https://commons.wikimedia.org/wiki/File:Lights_glowing_on_the_ALMA_correlator.jpg" rel="nofollow noreferrer">correlator computer</a>.</p>
<h2>Dish with a focal plane array</h2>
<p>Some radio telescope dishes are equipped with an array of receivers, analogous to a CCD array of pixels on an optical telescope, though much fewer pixels.</p>
<p>In this case the half-width of the field of view (in radians) is the half-width of the focal plane array divided by the effective focal length there, or more precisely the <em>arctangent</em> of that ratio.</p>
<p>For more on that see:</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/16237/7982">What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/21121/7982">How do ASKAP&#39;s focal plane phased array feeds interact with the entire array phasing?</a></li>
</ul>
<p><a href="https://i.sstatic.net/Rqq4X.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Rqq4X.jpg" alt="enter image description here" /></a></p>
<p>Cropped from <a href="https://en.wikipedia.org/wiki/File:CSIRO_ScienceImage_2161_Close_up_of_a_radio_astronomy_telescope_with_several_more_in_the_background.jpg" rel="nofollow noreferrer">CSIRO ScienceImage 2161 Close up of a radio astronomy telescope with several more in the background</a></p>

CC BY-SA 4.0
2020-07-03T00:09:13.490Z
2020-07-03T04:42:32.007Z
2020-07-03T04:42:32.007Z
uhoh
<p>There are 3 different calculations that can be made.</p>
<p>First is the apparent field of view and analogous to a optical telescope, it is given by:</p>
<p>AFOV=2⋅atan(0.5h/D)⋅180/π degrees</p>
<p>Where atan() is the inverse tangent, h is the length of the CMOS camera in a particular direction and D is the focal length.</p>
<p>Similarly, if a dipole of length λ/2 is kept at the focus, the AFOV for the radio telescope would be 2⋅atan(0.25λ/D)⋅180/π.</p>
<p>Assuming for small angles AFOV= (0.5λ/D)⋅180/π degrees where λ is the wavelength of radio frequency.</p>
<p>Second is the resolving power of the radio telescope which is as given,</p>
<p>θ=1.22λ/A where A is the aperture of the radio telescope.</p>
<p>Lastly, angular distance at the source between the two point sources is,</p>
<p>Angular distance between point sources=distance * resolving angle.</p>

CC BY-SA 4.0
2022-07-02T13:21:16.010Z
2022-07-06T09:34:50.007Z
2022-07-06T09:34:50.007Z
ranjit
<p>It’s my understanding that radio telescopes can only receive signals that hit the dish within a certain range of angles. How is that angular range determined i.e. measured and/or calculated?</p>

CC BY-SA 4.0
2020-07-02T21:57:55.027Z
A radio dish is a mirror like any other.  Aperture is measured the same way, by the dish's diameter.  Are you asking about the aperture, or what angle you need before the radio waves will bounce off the dish?
Kyle
<observational-astronomy><radio-astronomy><radio-telescope>
How is the field of view of a radio telescope determined?
PerplexedDimension
# How is the field of view of a radio telescope determined?

(Asked by: username_0 on 2020-07-02)

It’s my understanding that radio telescopes can only receive signals that hit the dish within a certain range of angles. How is that angular range determined i.e. measured and/or calculated?

**Tags:** observational-astronomy, radio-astronomy, radio-telescope
> **Comments:**
> **username_1**: A radio dish is a mirror like any other.  Aperture is measured the same way, by the dish's diameter.  Are you asking about the aperture, or what angle you need before the radio waves will bounce off the dish?  

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## 2 answers

### Answer by username_2 on 2020-07-03. Score: 5

## Single "pixel" dish

If you have a single point feed at the focal plane (one single-mode waveguide or one $\lambda/2$-sized antenna) then the antenna's reception pattern (or [radiation pattern](https://en.wikipedia.org/wiki/Radiation_pattern)) will be similar to it's [diffraction pattern](https://en.wikipedia.org/wiki/Diffraction). For a circular aperture and ignoring perturbations due to the small secondary or its supports this pattern will be the [Airy disk](https://en.wikipedia.org/wiki/Airy_disk).

The half-angle of the first minimum is given by the familliar equation used to express the resolution of any telescope, optical or radio:

$$\theta = 1.22 \frac{\lambda}{d}$$

where $\lambda$ is the wavelength and $d$ is the diameter of the telescope's aperture.

So if you have a 20 meter dish and you are receiving a signal at 3 GHz, your wavelength is 0.1 meters and the half-width of the angular acceptance is 0.006 radians or 0.3 degrees.

## Arrays of telescopes

If you have an array of radio telescopes and you are using interferometry to generate a synthetic aperture (like the [VLA](https://en.wikipedia.org/wiki/Very_Large_Array) or [ALMA](https://en.wikipedia.org/wiki/Atacama_Large_Millimeter_Array) or the [Event Horizons Telescope](https://en.wikipedia.org/wiki/Event_Horizon_Telescope) (EHT)) then your field of view can be much smaller but it is defined in the software running in the [correlator computer](https://commons.wikimedia.org/wiki/File:Lights_glowing_on_the_ALMA_correlator.jpg).

## Dish with a focal plane array

Some radio telescope dishes are equipped with an array of receivers, analogous to a CCD array of pixels on an optical telescope, though much fewer pixels.

In this case the half-width of the field of view (in radians) is the half-width of the focal plane array divided by the effective focal length there, or more precisely the *arctangent* of that ratio.

For more on that see:

* [What is the highest granularity focal-plane array on a dish radio telescope? Or is this the ONLY ONE?](https://astronomy.stackexchange.com/q/16237/7982)
* [How do ASKAP's focal plane phased array feeds interact with the entire array phasing?](https://astronomy.stackexchange.com/q/21121/7982)

[![enter image description here](https://i.sstatic.net/Rqq4X.jpg)](https://i.sstatic.net/Rqq4X.jpg)

Cropped from [CSIRO ScienceImage 2161 Close up of a radio astronomy telescope with several more in the background](https://en.wikipedia.org/wiki/File:CSIRO_ScienceImage_2161_Close_up_of_a_radio_astronomy_telescope_with_several_more_in_the_background.jpg)


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### Answer by username_3 on 2022-07-02. Score: 1

There are 3 different calculations that can be made.

First is the apparent field of view and analogous to a optical telescope, it is given by:

AFOV=2⋅atan(0.5h/D)⋅180/π degrees

Where atan() is the inverse tangent, h is the length of the CMOS camera in a particular direction and D is the focal length.

Similarly, if a dipole of length λ/2 is kept at the focus, the AFOV for the radio telescope would be 2⋅atan(0.25λ/D)⋅180/π.

Assuming for small angles AFOV= (0.5λ/D)⋅180/π degrees where λ is the wavelength of radio frequency.

Second is the resolving power of the radio telescope which is as given,

θ=1.22λ/A where A is the aperture of the radio telescope.

Lastly, angular distance at the source between the two point sources is,

Angular distance between point sources=distance \* resolving angle.


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