<p><a href="https://en.wikipedia.org/wiki/Gain_(electronics)" rel="nofollow noreferrer">Gain in electronics</a> can have different meanings:</p>
<blockquote>
<p>Gain is a measure of the ability of a [...] [device] to increase the power or amplitude of a signal from the input to the output port. [...] The term gain alone is ambiguous, and can refer to the ratio of output to input voltage (voltage gain), current (current gain) or electric power (power gain).</p>
</blockquote>
<p>For optics, I found the <a href="https://en.wikipedia.org/wiki/Optical_telescope#Light-gathering_power" rel="nofollow noreferrer">light-gathering power</a> the most appropriate measure, which relates the <a href="https://en.wikipedia.org/wiki/Aperture" rel="nofollow noreferrer">apparture</a> of the telescope with the one of the human eye:</p>
<blockquote>
<p>The light-gathering power of an optical telescope, also referred to as light grasp or aperture gain, is the ability of a telescope to collect a lot more light than the human eye. Its light-gathering power is probably its most important feature. [...]
The gathering power <span class="math-container">$P$</span> compared against a human eye is the squared result of the division of the aperture <span class="math-container">$D$</span> over the observer's pupil diameter <span class="math-container">$D_{p}$</span> [...] An example gathering power of [a telescope with] [...] <span class="math-container">$254\,{\rm mm}$</span> [aparture] compared to an adult pupil diameter being <span class="math-container">$7 \, {\rm mm}$</span> is given by:
<span class="math-container">$$P = \left(\frac {D}{D_{p}}\right)^2 = \left(\frac {254}{7}\right)^2 \approx 1316.7$$</span></p>
</blockquote>
<p><strong>Edit:</strong> After the question changed slightly, I guess I should add a bit on &quot;how to calculate gain from an image&quot;: Gain requires always <em>two</em> measurements, e.g. the image from your telescope and a base of comparison, e.g. the human eye, that's why I found the aperature-based-gain definition appealing.</p>
<p>If you have a datafile with intensities from your telescope - basically an image in grayscales depicting the spatial light intensity distribution - you would have to compile an image of the same area of the sky which serves you as a base of comparison.</p>

CC BY-SA 4.0
2021-03-23T09:44:23.233Z
2021-03-23T15:38:46.717Z
2021-03-23T15:38:46.717Z
B--rian
<blockquote>
<p>What is the gain of a telescope?</p>
</blockquote>
<p>I'll answer this from a different perspective. The <a href="https://en.wikipedia.org/wiki/Antenna_gain" rel="nofollow noreferrer">gain of an antenna</a> such as a radio telescope can be defined as the ratio over the gain of an <a href="https://en.wikipedia.org/wiki/Isotropic_radiator#Antenna_theory" rel="nofollow noreferrer">isotropic antenna</a>. For example, a simple, bare, <a href="https://en.wikipedia.org/wiki/Dipole_antenna#Radiation_pattern_and_gain" rel="nofollow noreferrer">half-wave dipole has a gain of 1.6 dBi</a> over isotropic.</p>
<p>From <a href="https://space.stackexchange.com/a/35709/12102">this answer to <em>Quantitatively, why will optical communication be better than X-band for deep-space communications?</em></a>:</p>
<h2>Link Budget</h2>
<p>From <a href="https://space.stackexchange.com/a/24343/12102">this answer</a>:</p>
<blockquote>
<p><span class="math-container">$$ P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX} $$</span></p>
<ul>
<li><span class="math-container">$P_{RX}$</span>: received power on Earth</li>
<li><span class="math-container">$P_{TX}$</span>: transmitted power by Voyager</li>
<li><span class="math-container">$G_{TX}$</span>: Gain of Voyagers transmitting antenna (compared to isotropic)</li>
<li><span class="math-container">$L_{FS}$</span>: Free space Loss, what we usually call <span class="math-container">$1/r^2$</span></li>
<li><span class="math-container">$G_{RX}$</span>: Gain of Earth's receiving antenna (compared to isotropic)</li>
</ul>
<p><span class="math-container">$$G \sim 20 \times \log_{10}\left( \frac{\pi d}{\lambda} \right)$$</span></p>
<p><span class="math-container">$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right).$$</span></p>
</blockquote>
<ul>
<li><a href="https://space.stackexchange.com/search?q=user%3A12102+link+budget">further reading</a></li>
</ul>
<p>So at a wavelength of 10 centimeters, a ten meter dish will have a gain of about 70 dBi.</p>
<p>By extension, at a wavelength of 0.5 microns a ten centimeter amateur telescope will have a gain of 116 dBi and a 6 mm human pupil will have a gain of 91.5 dBi</p>
<p>That means that the telescope has a gain of 24.5 dB compared to the eye.</p>
<p><span class="math-container">$10^{24.5/10}$</span> is (within rounding) the same as <span class="math-container">$(10/0.6)^2$</span>.</p>
<p>At first you may think it strange to think of antennas for light, but this is exactly where things are going in photonics; light receivers the size of half-wave dipoles, and diodes and other circuit elements to convert light oscillations to signals that can be treated as electrical signals. This is outside the scope of Astronomy though.</p>

CC BY-SA 4.0
2021-04-01T07:47:17.873Z
Great answer, in particular that you are giving back dB values.
B--rian
CC BY-SA 4.0
2021-03-24T02:56:13.727Z
2021-03-24T02:56:13.727Z
uhoh
<p>What is the gain of a telescope? How would I go about calculating the gain from a flatfield image?</p>

CC BY-SA 4.0
2023-01-17T07:14:53.960Z
Gain=Power(output)/Power(input)
db=10log(Gain)
Hope it helps!
Dinesh Maheswari Vaittinadane
CC BY-SA 4.0
2023-01-17T08:32:24.227Z
Do you mean the gain of the detector? i.e. Number of electrons per ADU?
ProfRob
<telescope><photometry>
What is the gain of a telescope?
nocappp
# What is the gain of a telescope?

(Asked by: username_0 on 2021-03-18)

What is the gain of a telescope? How would I go about calculating the gain from a flatfield image?

**Tags:** telescope, photometry

> **Comments:**
> **username_1**: Gain=Power(output)/Power(input) db=10log(Gain) Hope it helps!  
> **username_2**: Do you mean the gain of the detector? i.e. Number of electrons per ADU?  

---

## 2 answers

### Answer by username_3 on 2021-03-23. Score: 2

[Gain in electronics](https://en.wikipedia.org/wiki/Gain_(electronics)) can have different meanings:

> Gain is a measure of the ability of a [...] [device] to increase the power or amplitude of a signal from the input to the output port. [...] The term gain alone is ambiguous, and can refer to the ratio of output to input voltage (voltage gain), current (current gain) or electric power (power gain).

For optics, I found the [light-gathering power](https://en.wikipedia.org/wiki/Optical_telescope#Light-gathering_power) the most appropriate measure, which relates the [apparture](https://en.wikipedia.org/wiki/Aperture) of the telescope with the one of the human eye:

> The light-gathering power of an optical telescope, also referred to as light grasp or aperture gain, is the ability of a telescope to collect a lot more light than the human eye. Its light-gathering power is probably its most important feature. [...]
> The gathering power $P$ compared against a human eye is the squared result of the division of the aperture $D$ over the observer's pupil diameter $D\_{p}$ [...] An example gathering power of [a telescope with] [...] $254\,{\rm mm}$ [aparture] compared to an adult pupil diameter being $7 \, {\rm mm}$ is given by:
> $$P = \left(\frac {D}{D\_{p}}\right)^2 = \left(\frac {254}{7}\right)^2 \approx 1316.7$$

**Edit:** After the question changed slightly, I guess I should add a bit on "how to calculate gain from an image": Gain requires always *two* measurements, e.g. the image from your telescope and a base of comparison, e.g. the human eye, that's why I found the aperature-based-gain definition appealing.

If you have a datafile with intensities from your telescope - basically an image in grayscales depicting the spatial light intensity distribution - you would have to compile an image of the same area of the sky which serves you as a base of comparison.


---

### Answer by username_4 on 2021-03-24. Score: 1

> What is the gain of a telescope?

I'll answer this from a different perspective. The [gain of an antenna](https://en.wikipedia.org/wiki/Antenna_gain) such as a radio telescope can be defined as the ratio over the gain of an [isotropic antenna](https://en.wikipedia.org/wiki/Isotropic_radiator#Antenna_theory). For example, a simple, bare, [half-wave dipole has a gain of 1.6 dBi](https://en.wikipedia.org/wiki/Dipole_antenna#Radiation_pattern_and_gain) over isotropic.

From [this answer to *Quantitatively, why will optical communication be better than X-band for deep-space communications?*](https://space.stackexchange.com/a/35709/12102):

## Link Budget

From [this answer](https://space.stackexchange.com/a/24343/12102):

> $$ P\_{RX} = P\_{TX} + G\_{TX} - L\_{FS} + G\_{RX} $$
>
> * $P\_{RX}$: received power on Earth
> * $P\_{TX}$: transmitted power by Voyager
> * $G\_{TX}$: Gain of Voyagers transmitting antenna (compared to isotropic)
> * $L\_{FS}$: Free space Loss, what we usually call $1/r^2$
> * $G\_{RX}$: Gain of Earth's receiving antenna (compared to isotropic)
>
> $$G \sim 20 \times \log\_{10}\left( \frac{\pi d}{\lambda} \right)$$
>
> $$L\_{FS} = 20 \times \log\_{10}\left( 4 \pi \frac{R}{\lambda} \right).$$

* [further reading](https://space.stackexchange.com/search?q=user%3A12102+link+budget)

So at a wavelength of 10 centimeters, a ten meter dish will have a gain of about 70 dBi.

By extension, at a wavelength of 0.5 microns a ten centimeter amateur telescope will have a gain of 116 dBi and a 6 mm human pupil will have a gain of 91.5 dBi

That means that the telescope has a gain of 24.5 dB compared to the eye.

$10^{24.5/10}$ is (within rounding) the same as $(10/0.6)^2$.

At first you may think it strange to think of antennas for light, but this is exactly where things are going in photonics; light receivers the size of half-wave dipoles, and diodes and other circuit elements to convert light oscillations to signals that can be treated as electrical signals. This is outside the scope of Astronomy though.

> **Comments:**
> **username_3**: Great answer, in particular that you are giving back dB values.  

---

