<p>I've just rewritten this answer - @MikeG <a href="https://astronomy.stackexchange.com/questions/14663/pinhole-projector-for-the-transit-of-mercury/14664#comment22309_14664">caught</a> a glaring error by pointing out a really basic handy relationship called the <a href="https://en.wikipedia.org/wiki/Angular_resolution" rel="nofollow noreferrer">Rayleigh criterion</a>. </p>

<p>\begin{align} {\theta}_R \approx1.22 \frac{\lambda}{D}. \end{align}</p>

<p>It's better to read the (or any) article, but very briefly, the angular resolution is roughly the ratio of the wavelength to the diameter of a circular aperture. You can apply this equally well to pinhole-only imaging, or to a system which images by focusing with curved mirrors or lenses.</p>

<p>Mercury's diameter is about 4900 km and since it will be on the line between the sun and the earth, the distance will be about 150,000,000 minus 58,000,000 or 92,000,000km. In that case the angular width of Mercury will be about:</p>

<p>\begin{align} {\theta}_{merc} \approx \frac{4.9 \times 10^3\ \mathrm{km}}{9.2 \times 10^7\ \mathrm{km}} \approx 0.000052 \ \mathrm{rad} \approx 0.0031°  \approx 11 \ \mathrm{arcsec}.  \end{align}</p>

<p>So to even poorly resolve Mercury as a dark fuzzy dot, you'd like the diffraction width to be equal or less than the angular width. If you set the two angles equal and let $\lambda$ = 580 nm, you get </p>

<p>\begin{align} D \approx 1.22\ \frac{5.8 \times 10^{-7}\ \mathrm{m}}{\theta_R}  \approx \frac{7.1 \times 10^{-7}}{5.2 \times 10^{-5}}\ \mathrm{m}   \approx 14\ \mathrm{mm}\ (minimum)   \end{align}</p>

<p>Since the light is essentially parallel, your geometrical resolution on the screen is about the diameter of the pinhole. To make a 5.2E-05 radian object 14mm on a screen the screen would have to be VERY far away:</p>

<p>\begin{align} L_{to screen} \approx \frac{D}{\theta_{merc}}   \approx  \frac{0.014}{0.000052} \approx 270 \ meters! \end{align} </p>

<p>You can try that, use a 14 or 20mm "pinhole" at 270 meters away, but I think the light will be far to faint to see. I once did something similar to see a solar eclipse. It may have been a 10mm "pinhole" but I'm sure I wasn't <em>that</em> far away. I used household mirrors to bring the light indoors to a <em>very</em> dark area, and it worked great.  But that was at most only abour 30 meters!</p>

<p>If you would really like to try it, here are some tests you can do ahead of time:</p>

<ol>
<li>Search to see if someone has done it and given details</li>
<li>Try some simple experiments using the full sun. If the fuzziness at the edge of the sun's image really seems to be 300 times smaller than the sun's image, then that's a <em>suggestion</em> that it <em>might</em> work.</li>
<li>Check the internet for daily sun images and see if you can find a sunspot. Then test your setup and see if the sunspot suggests your resolution looks as good as 1/300.</li>
</ol>

CC BY-SA 3.0
2016-04-25T17:29:15.687Z
I don't think we can ignore diffraction. If Mercury is 12 arcsec (5.8e-5 radian) wide, the [Rayleigh criterion](https://en.wikipedia.org/wiki/Angular_resolution) $\theta = 1.2 \lambda / D$ suggests a minimum aperture of 12 mm for 580 nm yellow light.
Mike G
CC BY-SA 3.0
2016-04-25T17:58:12.490Z
@MikeG Yikes I can't figure out why I even would have written that - OK I'll do it again and show my work.
uhoh
CC BY-SA 3.0
2016-04-25T19:44:19.100Z
@MikeG Thank you for catching that! I'm still not able to concentrate, if you or someone can check my arithmetic and you find something else, let me know!
uhoh
CC BY-SA 3.0
2016-04-26T08:06:48.103Z
So, for all extents and purposes, is a pinhole camera is pretty futile for Mercury?
MichaelJRoberts
CC BY-SA 3.0
2016-04-26T08:20:17.580Z
@MichaelJRoberts I purposely didn't say either way. I'll think about it more, but that is *specifically because* it sounds so incredibly difficult! Maybe there's a world record out there to be won! But it would be much easier to use [eyepiece projection](http://www.skyandtelescope.com/online-gallery/eyepiece-projection-of-partial-solar-eclipse/) and [here](https://bobmoler.wordpress.com/tag/pinhole-projection-eyepiece-projection/) and [here](http://solar-center.stanford.edu/observe/) but never NEVER **EVER** anybody ever look through it!!
uhoh
CC BY-SA 3.0
2016-04-26T08:34:28.660Z
@MichaelJRoberts the best advice I never took but always wish I had *during* eyepiece projection with random binoculars or small telescopes is to try it once or twice ahead of time, see how big you can make the sun and how sharp the edges (or sunspots) are defined, how shadowed or dark you can make the area where you are projecting, and how easy it is to hold a camera and a piece of paper steady and both in focus to snap a few photos! A rehearsal would be really helpful, and putting some marks or lines on the paper sometimes helps your camera autofocus if you need it to.
uhoh
CC BY-SA 3.0
2016-04-26T08:35:41.113Z
Absolutely, I think practise would make perfect. I'm hardpressed for time as I currently revising for my finals so this transit will pass me by, but hopefully for the future it will certainly be best to practise.
MichaelJRoberts
CC BY-SA 3.0
2016-04-25T07:18:53.433Z
2016-04-25T21:20:35.577Z
2017-04-13T12:59:06.910Z
uhoh
<p>Very quick and simple one today. What would be the best/optimal pinhole size for a pinhole projector to observe the transit of Mercury on May 9th?</p>

<p>I want to get the optimum between resolution and brightness...has anyone any experience in this area?</p>

CC BY-SA 3.0
2016-04-25T12:35:03.633Z
See also http://astronomy.stackexchange.com/questions/14635/watching-the-mercury-transit-with-improvised-devices
Sir Cornflakes
<observational-astronomy><amateur-observing><mercury><planetary-transits>
Pinhole projector for the Transit of Mercury
MichaelJRoberts
# Pinhole projector for the Transit of Mercury

(Asked by: username_0 on 2016-04-25)

Very quick and simple one today. What would be the best/optimal pinhole size for a pinhole projector to observe the transit of Mercury on May 9th?

I want to get the optimum between resolution and brightness...has anyone any experience in this area?

**Tags:** observational-astronomy, amateur-observing, mercury, planetary-transits
> **Comments:**
> **username_1**: See also http://astronomy.stackexchange.com/questions/14635/watching-the-mercury-transit-with-improvised-devices  

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## 1 answers

### Answer by username_2 on 2016-04-25. Score: 5 | Accepted answer

I've just rewritten this answer - @MikeG [caught](https://astronomy.stackexchange.com/questions/14663/pinhole-projector-for-the-transit-of-mercury/14664#comment22309_14664) a glaring error by pointing out a really basic handy relationship called the [Rayleigh criterion](https://en.wikipedia.org/wiki/Angular_resolution).

\begin{align} {\theta}\_R \approx1.22 \frac{\lambda}{D}. \end{align}

It's better to read the (or any) article, but very briefly, the angular resolution is roughly the ratio of the wavelength to the diameter of a circular aperture. You can apply this equally well to pinhole-only imaging, or to a system which images by focusing with curved mirrors or lenses.

Mercury's diameter is about 4900 km and since it will be on the line between the sun and the earth, the distance will be about 150,000,000 minus 58,000,000 or 92,000,000km. In that case the angular width of Mercury will be about:

\begin{align} {\theta}\_{merc} \approx \frac{4.9 \times 10^3\ \mathrm{km}}{9.2 \times 10^7\ \mathrm{km}} \approx 0.000052 \ \mathrm{rad} \approx 0.0031° \approx 11 \ \mathrm{arcsec}. \end{align}

So to even poorly resolve Mercury as a dark fuzzy dot, you'd like the diffraction width to be equal or less than the angular width. If you set the two angles equal and let $\lambda$ = 580 nm, you get

\begin{align} D \approx 1.22\ \frac{5.8 \times 10^{-7}\ \mathrm{m}}{\theta\_R} \approx \frac{7.1 \times 10^{-7}}{5.2 \times 10^{-5}}\ \mathrm{m} \approx 14\ \mathrm{mm}\ (minimum) \end{align}

Since the light is essentially parallel, your geometrical resolution on the screen is about the diameter of the pinhole. To make a 5.2E-05 radian object 14mm on a screen the screen would have to be VERY far away:

\begin{align} L\_{to screen} \approx \frac{D}{\theta\_{merc}} \approx \frac{0.014}{0.000052} \approx 270 \ meters! \end{align}

You can try that, use a 14 or 20mm "pinhole" at 270 meters away, but I think the light will be far to faint to see. I once did something similar to see a solar eclipse. It may have been a 10mm "pinhole" but I'm sure I wasn't *that* far away. I used household mirrors to bring the light indoors to a *very* dark area, and it worked great. But that was at most only abour 30 meters!

If you would really like to try it, here are some tests you can do ahead of time:

1. Search to see if someone has done it and given details
2. Try some simple experiments using the full sun. If the fuzziness at the edge of the sun's image really seems to be 300 times smaller than the sun's image, then that's a *suggestion* that it *might* work.
3. Check the internet for daily sun images and see if you can find a sunspot. Then test your setup and see if the sunspot suggests your resolution looks as good as 1/300.

> **Comments:**
> **username_3**: I don't think we can ignore diffraction. If Mercury is 12 arcsec (5.8e-5 radian) wide, the [Rayleigh criterion](https://en.wikipedia.org/wiki/Angular_resolution) $\theta = 1.2 \lambda / D$ suggests a minimum aperture of 12 mm for 580 nm yellow light.  
> **username_2**: @MikeG Yikes I can't figure out why I even would have written that - OK I'll do it again and show my work.  
> **username_2**: @MikeG Thank you for catching that! I'm still not able to concentrate, if you or someone can check my arithmetic and you find something else, let me know!  
> **username_0**: So, for all extents and purposes, is a pinhole camera is pretty futile for Mercury?  
> **username_2**: @MichaelJRoberts I purposely didn't say either way. I'll think about it more, but that is *specifically because* it sounds so incredibly difficult! Maybe there's a world record out there to be won! But it would be much easier to use [eyepiece projection](http://www.skyandtelescope.com/online-gallery/eyepiece-projection-of-partial-solar-eclipse/) and [here](https://bobmoler.wordpress.com/tag/pinhole-projection-eyepiece-projection/) and [here](http://solar-center.stanford.edu/observe/) but never NEVER **EVER** anybody ever look through it!!  
> **username_2**: @MichaelJRoberts the best advice I never took but always wish I had *during* eyepiece projection with random binoculars or small telescopes is to try it once or twice ahead of time, see how big you can make the sun and how sharp the edges (or sunspots) are defined, how shadowed or dark you can make the area where you are projecting, and how easy it is to hold a camera and a piece of paper steady and both in focus to snap a few photos! A rehearsal would be really helpful, and putting some marks or lines on the paper sometimes helps your camera autofocus if you need it to.  
> **username_0**: Absolutely, I think practise would make perfect. I'm hardpressed for time as I currently revising for my finals so this transit will pass me by, but hopefully for the future it will certainly be best to practise.  

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