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?
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?
I've just rewritten this answer - @MikeG caught a glaring error by pointing out a really basic handy relationship called the Rayleigh criterion.
\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: