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I've been trying to estimate the contrast of solar images projected on the wall through (large) pinholes, and I seem to be missing something.

According to the internet an unlit room will be around 100 lux, maybe less depending on the windows, etc.

https://learn.microsoft.com/en-us/windows/win32/sensorsapi/understanding-and-interpreting-lux-values

Now if we have a pinhole with a 1cm area, then it would let through 1/10000 of the brightness of the source. In this case the sun which has ~120000 lux in brightness.

This means that the pinhole will emit ~12 lux of light, which is around 1/10th of the ambient light level. Much less if we let the spot actually become bigger, say 10cm in diameter. This would make it pi*5^2 ~ 80 x dimmer.

This means that the spot is roughly 0.15 lumen, where the wall is ~100.

Why can I clearly see the spot? Does this have to do with the logarithmic nature of the eyes, or am I making a mistake here? Maybe with ambient light vs direct light?

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  • $\begingroup$ The data your link gives are very strange, for dark rooms at least. Better look at en.wikipedia.org/wiki/Lux $\endgroup$
    – trula
    Commented Jul 29, 2023 at 16:44

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Lux is a unit of illuminance, i.e. luminous flux per unit area, so it is not reduced by passing through a large aperture. For example, if there is $100000$ lux of sunlight outdoors, then there will still be $\sim100000$ lux indoors over the area illuminated by a large open window.

A pinhole does reduce the luminous flux per unit area for sunlight because the light passing through the small area of the pinhole is spread out because of diffraction and the angular size of the sun.

The angular size of the Sun is $\theta_\mathrm{Sun}\approx 0.54$ degrees or about $9.4$ milliradians, so its geometric projection on a wall $f=2$ metres away has a diameter $D_{\mathrm{geom}}\approx \theta_\mathrm{Sun} f \approx 2$ cm. For sunlight with a mean wavelength of $\lambda=550$ nm passing through a $d_\mathrm{hole}=1$ cm diameter hole, the smearing due to diffraction is roughly given by size of the Airy disk, i.e. $$D_{\mathrm{diff}}\approx f2\theta\approx 2.44 f\lambda/d_\mathrm{hole} = 0.27\,\mathrm{cm}$$ This is neglible compared to the geometric diameter, so for 100000 lux of sunlight projected through a 1 cm diameter "pinhole" on a wall $2$ metres away, the illuminance would simply be about $100000\times(1\mathrm{cm}/2\mathrm{cm})^2=25000$ lux, which is very easily seen. Of course, a $1$ cm hole is too large to image the sun, so this would just be a sunbeam.

The optimal size of a pinhole for maximum resolution to actually image the Sun is when the angular size of the pinhole as seen from the projecting surface is roughly equal to the diffraction spread, i.e. $d/f=2.44 \lambda/d$, giving $$d_\mathrm{optimal}=\sqrt{2.44f\lambda}$$ This is about $2$ mm for sunlight projected over $2$ metres. This is much less than the geometric spread, so the image size will still be about $2$ cm.

For such an optimal pinhole, the projected illuminance $E_v\mathrm{(projected)}$ will be determined by the ratio of the areas of the hole and the projected image: $$\frac{E_v\mathrm{(projected)}}{E_v\mathrm{(sunlight)}}\approx\frac{d_{optimal}^2}{D_\mathrm{geom}^2} =\frac{2.44\lambda}{\theta_\mathrm{Sun}^2 f}$$ For $\sim550\,\mathrm{nm}$ $100000$ lux sunlight projected on a surface $2$ m away, this ratio is about $0.008$, so the projected spot will have an illuminance of about 800 lux. This should be easily visible in any reasonably dim room, even if the sunlight is closer to the lower end of typical values, e.g. 30000 lux.

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