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How many photons enter our eyes per second when looking at the blue sky on a sunny day? Say the sun is directly over head and you are looking at the blue sky on the horizon. Say that the pupil is 2mm in diameter. I'm looking for an order of magnitude calculation here.

Update: the light hitting your eyes would only include the blue light scattered by the atmosphere. Not the directional light that is hitting the top of your head.

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    $\begingroup$ Try calculating the total radiation power entering your eye at each frequency and apply $E=\hbar \omega$. $\endgroup$ – Mo_ Apr 30 '17 at 11:19
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    $\begingroup$ Are you looking for an answer on how many photons are absorbed by photoreceptors or just how many photons enter a circle with 2mm diameter? $\endgroup$ – macco Apr 30 '17 at 11:41
  • $\begingroup$ Relevant: en.wikipedia.org/wiki/Solar_constant $\endgroup$ – John Dvorak Apr 30 '17 at 11:50
  • $\begingroup$ Perhaps more relevent: en.wikipedia.org/wiki/Diffuse_sky_radiation $\endgroup$ – garyp Apr 30 '17 at 12:47
  • $\begingroup$ Hi sebastianspiegel. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Apr 30 '17 at 20:26
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The surface brightness of the Sun is -10.6 mag per square arcsecond.

The full moon on the other hand is about 14.5 (astronomical) magnitudes fainter than the Sun, has a similar apparent angular size and is just visible in a bright daytime sky.

The flux from the daylight sky incident upon the eye is therefore around $10^{14.5/2.5}$ times less than the solar constant. i.e. About $2\times 10^{-3}$ W/m$^2$.

The pupils of the eye might have a 2mm diameter in bright light, so receive around $6.2\times 10^{-9}$ W.

Let's assume that the average blue sky photon is at 400 nm with an energy of 3.1 eV, then you receive about $10^{10}$ per second (in each eye).

Ah, but this would be correct for a small patch of blue sky with the same angular extent as the full moon (about $7\times 10^{-5}$ steradians). The eye actually collects light from $\sim \pi$ steradians, but then the projected area of the pupil is reduced by a small factor (I think 0.75) because of the $\cos \theta$ term.

So the final result is $3\times 10^{14}$ photons per eye.

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    $\begingroup$ Arguing from the known brightness of the moon is clever! $\endgroup$ – tfb Apr 30 '17 at 22:38
  • $\begingroup$ @tfb Note the revision! $\endgroup$ – Rob Jeffries Apr 30 '17 at 23:03
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Let's gather some sources:

The Wikipedia article on Solar constant says:

the solar energy arriving at the surface can vary from 550 W/m² with cirrus clouds to 1025 W/m² with a clear sky.

Since you specify a clear sky, we'll use the upper figure. If my reading is correct, the figure is for the Sun being directly overhead.

The next factor is that you're looking toward the horizon rather than directly upwards. Only 50% of your field of vision is covered by sky, the other being ground. The ground relative brightness can vary (a vantablack floor will reflect less than fresh snow), but the overall difference between the two extremes is just 0.3 orders of magnitude. Let's go aged concrete, albedo 0.2-0.3. The human eye field of vision gives us another factor which I'll approximate to 50%. So, let's say that the energy that enters the eye is about 30% of the energy that passes through a pupil-sized circle on the ground, half an order of magnitude. Although, most of the sunlight comes from the Sun, which will be out of your field of view when it's directly overhead and you're looking at the horizon...

Then we have the pupil size. This figure has been given by the asker, 2mm diameter. $\pi r^2 = 12.57\ mm^2$.

Multiplying these three together gives us 3.8 mW of solar energy entering the human eye on a sunny day on the equator looking at the horizon over a field of aged concrete. But really, our sources don't even warrant the second digit of precision.

How many photons is that? Visual inspection of the red area at https://en.wikipedia.org/wiki/Solar_irradiance#/media/File:Solar_Spectrum.png tells us that most of the energy (55% according to Gimp) is in the visual range and most of the rest is infrared. Let's say that your typical solar photon is 700 nm. Coincidentally, it's the same photon energy that plants use for photosynthesis, and Wikipedia tells us these photons have an energy of $3*10^{-19}\ J$ each.

Dividing those two gives us the upper estimate of $1.3 * 10^{16}$ photons entering the eyeball per second, although you will need to tilt your head a little and look at the Sun (not recommended) to achieve this value. $1*10^{16}$ might be a good estimate for a sunny day. Now, if there are clouds - or more simply if you decide not to look directly at the Sun - my even vaguer guesstimate says $1*10^{15}$ might be a more common value for your day to day outdoor stay.

To compare to a case where the power involved is known, you're estimating about 4 mW power entering the eye. However a typical 5 mW laser pointer says "danger: pointing directly into eye risks permanent damage." Generally a five milliwatt beam is near the boundary where your blink reflex is still effective. – rob♦

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  • $\begingroup$ I don't believe that you can use half the suns energy like that. Most of the suns rays are hitting the top of your head, not your eyes. Only the blue light that has been scattered will reach your eyes. $\endgroup$ – sebastianspiegel Apr 30 '17 at 20:16
  • $\begingroup$ Like I said in the answer, you'll have to tilt your head up a little to actually see the sun if you want to hit the upper estimate. But looking at the sky vs. at the ground makes less than an order of magnitude of difference $\endgroup$ – John Dvorak Apr 30 '17 at 20:18
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    $\begingroup$ The kilowatt-per-square-meter figure is if you're looking directly at the Sun. If you're looking at light that's been scattered 90$^o$ by the air, you're down by several orders of magnitude, because the atmosphere is mostly transparent. $\endgroup$ – rob Apr 30 '17 at 21:03
  • $\begingroup$ Air is mostly transparent, but there's a lot of it. Otherwise I'd expect the sky to look much darker than the ground, which it doesn't. There will be some drop-off (mentioned in the answer), but I don't think it's "several orders of magnitude". The Sun is bright, but relatively small in the sky. $\endgroup$ – John Dvorak Apr 30 '17 at 21:06
  • $\begingroup$ Unless I'm misunderstanding how albedo works, that is? $\endgroup$ – John Dvorak Apr 30 '17 at 21:11
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Try to remember that a blue sky on a sunny day is a visual representation that exists in your head. What we perceive in this representation is objects which emit, reflect, refract or absorb light. Brightness is a representation of more photons striking an area per second. This is just visible light. Most photons we don't even detect. Your best guesstimate would be to calculate the volume of matter within your visual range which emits, reflects and refracts light and multiply by how many photons each emits, reflects or refracts per second. Good luck!

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