How many photons enter our eyes per second when looking at the blue sky on a sunny day?

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Solution 1

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.

Solution 2

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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Updated on August 01, 2022

Comments

  • sebastianspiegel
    sebastianspiegel over 1 year

    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.

    • Mostafa
      Mostafa over 6 years
      Try calculating the total radiation power entering your eye at each frequency and apply $E=\hbar \omega$.
    • macco
      macco over 6 years
      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?
    • John Dvorak
      John Dvorak over 6 years
    • garyp
      garyp over 6 years
    • Qmechanic
      Qmechanic over 6 years
      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.
    • peterh
      peterh over 6 years
      @mocco I think the second.
    • peterh
      peterh over 6 years
      I would be so happy to make this interesting calculation, but now I don't have time, and as I will, this question will be long closed :-(
  • sebastianspiegel
    sebastianspiegel over 6 years
    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.
  • John Dvorak
    John Dvorak over 6 years
    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
  • rob
    rob over 6 years
    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.
  • John Dvorak
    John Dvorak over 6 years
    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.
  • John Dvorak
    John Dvorak over 6 years
    Unless I'm misunderstanding how albedo works, that is?
  • rob
    rob over 6 years
    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.
  • John Dvorak
    John Dvorak over 6 years
    Good comparison - staring directly into the Sun does risk permanent damage, so that seems about right. I think adding "and you're not staring directly into the Sun" to the last paragraph should do the trick?
  • rob
    rob over 6 years
    One way to estimate sky brightness is to know that it's possible during the daytime to see the brighter phases of Venus, if you know exactly where to look, but not Sirius. So the apparent magnitude of a patch of blue sky corresponds to somewhere between $0$ and $-3$. The Sun's apparent magnitude is $-26$. A difference of five magnitudes is a factor of 100 in brightness, so the blue sky is apparently about $10^{-10}$ times the brightness of the Sun. The full Moon, magnitude $-12$, is about a million times dimmer than the Sun.
  • rob
    rob over 6 years
    There may be some "apparent visible area" that I'm missing. A rate of $10^5$ photons/second sounds reasonable for look at Venus, but not for looking at a swathe of blue sky that's the same brightness as Venus. A rate $10^{10}$ photons/second sounds reasonable for a full Moon, which is brighter than the blue sky around it but occupies about $10^{-5}$ of the sky.
  • John Dvorak
    John Dvorak over 6 years
    It must be noted that Earth's atmosphere has a much bigger apparent size (180°) than Sun does (0.5°). I'm a little rusty in geometry, but five to six orders of magnitude (in the decimal logarithm sense) seems about right.
  • John Dvorak
    John Dvorak over 6 years
    @rob I will be glad to see your estimates expanded upon in an answer of yours. Please let me know when you post it.
  • crazyGuy
    crazyGuy over 6 years
    Arguing from the known brightness of the moon is clever!
  • ProfRob
    ProfRob over 6 years
    @tfb Note the revision!