Sunny 16, Looney 11 and diffraction

Discussion in 'Open Discussion' started by dav1dz, Apr 23, 2013.

  1. dav1dz

    dav1dz Mu-43 Top Veteran

    Nov 6, 2012
    Should I be concerned about diffraction at f/16? A friend who's been interested in m4/3 seems to be. But I've shot some following the Sunny 16 rule and I can't really tell.

    Do I need to be pixel peeping beyond 100% to see a difference?

    Disclaimer: I'm not a nut landscaper, but he is.

    My interpretation of the respective rules: 1/ISO speed in bright sun @ f/16 and 1/ISO speed in moonlight @ f/11. Right...?
  2. MAubrey

    MAubrey Photographer

    Jul 9, 2012
    Bellingham, WA
    Real Name:
    Mike Aubrey
    If you don't pixel peep and don't do big prints, there's really nothing to be concerned about at all.

    Also, interestingly enough, images that are super soft from diffraction allow for far larger amounts of sharpening than those that don't for some reason. - Overcoming My f / Entekaphobia
  3. ~tc~

    ~tc~ Mu-43 Hall of Famer

    Oct 22, 2010
    Houston, TX
    You can always program shift the rule - you can still use it at f/8!
  4. kwalsh

    kwalsh Mu-43 Top Veteran

    Mar 3, 2012
    Baltimore, MD
    Erm, that moonlight figure isn't right, though it would be an OK setting for taking a picture of the moon itself.

    Diffraction becomes noticeable past about F/8 in m43 if you view at 100%. However, diffraction also sharpens quite nicely as the lens rentals link shows. Personally I rarely shoot past F/8 as all lenses do best there or wider. But if DoF is a concern F/11 seems fine and I suspect F/16 with sharpening as well. Really for m43 I'd use F/8 and 1/2*ISO for sunlight as a default.

    EDIT: That should be F/8 and 1/4*ISO of course!
  5. David A

    David A Mu-43 All-Pro

    Sep 30, 2011
    Brisbane, Australia
    I thought the moon rule was for F/8, not F/11, and the moon rule is for photographing the moon itself, not for taking photos of subjects lit by moonlight. You don't need to adjust the moon rule for less than full moons either since it's intended to capture the bright part of the moon correctly. If the moon is only part full, the lit part will be exposed correctly and the unlit part will be deep shadow to black.

    I think kwalsh has the exposure for sunlight at F/8 wrong. F/8 is 2 stops faster than F/16 and that means 4 times more light so the shutter speed should be 1/4*ISO, not 1/2*ISO.
  6. kwalsh

    kwalsh Mu-43 Top Veteran

    Mar 3, 2012
    Baltimore, MD
    Doh! Don't post when tired I guess! Edited my post with the correction.

    As to the moon I've never heard a hard and fast rule. Really if you want an "accurate" exposure the correct exposure for the moon is Sunny F/16 - it is an object directly lit by the sun after all. But the surface of the moon is actually quite dark (check out all those photos of astronauts in white space suits standing on its surface for an idea) and since we see the moon at night in a dark sky we perceive it as bright white even though it is actually a dark gray. So often we "over expose" the moon to match this perception. By how much is probably a "to taste" kind of thing.

    Thanks again for fixing my bad math!
  7. David A

    David A Mu-43 All-Pro

    Sep 30, 2011
    Brisbane, Australia
    I don't know where I heard about the "moonie 8" rule but it does seem to work.

    I think there is a difference between the moon as an object lit by the sun, and things on earth lit by the sun, and that difference is distance. We're talking reflected light, and light reflected from a roughly spherical object at that so that light is being reflected in a spherical wave front which ensures that the amount of light energy per unit area surface at the wave front is reducing as the radius of the sphere increases. The radius of the sphere, for a photograph of the moon, is the distance from the moon to earth which is a lot larger than the distance of any terrestrial object to our cameras. I think we get some loss of light reflected from the moon due to distance.

    Of course I could be wrong. I'm basing my guess on the behaviour of sound radiated from a point, something I'm a little more familiar with but still nowhere near an expert on, but sound and light both behave as waves and sound reflecting from a spherical surface is diffused, resulting in less sound being reflected to an ear or microphone some distance from the spherical surface than would be the case if the sound were reflected from a flat surface to an ear or mic the same distance from the flat surface.

    So I think we won't get equivalent reflectance of light from the spherical surface of the moon as we do from terrestrial surfaces, partly due to the nature of the surfaces and partly due to distance.
  8. kwalsh

    kwalsh Mu-43 Top Veteran

    Mar 3, 2012
    Baltimore, MD
    Do buildings get dimmer when you photograph them from twice the distance away? No. If you photograph a spherical object (say a radome) from twice the distance does it get dimmer? No. If there were any sort of square law rule for light scattered from a surface what would we expect the brightness of the moon to be? Say objects on earth are 1 mile away, moon is 240,000 miles away so 240000^2- that would be 57.6 billion times less light. Or 35 stops. So some quick thought experiment makes it pretty clear distance can't be the cause nor can the moon being a sphere be the cause.

    For an extended object (not a point source like a star) viewed at twice the distance the flux from an individual point is one fourth, but within the same solid angle from the vantage point there is also now four times as much surface reflecting. The two cancel out resulting in the same apparent surface luminance.

    The albedo of the moon in the visible band is 7%. We normally meter scenes with an 18% grey card. Thus the surface of the moon is a very dark grey, 1.3 stops less reflective than an 18% grey card. If we to want it to appear as "middle grey" (18%) we need to over expose by 1.3 stops compared to Sunny F/16. Want it to be a bit brighter than "middle grey" (most people do) and you've got Loony F/8. It works well exactly because the moon is lit by the sun (so use Sunny F/16) but we need to over expose by two stops to get the moon to appear lighter in tone than middle grey.

    Be careful of point source vs. extended source.

    Wavelength is the problem here. Walls and other surfaces appear as mirrors to sound because the surface roughness of the wall is much much less than the wavelength of the sound. So the sound reflects rather than scattering. For light anything but a polished surface has a surface roughness much greater than the wavelength of the light so the result is that the light scatters rather than reflecting.

    So the sound analogy would only be valid if the moon were a giant polished spherical mirror (which would be pretty awesome actually).

    It is really easy to get confused with sound/light analogies because of this difference - most of the time sound reflects from things but light usually scatters. This is because it is so easy to make a surface "smooth" to sound (wavelength at 3KHz is about 0.1m) that almost every surface - even natural ones like cliff faces - appear to be "smooth" and thus mirror like. For light on the other hand (wavelength for green is about 0.0000006m) almost every surface we see is "rough" and scatters unless you go to great lengths to polish a surface smooth (half way decent mirror is smooth to 1/4 wavelength).

  9. David A

    David A Mu-43 All-Pro

    Sep 30, 2011
    Brisbane, Australia
    I'm out-argued :)

    As a general principle, I think you're wrong on distance not having an effect. If that were true then any star in the night-time sky that was brighter than our sun should appear brighter than our sun, and the light from the Milky Way should be many, many times brighter than the light from the sun. That's not the case and albedo doesn't count because we're not comparing reflectance here, just direct light over astronomical differences.

    I do think you're absolutely right about the issue with our moon exposures being related to the moon's albedo and how bright we want the moon to appear in our images.

    So why, if distance does diminish light intensity as I am suggesting, does it not do so in the case of the moon?

    We're comparing the intensity of light reflected from a terrestrial object to that of light reflected from the moon. You mention an inverse square law but I think we're talking an inverse cube law if we assume a point source. Let's assume reflectance from a perfect mirror (no losses on reflection). The light source to mirror distance is roughly 93 million miles (earth to sun), and the mirror to viewer distance is a mile or so for the terrestrial object and 240,000 miles for the moon. There's an infinitesimally small difference between the cube root of 93million (ignoring the mile in the terrestrial case) and the cube root of 93.24 million. We have to consider the distance the light travels from source to viewer, not from reflecting surface to viewer if we have a perfect mirror. The distance from earth to moon has virtually no effect at all under these circumstances because it's a small fraction of the total distance. Of course neither the terrestrial object nor the moon are perfect mirrors but the impact that the differences in their reflectivity will have, plus how bright we want their images to be in our photos, is going to be what accounts the difference in exposure and that is covered in your paragraph on albedo.

    I think we both made errors but you win the cigar for explaining the exposure difference between the sunny 16/moonie 8 rules.
  10. dhazeghi

    dhazeghi Mu-43 Hall of Famer

    Aug 6, 2010
    San Jose, CA
    Real Name:
    Yes, and unless you're at particularly long focal lengths or shooting macros, it isn't as if f/16 gains you much. Higher shutter speeds on the other hand never hurt and often help.
  11. kwalsh

    kwalsh Mu-43 Top Veteran

    Mar 3, 2012
    Baltimore, MD
    Point source vs. extended source is the difference here. For extended sources distance doesn't matter, for point sources it does. What determines what is a point source and what is an extended source? For light and optics it is whether the object has an angular extent larger than the point spread function of the imaging system. A classic example of the dividing line between these would be Mars. When close to the earth to our eye it appears brighter in the sky than when it is distant. Our eyes have too small an aperture to resolve it and so it is a point source. On the other hand, for a telescope the exposure for Mars remains identical when close or far but what happens is Mars appears larger on the exposed sensor or plate. And that's the crux of the difference - when there is an extended source the increased flux received is represented by the image becoming larger while the surface has the same brightness. For a point source the point just becomes brighter as it gets closer.

    See above - moon is an extended source. Similarly, the sun when looked at directly from Mars is no dimmer - just smaller than on Earth. Being smaller that does mean that less light falls on the surface of Mars than on earth.

    Most definitely square law.

    Point source - Wikipedia, the free encyclopedia

    Thinks about it, flux is measured in power/area. Since energy must be conserved, then power is conserved. That means (area of a sphere) * flux must be constant for any radius. Area of sphere is square law, not cube.

    No, not a valid assumption. These aren't mirrors - diffuse reflection and specular reflection are different things with entirely different consequences:

    Diffuse reflection - Wikipedia, the free encyclopedia

    Anyway, got to run at the moment! Give it some thought and follow a few Wikipedia links and things should be more clear.

    Thanks for the thought provoking discussion so far!