Not Your Everyday Banana

by Bart Arondson

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In this picture, we see that the sun comes out as a hexagon. I am sure it is not arbitrary. What does the hexagon tell us about the instrument that captured the image?

sunset @ pier39

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This question is also answered, with additional illustrations and explanations of the phenomenon, at photo.stackexchange.com/q/6605/1356 . –  whuber Mar 9 '11 at 17:26
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This has also been answered at the physics.SE: physics.stackexchange.com/q/9899/869 –  Colin K May 16 '11 at 15:54
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3 Answers 3

up vote 38 down vote accepted

It tells us that the aperture contains either three or six blades and that where these blades meet there is a corner which results in Fraunhofer diffraction. This is discussed mathematically in Physics SE.

It also tells us that the lens was stopped down, as if it were wide open there would be no corners to cause diffraction, regardless of the number of aperture blades.

Incidentally the number of (distinct) points to the star is equal to double the total number of unique orientations* in the sides of the aperture shape i.e. three blades would be six points, six blades would also be six points as a hexagon has only three unique orientations in its sides.

* a hexagonal aperture has six sides but only three unique orientations as there are three pairs of parallel sides.

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Spot on! Interestingly, when there are an odd number of blades, the star has double the points - because the diffraction goes in both directions. Ie, 6 blades = 6 point star, 7 blades = 14 point star, 8 blades = 8 point star, etc. –  thomasrutter Mar 9 '11 at 16:11
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@Thomas Actually, 6 blades = 12 point star, but the points come in overlapping pairs. Each blade throws points to both sides but the blades come in parallel sets. I'm sure that a careful measurement would show the points for the even-bladed apertures are exactly twice as bright as the points for the odd-bladed apertures. –  whuber Mar 9 '11 at 17:28
    
I've edited the answer in an attempt to be unambiguous regarding blades and points! –  Matt Grum Mar 9 '11 at 17:39
    
Could you describe what you mean by "unique orientations"? I think I may have figured it out based on the comments, but I'm notcertain. –  lindes Mar 9 '11 at 19:17
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@lindes, it means that when there is an even number of blades, the edge of every blade faces in the same direction as the edge of the blade opposite, so for example when there are six edges there are only 3 unique orientations, with every two opposite blades sharing the same orientation. –  thomasrutter Mar 10 '11 at 1:00
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The shape of the lens flare is related to the shape of the aperture while the characteristics of the flare as a whole have more to do with the elements used in the lens.

The lens in that image would be using a six blade aperture.

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I don't know that you'd technically call this lens flare as it's caused by diffraction at the edge of the aperture blades. –  thomasrutter Mar 9 '11 at 16:12
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The lens is using an aperture with six blades (or, theoretically, three - see comments); most probably six, since there are very few, if any, lenses with three aperture blades.

The lens is stopped down, and the aperture blades aren't rounded (or not enough for this aperture setting).

OR someone is using a star filter (though probably not, they are not very widely used).

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You always get two streaks per edge, it just that with an even number of edges streaks from parallel edges overlap and constructively interfere. This means the aperture has exactly six blades in this instance, however you wouldn't be able to tell the difference easily between a 5 and 10 blade aperture. –  Matt Grum Mar 9 '11 at 15:21
    
ps good point about the starburst filter, I used to have one of those in the film days. –  Matt Grum Mar 9 '11 at 15:22
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You're right, I've had it wrong: I didn't mean a multiple of six blades, but actually an even fraction. Theoretically, three blades would also produce six streaks, but I don't know any lenses with three aperture blades. –  fzwo Mar 9 '11 at 15:25
    
+1 for mentioning roundening of aperture blades (the lack thereof, in this example) –  jdv-Jan de Vaan Mar 10 '11 at 10:12
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