67
$\begingroup$

I recently stumbled upon this interesting image of a wasp, floating on water:

enter image description here

Assuming this isn't photoshopped, I have a couple of questions:

  1. Why do you see its image like that (what's the physical explanation; I'm sure there is an interesting one)?

  2. Why are the spots surrounding the wasp's legs circle shaped? Would they be square shaped if the 'feet' of the wasp were square shaped?

Improve this question
$\endgroup$
2
  • $\begingroup$ I didn't know exactly what tags would be suitable here, I wanted to give it the 'phenomenons' tag, but that one has been recently removed, so feel free to edit the tags. $\endgroup$ – OmnipresentAbsence Mar 4 '13 at 15:45
  • 2
    $\begingroup$ It's definitely not photoshopped. Or if it is, real pictures like it can be taken. One sees this often in summertime in strong sunlight when sitting next to shallow, clear water at the edges of creeks with very clean riversand on their bottoms. But, most wonderfully beautiful picture nonetheless. $\endgroup$ – Selene Routley Sep 6 '13 at 7:41
61
$\begingroup$

  1. The mechanism at play here is surface tension. The cohesion of the molecules of water is what keeps the wasp afloat. Due to this cohesion, the surface of the water behaves like a membrane and is curved inwards. The light rays that would be refracted from the perfectly flat surface are now incident at an altered angle and are reflected or refracted by altered angles around the tip of the wasp's legs, hence the shadow.

  2. The curvature of the surface traces the shape of the object that touches the surface. As you can see though, the area of the shadow is much larger than the wasp's legs' tips. The shape of the shadow will therefore always be rounded. The radii of the curvature can also be calculated, given the difference in pressure between air and water.

Improve this answer
$\endgroup$
2
  • 9
    $\begingroup$ +1 I would also think that the bright margins around the "feet" shadows would be due to the curved water focusing the light. $\endgroup$ – DQdlM Mar 4 '13 at 18:38
  • 6
    $\begingroup$ See also, Caustics. $\endgroup$ – Geoff Reedy Mar 4 '13 at 23:07
19
$\begingroup$

That is a truly amazing picture! I am by no means an expert, but I have an idea.

When the wasp stands on the water, it is curved down slightly. The light that hits these parts will then be bent more outwards than if it just hit regular water. This happens at every side of the circle, so the light is always bent out, and doesn't reach the bottom at those points.

Therefore, you get this effect.

Also, to answer your second question, the feet are so small compared to the shadow that they play no big role in the shape.

Improve this answer
$\endgroup$
10
$\begingroup$

This is a great example of how nice it can be to reason about refraction stuff using Fermat's principle.

Let's reduce all this to 2 dimensions. The surface tension produces something like this:2Dschem-plain

Now if we want to know where a light “ray” needs to go to get from some light source, we just need to find the way that takes it the least time. Light is slower in water, so it wants to go as far as possible in air – of course, only if that's not too much longer to go. So far from the insect, a light ray would just enter the water straight perpendicular, since that minimises both the total waylength and the path in water. 2Dschem-straightray

However, right below the insect foot, that won't work – the foot itself is not translucent† – and, more importantly, a little way left or right from right below the foot the quickest path will still be right through the foot, since any other path will require the light to travel substantially more through water, while the total path length is only a little shorter. 2Dscheme-forbiddencurveray

so all these rays are “invisible”. Whether that works out this way depends on how far we are away from right-below-the-foot, so that makes a circular shadow, even when the foot itself has another shape.

†Actually, it is kind of translucent I suppose, but we know the small foot will only get hit by a tiny amount of light. So if that bit of light has to be spread over a whole lot of ground, there won't be much intensity down there.

Improve this answer
$\endgroup$
6
  • 3
    $\begingroup$ IMO, what happens at the point the insect's foot touches the water is not a very interesting nor enlightening. Why would it bend left, and not right? It would be more clearer to draw what happens to the refracted light as it got nearer and nearer to the foot. Due to the curve on the water, light are refracted away and awayer from the straight line, it's not the foot that are casting shadow, it's the water that are refracting light away from the point where the feet touches the water. Whether the foot is translucent or not is irrelevant. A similar effect called caustic happens without any foot. $\endgroup$ – Lie Ryan Mar 5 '13 at 11:08
  • $\begingroup$ I disagree: "as it gets nearer to the foot" would lead to some messing around with Snell's law, which is physically far less enlightening than Fermat's principle. — Right – as I said, the foot itself doesn't matter. What matters is that there's a "pike" where a whole bunch of light rays have their quickest path (and therefore none of them gets much energy). As for "why would it bend left", I think it's clear that this is just an example ray, like the straight one. $\endgroup$ – leftaroundabout Mar 5 '13 at 12:17
  • $\begingroup$ Yeah, but the pike itself only plays a small part on the whole process since the majority of the refraction is done by the rest of the curves. $\endgroup$ – Lie Ryan Mar 5 '13 at 12:32
  • $\begingroup$ Refraction is merely a special way to look at photon propagation that only works for smooth (differentiable) surfaces. To describe something like this phenomenon you first have to do some ugly limit discussion (of course, most physicists would not make that step explicitly). On the other hand, Fermat's principle works directly (as it corresponds to the more fundamental Feynman-path propagation of the photons) and can thus be used to describe the circular shadow right away. $\endgroup$ – leftaroundabout Mar 5 '13 at 12:50
  • $\begingroup$ But even if you smooth up the pike slightly, the shadow would still be very visible. The pike is not a necessary nor the major cause of the large shadow. This answer, while good, IMO focuses too much on the discontinuity, which I believe contributes very little to the shadow than the rest of the curve. $\endgroup$ – Lie Ryan Mar 5 '13 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.