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I am sure it's just your eyelashes creating a filtering effects, but if you look a a bright(ish) light source such as a lightbulb while squinting, if looks like you are seeing straight light rays emitting from the lightbulb, I assume you can't see an individual ray of light like this?

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  • $\begingroup$ i've downvoted because your question might encourage readers to stare at bright objects. "Staring directly at the sun can permanently scar the retina, the area at the back of the eye responsible for vision." $\endgroup$ – innisfree Mar 21 '14 at 14:01
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    $\begingroup$ @innisfree Lightbulb $\neq$ Sun. To the OP: diffraction is a relevant factor. $\endgroup$ – jinawee Mar 21 '14 at 14:22
  • $\begingroup$ @jinawee I wouldn't recommend staring into a light bulb. Although I admit that it would be unlikely to alter your vision for more than a few minutes afterwards, it still seems unwise to me. $\endgroup$ – innisfree Mar 21 '14 at 14:26
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    $\begingroup$ Related: physics.stackexchange.com/q/34222/2451 $\endgroup$ – Qmechanic Mar 21 '14 at 15:32
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As Qmechanic noted, a question pretty similar to this has already been asked, but I figured you might be interested in a non-diffractive answer.

As is obvious from women's hair-dying commercials, human hair has a small degree of reflective sheen to it, and so when you squint, it seems plausible that as the eyelashes intermesh over each other, some light which otherwise would not enter your pupil will strike your eyelashes, be reflected, and enter your eyes from a different angle than the original light source.

One simple way to model this is to model the eyelash mesh as a semi-transparent isotropic diffusive scattering surface with absorption. An ideal example of this would be a thin sheet of plastic with absorbing dye and scattering $\text{TiO}_2$ particles dispersed throughout it.

In this model, a fraction $T<1$ of the light is transmitted without any collision or angular deviation, a fraction $A$ is absorbed by the eyelashes, and a fraction $S$ is scattered isotropically from the scattering plane, with $$T+A+S=1$$.

For concreteness, let the $YZ$-plane be the scattering surface, let the eye (modeled as a tiny square of area $\alpha$ in the YZ plane) be located at $(-d,0,0)$ and put a point source of light at $(r,0,0)$ with emission intensity $\mathcal{I}r^2$ per steradian (so that the light intensity remains constant regardless of distance $r$). Then from the light due to the point source which passes through without being absorbed or scattered, the eye receives an amount of light $$P_\text{source}=T\mathcal{I}r^2\frac{\alpha}{(r+d)^2}\rightarrow T\mathcal{I}\alpha$$ in the limit $r\rightarrow\infty$ (ie, as the point source is moved to infinity but its apparent brightness is kept constant).

Meanwhile, a patch of diffusing surface located at $(0,y,z)$ with unit area will receive an amount of light $$\mathcal{I}r^2\frac{r}{\left(r^2+y^2+z^2\right)^{3/2}}\rightarrow\mathcal{I}$$ which will then be isotropically reradiated with intensity $\frac{S\mathcal{I}}{4\pi}$ per steradian. From this, the eye will detect an amount $$P_\text{diffuse}=\alpha\frac{S\mathcal{I}}{4\pi}\frac{d}{\left(d^2+y^2+z^2\right)^{3/2}}.$$

Now all that remains is to convert these to solid angle intensities as detected by the eye. For the diffuse light, note that when looking at the point $(0,y,z)$ on the diffusing panel, a solid angle $d\Omega$ looks at a patch of area $$a=4\pi d\Omega\frac{\left(d^2+y^2+z^2\right)^{3/2}}{d}$$ and thus the apparent visual brightness due to the diffusing surface $I_\text{diffuse}$ is given by $$I_\text{diffuse}d\Omega=aP_\text{diffuse}=\alpha S\mathcal{I}d\Omega$$ which is exactly the apparent brightness of a Lambertian surface, and is independent of viewing angle $\Omega$.

Meanwhile, the apparent brightness of the point source is given by $$I_\text{source}(\Omega)=P_\text{source}\delta(\Omega)=\alpha T\mathcal{I}\delta(\Omega)$$ where the angular coordinates have been chosen so that $\Omega=0$ points towards the light source and where $\delta$ is the Dirac delta function.

It is the $I_\text{diffuse}$ which gives rise to the light halo which surrounds bright objects when you squint at them. In particular, by convolving the object's visual profile with the response function previously obtained, one sees that you visualize the original object (with brightness reduced by a factor of $T$) along with a uniformly smeared-out halo.

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As the linked question (Qmechanic) suggests, diffraction effects cause some of that streaking. There are other cool things that can happen -- the same link mentions eyelash filter effects. I remember observing that back in the analog TV days. I could squint at a slightly noisy TV image and actually clean up the image, because I'd filtered out the high-frequency components which were due to background noise.

BTW, an "individual ray of light" is not something you could see as a line, because it'd be pointing straight at your eyeball. The fact that you see a streak means you're perceiving light arriving from a continuous range of angles.

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Okay so I just witnessed this phenomenon and I'm not 100% convinced its entirely from diffraction, because, like Carl Witthoft said, "...an individual ray of light is not something you could see as a line, because it'd be pointing straight at your eyeball."

I was looking at an orange colored light. I noticed that when my eye focused at a certain distance away from my body, (eyes) not directed at the light source, I could see what appeared to be focused 'lines' of light like a laser would shine.

I assume since the light is coming from a bulb which has a curved surface that I was viewing light that was diffracted by particles in the air at a certain radial distance from my eyeball. Since my vision was not focused on the light source I don't see how DumpsterDoofus's explanation of the diffuse halo which surrounds light sources is a well defined explanation of what I or Kyle Kanos was seeing.

It looks like diffracted light that you can focus on; like being able to focus a microscope on the different well-defined additive light wave interference. The key is that instead of focusing my vision on the effect of diffraction leading to local collections of additive interfering light reflected off of a surface perpendicular to my eye, my eye(s) (I could close one eye and focus even better on this phenomenon, without squinting) were focusing on light that was being emitted from the source at an angle with respect to my eyeball. Instead of this: https://upload.wikimedia.org/wikipedia/commons/f/f1/Wavepanel.png

I was seeing something more like this: http://reednightingale.com/projects/physical/laserSpirograph/DSCN0025.JPG

So, I suppose that I was seeing light perpendicular to my eye that was being refracted off of particles in the air after that light was diffracted by particles in the air at a different position (than that of the refracting particles). The latter position being the focal distance of my vision which led to me being able to 'see' the 'rays of light' which were being emitted at an angle from the light source w/r/t my eyeball.

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  • $\begingroup$ The light was probably reflected off of the particles in the air, not refracted. Sorry. Replace every instance of the word refraction with the word reflection. $\endgroup$ – Cody Hildebrand Aug 1 '15 at 4:30
  • $\begingroup$ Cody, you can edit your own post :-) $\endgroup$ – Brandon Enright Aug 1 '15 at 7:05

protected by Qmechanic Aug 1 '15 at 4:49

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