Viewing 1k x 1k slab of desert sand during extreme cold/hot temps?

In summary, viewing a 1k x 1k slab of desert sand during extreme temperatures reveals the stark contrasts in the desert environment. During extreme cold, the sand may appear frozen and less active, reflecting light differently. In hot conditions, the sand becomes dry and shifts dramatically, showcasing its dynamic nature. Both scenarios highlight the unique characteristics and resilience of desert landscapes under temperature extremes.
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ndvcxk123
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TL;DR Summary
Epsilon permittivity causes temperature-specific bending of light as it enters translucent medium, causing reliable targeting of the light toward separate retinal cells..so why no major image change during major temp. change ?
So, non-quartz sand has 1/2 mm kernels, they can be seen as individual objects from about 60cm distance. Standing in front of it, about a half of the individual sand kernels beam each separately into your pupil, so 500,000 discrete beams, each covers almost the entire pupil. Yet, you see a precise, differentiated image, the black one there, here the brown one, and so on. If we focus now now on just one micron spot on your pupil, lots of beams are hitting it concurrently, and yet they are neatly relayed to the "appropriate" (or should we say "average-temperature-positioning-angle" ?? )retinal cells. And this is the question, - if the permittivity is temperature dependent, should there not be a major difference in the image between a -5 celsius + a +28C day, always assuming you've spend 30 min. outdoors to cool/heat your pupil ? Is the answer that a 33 degree change is simply not enough ?

Note: My own experiment in the sand dunes located east of S.Diego was brief loss of 3D vision when no other visual cue was there, and one stood next to a house-sized wall of sand. (This also happens during dense, fast snowfall while skiing if no other objects are visible.) Thx.
 
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ndvcxk123 said:
And this is the question, - if the permittivity is temperature dependent, should there not be a major difference in the image between a -5 celsius + a +28C day, always assuming you've spend 30 min. outdoors to cool/heat your pupil ? Is the answer that a 33 degree change is simply not enough ?
Um, you are a warm-blooded human, no? Does your temperature swing 33 degrees in those circumstances? :wink:
 
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thx for replying. The pupil itself (unlike the eye) just the lens, has no vasculature, Its temperature drops in fact. But it is responsible for controlling the bending of the angle of the arriving light, so just a mm error would mean it is hitting a retina cell normally being hit w. light from another area - so hence the expecation of distortion.
 
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ndvcxk123 said:
thx for replying. The pupil itself (unlike the eye) just the lens, has no vasculature, Its temperature drops in fact. But it is responsible for controlling the bending of the angle of the arriving light, so just a mm error would mean it is hitting a retina cell normally being hit w. light from another area - so hence the expecation of distortion.
Quoting from this 2015 reference: Correlation between corneal and ambient temperature with particular focus on polar conditions
"In healthy volunteers, the cornea required at least 20–30 min to adapt to change in ambient temperature. The relationship between corneal and external temperature was relatively linear. At the two extremes, +83°C and −40°C, the corneal temperature was +42°C and +25.1°C, respectively. In the experimental setting, corneal temperature was +24.3°C at air temperature −40°C."
So an air temperature swing of 123°C results in a corneal temperature range of just 17°C. How much optical distortion do you expect that to induce in the eye?
 
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  • #5
ndvcxk123 said:
TL;DR Summary: Epsilon permittivity causes temperature-specific bending of light as it enters translucent medium, causing reliable targeting of the light toward separate retinal cells..so why no major image change during major temp. change ?
What sort of "image change" do you expect to happen?

Also, note that the rods and cones of your eye (and vision system in general) function nothing like a pixelated solid-state detector array.
 
  • #6
ndvcxk123 said:
Standing in front of it, about a half of the individual sand kernels beam each separately into your pupil, so 500,000 discrete beams, each covers almost the entire pupil.
Technically, each point in object space sends a cone of light that fills the entirety of the pupil. Individual sand particles do not send beams of light to different parts of the pupil. Every beam/cone overlaps every other beam/cone in the pupil.
ndvcxk123 said:
And this is the question, - if the permittivity is temperature dependent, should there not be a major difference in the image between a -5 celsius + a +28C day, always assuming you've spend 30 min. outdoors to cool/heat your pupil ? Is the answer that a 33 degree change is simply not enough ?
Heating or cooling a lens or mirror tends to mostly change the optical power of the system (the lens/mirror gets bigger or smaller, but stays the same shape, along with the refractive index slightly changing), which in turn changes the focal length. But the change in the focal length for the eye is very small, thanks to the eye's already very short focal length. In practice the change is not noticeable, especially given various other optical aberrations that the average person already has.
 
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For more detail:

The change in the index of refraction for a lens is given by: ##n' = n + \frac{dn}{dT}ΔT##
Where ##n'## is the refractive index at the new temp, ##n## is the original refractive index, ##\frac{dn}{dT}## is the thermal-optic coefficient, and ##ΔT## is the change in the temperature.

For a lens made of BK7 glass, ##n## is 1.51872 at 546.1 nm, with a coefficient of 3x10-6/K. A 17 K change in temp results in a change in ##n## of only 0.000051.

Quick and dirty thin lens calculations:

A symmetrical lens made of BK7 glass, with n = 1.51872 and an 8mm radius of curvature, has a focal length of 7.71128 mm.
The same lens with n = 1.518771 has a focal length of 7.710531 mm.

That's a difference of only about 0.00076 mm, or 0.76 microns.

Granted, this is a rough calculation using the thin lens formula and the eye isn't a simple lens made up of BK7 glass, but it should give you at least a very rough idea of the scale of change here.

Source: https://wp.optics.arizona.edu/optom...s/sites/53/2016/08/19-Optical-materials-1.pdf
 
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ndvcxk123 said:
so hence the expecation of distortion.
but there is focusing feedback? How else would you accommodate?
 
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