If we see wavelengths from 0.4 mm (blue) (not 400nm) and 0.7 mm(red) (not 700nm)

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In summary, the conversation discusses the concept of having a different visual range where the human eye is sensitive to wavelengths of 0.4 mm to 0.7 mm instead of the typical range of 400 nm to 700 nm. This would allow the individual to see objects that emit infrared light in a range of 1.5 x 10^-5 m to 10^-3 m, but they would not be able to see visible light or human bodies. The conversation also touches on Wien's law to calculate the wavelength of light emitted by different temperatures, and the idea of an "everything-scope" that assigns a color to every wavelength of electromagnetic radiation.
  • #1
physics2460
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Hello everyone, here's a question that I have:

Imagine your eye was sensitive in wavelengths of 0.4 mm to 0.7 mm rather than 400 nm to 700 nm. In this case, your mind would identify light waves that were 0.4 mm as blue and 0.7 mm as red. What would your world look like? How would it be different that what you see now?My attempt at answering this:

In the electromagnetic spectrum the light with the wavelengths from 0.4 x 10^(-3) m
To 0.7 x 10^(-3) m corresponds to the Infrared Light whose wavelength can be from 7.5 x 10^(-7) to 10^(-3) m (750 nm to 1 mm).

If our eyes can see infrared light with its own spectrum from blue to red, the blue and red colors that we will see will correspond to different intensity of Infrared Light. There are different regions in the IR spectrum. The wavelengths from 0.4 x 10^(-3) m to 0.7 x 10^(-3) m falls into the far-infrared spectrum that ranges from 1.5 x 10^(-5) m to 10^(-3) m (15 to 1000 micrometers) which is progressively far from the visible regime.

So it is clear to me that if we see this kind of light in colors, so we'll be able to see objects that emit this kind of light with different intensity and we'll see them in colors as of regular visible light. That said, we will not see the visible light any more, only the objects that radiate the light within this range will only be available to our sight. So, what will we see then?

What emits this kind of infrared light?

Thanks a lot,
Dina
 
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  • #2
You seem to have got mm and nm mixed up in the text - so I'm not sure exactly waht you are saying.
If our eyes worked form 400um to 700um rather than 400-700nm we wouldn't see very much at all.
The peak wavelength emitted by a hot object follows a very simple law (wien's law)
wavelength = 3mm K / T
So a wavelength of 0.4mm would be emitted by something at 10K which is a very low temperature, so we wouldn't really be able to distinguish different objects by the amount of light they emit.

If you wanted heat-vision to see everyday objects by their different temperatures you want a range of around 2-10um.
 
  • #3
mgb-phys: thanks so much.

But look what I found on wikipedia:

We will not see human bodies because at a temperature of 37 degrees C, our bodies radiate with a peak intensity near 900 nm (9 x 10^(-7) m) which is much shorter than what we can see with our new ability to see long infrared light.

well, this really confirms what you suggested: we cannot see bodies with low T. But what can we see? What are those objects that emit enough of infrared light for us to see?

How can I derive the wavelength from knowing the temperature, or visa-versa? Maybe I don't even need to name the objects that I'll see, but just their temperatures that corresponds to the given range of wavelengths: 0.4 x 10^(-3) to 0.7 x 10^(-3)m, i.e., 0.4 mm to 0.7 mm.
 
  • #4
I think you got the figures slighty wrong - the peak emmision for a person is around 10um

The blackbody curve is not symetric it has a long tail at the red end.
So in a dark room with our eyes you would not see an object at 300K because it's peak emmision is at 10um but you would see an object at 10,000K (like a flash bulb) because although it's peak emmsion is at 300nm (in the UV) it emits a tail of longer wavelength radiation into the visible.
 
  • #5
Dear mgb_phys,

let's try it one more time, here's what I understood:

if we see only the wavelengths that range from 0.4 x 10^(-3) m to 0.7 x 10^(-3)m (o.4mm to 0.7 mm) that means that we can't see visible light that is reflected off the objects around us. We also cannot see human bodies even if we do see the infrared light
of hot objects, the intensity of the infrared light isn't strong enough, I mean the wavelength of this infrared is too short for someone who sees in the range of 0.4mm to 0.7 mm.
The wavelength of the light emitted by a human body is 9500 nm (9.5 x 10^(-6) m) - practicing my conversions) which is too short, therefore we won't see it.

The wavelength of any other light emitted by a body we can calculate using Wien's constant:
b = 2.897768 5(51) × 106 nm K,

wavelength = b/T

then we take the result and compare the wavelength with the one we can see. If it shorter or longer, that means we won't be able to see it.

Let's try this in practice with the light bulb example:

wl = 2.897 768 × 10–3 m K / 10,000 K = 2.9 x 10^(-7) m - so as you said it's too short for us ( me and you and those with 0.4mm to 0.7mm wl vision!) to see, but because of the fact that it emits a tail of longer wavelength radiation into the visible, those with the special vision will be able to see it.

OK, here's what I don't understand: how do I know whether or not the hot object emits "a tail of longer wavelength into the visible"?
And what other object the person with the special vision will see?

Sorry, it might seem a dumb question, but I'm trying to understand it.
 
  • #6
Yeah, it's similar to a thought experiment of mine - the "everything-scope". Imagine assigning a colour to every single wavelength of EM radiation (obviously with limits on either side of the spectrum). Better, take the visible wavelenth "bracket", i.e. that specific range of lengths) and shift it to a different set of lengths. Then in our everything-scope, you shift a slider which moves the position of that range and gets a false colour image of everything there.

Is there any ONE mechanism that can do that? You have x-ray machines, electron microscopes, IR imagers, visible microscopes, radio telescopes - but can one single "thing" do that? What properties would be needed? For the opposite (emission), you would need a material with a variable work function with respect to some other variable (heat, current, etc). Any such thing?

In other words:

ANY wavelength of EM in => Machine => False colour image out
 
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  • #7
Wien's law gives the peak of the blackbody radiation. But it doesn't just emit at that wavelength, it emits small amounts at shorter wavelengths ( the curve falls off very steeply bluewards of the peak) but emits more radiation at longer wavelengths ( the curve falls off slowly redwards ) - you can see here http://en.wikipedia.org/wiki/Image:Wiens_law.svg

The sun for instance has a peak temperature around 6000K and a peak emmision near 500nm but still emits a lot of far infrared light. Once you get beyond 1mm the blackbody temperate would be too cold and radiation longer than this isn't emitted by a blackbody but by other mechanisms.

Also the blackbody curve is purely statistical, as you get further from the peak the number of photons with that wavelength drops but classically never to 0, eg. you can calculate that a room temperature blackbody will emit a visible wavelenght photon every 1000years / m^2.
 
  • #8
dst said:
but can one single "thing" do that?What properties would be needed?
A semiconductor - if you can make the band gap small enough.
A practical problem is that as make the band gap smaller the number of electrons from thermal events goes up so generally as you go to longer infrared wavelengths you have to cool the detector more.

An alternative approach if you don't need sensitivty is a bolometer - since all photons have energy you just need to let them hit a piece of material and measure the change in temperature. Works well with short wavelengths but you don't get much energy from a long-wave radio photon!
 
  • #9
Is there any ONE mechanism that can do that? You have x-ray machines, electron microscopes, IR imagers, visible microscopes, radio telescopes - but can one single "thing" do that? What properties would be needed? For the opposite (emission), you would need a material with a variable work function with respect to some other variable (heat, current, etc). Any such thing?

In other words:

ANY wavelength of EM in => Machine => False colour image out[/QUOTE]Ok, I understand the machine concept, but my machine sees the wavelengths from 0.4mm to 0.7mm. In fact, let's imagine that humans ONLY see these wavelengths. What will their world look like than? What will they see and what they won't?

We already figured out that they won't see human bodies, they will see light bulbs. I guess they will see the sun, but won't see the regular visible spectrum, they won't see the reflected light.

Can we come up with something else? Is there a simple answer?
 
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  • #10
mgb_phys said:
A semiconductor - if you can make the band gap small enough.
A practical problem is that as make the band gap smaller the number of electrons from thermal events goes up so generally as you go to longer infrared wavelengths you have to cool the detector more.

An alternative approach if you don't need sensitivty is a bolometer - since all photons have energy you just need to let them hit a piece of material and measure the change in temperature. Works well with short wavelengths but you don't get much energy from a long-wave radio photon!

Good thinking. But what I was thinking is a mechanism, as opposed to a material coupled with a mechanism (i.e. we want a pendulum clock as opposed to a quartz clock) - it just happens to be simpler, less static, and so on. Let's say, in a hypothetical world, the faster something is, the more red-shifted light from it becomes - directly and on a small enough scale to be observed. Then you can accelerate it through a magnetic field or some other and reproduce exactly the same wavelength each time.

That's obviously replicating instead of observing, but the logical process is what I'm getting at - a magnetic field (or anything really) of which we can just vary the strength to get any result and hence do it at any old precision we like, giving us an "image of everything". Your "any-wavelength-laser" would be able to create any frequency of light just by varying the voltage applied - simple as that.


physics2460 said:
Ok, I understand the machine concept, but my machine sees the wavelengths from 0.4mm to 0.7mm. In fact, let's imagine that humans ONLY see these wavelengths. What will their world look like than? What will they see and what they won't?

We already figured out that they won't see human bodies, they will see light bulbs. I guess they will see the sun, but won't see the regular visible spectrum, they won't see the reflected light.

Can we come up with something else?

Since neither of us likely understands how colour is assigned by the brain based on wavelength, or if the colours are entirely random (i.e. "this shade" = "this wavelength" instead of "shade = f(wavelength)") and share no relation with wavelength, do we have a hope of answering that? Humans can discern between only a certain number of colours, estimated at 10 million I think? Then that would imply something like "f(wavelength rounded to nearest neat number) = shade".

Your range is between far IR and microwaves. So anything emitting those would show up. You could get a "true" image (i.e. all the information is presented without any guesswork) by taking a brightness map of everything (i.e. a monochrome bitmap image), and you could assign false colour to it which wouldn't be a "true image" but quite good nonetheless. How would it look? Set the slider on our "everything-scope" to that range and find out. :biggrin:
 
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  • #11
[Since neither of us likely understands how colour is assigned by the brain based on wavelength, or if the colours are entirely random (i.e. "this shade" = "this wavelength" instead of "shade = f(wavelength)") and share no relation with wavelength, do we have a hope of answering that? Humans can discern between only a certain number of colours, estimated at 10 million I think? Then that would imply something like "f(wavelength rounded to nearest neat number) = shade". Quote]Right, but I assign to 0.4mm the color blue and the 0.7mm color red. My range between far IR adn microvawes is going to be perceived like our regular visible light. You are right, those are false colors, but it's all imaginary anyway, I have to pretend they are real and try to see the world with those glasses on.

So what will appear blue to me and what will appear red? :) Or green, or yellow, etc. What objects will show up in my view? If Wien's law works only for black bodies, how can I calculate the wavelength knowing the temperature of other objects?
 
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  • #12
taking a brightness map of everything (i.e. a monochrome bitmap image), and you could assign false colour to it which wouldn't be a "true image" but quite good nonetheless. How would it look? Set the slider on our "everything-scope" to that range and find out. :biggrin:[/QUOTE]


OK, how can I do that? (sorry guys, I'm in physics for poets class, I guess there's a difference between homework postings and more advanced stuff, I should've posted my question in homework section in order not to provoke such deep answers that I can't understand) Thanks anyway to you all.

So, how can I do that? What slider are we talking about?
 
  • #13
A hypothetical slider on a hypothetical machine.

But in your case, you can do that right now assuming you have a nice fat budget and the knowledge. Just make some sort of array of pixels that each vary one quantity (voltage, for instance) based on a variation in your quantity (brightness) and has multiple sensors for each "colour". I imagine there must be a CCD tuned to your wavelengths floating around somewhere (extracting information from that might be slightly difficult).

Then you can simply take all the brightness information and assign it to whichever pixel you got it from (exact same process as making a digital 'black and white' photograph of something) or do the same and add false colour. It's a technique routinely applied in astronomy.

For instance: http://www.allthesky.com/articles/imagecolor.html
 
  • #14
I've only read the first couple posts, and I wanted to make the following point: For the most part, we don't see objects due to their black-body radiation, but instead, due to their reflection spectra. Of course, there is one body whose emission spectrum is important - the sun!
 
  • #15
Gokul43201 said:
I've only read the first couple posts, and I wanted to make the following point: For the most part, we don't see objects due to their black-body radiation, but instead, due to their reflection spectra. Of course, there is one body whose emission spectrum is important - the sun!

Does the sun emit enough IR to "illuminate" things at the wavelengths given by the OP?
 
  • #16
Yes ,it puts out quite a lot of submilimetre but a lot is absorbed by atmospheric water vapour. You can do submilimetre from the ground on a good site like Mauna Kea.
You can do some work during the day because although the sun is a strong source mm wavelengths don't get scateered very much so as long as you aren't looking near the sun the sky is pretty dark.
 
  • #17
cepheid said:
Does the sun emit enough IR to "illuminate" things at the wavelengths given by the OP?


It looks like we will not even be able to see the sun as the wavelength of its radiation will be too short for us to perceive (5 x 10^(-7) m to be exact.) The wavelength for black bodies can be calculated using Wien’s desplacement law: wavelength = b / T where b is a constant (2.9 x 10^(-3) m K) and T is the body’s temperature in K.

Is it correct?
 
  • #18
physics2460 said:
It looks like we will not even be able to see the sun as the wavelength of its radiation will be too short for us to perceive (5 x 10^(-7) m to be exact.) The wavelength for black bodies can be calculated using Wien’s desplacement law: wavelength = b / T where b is a constant (2.9 x 10^(-3) m K) and T is the body’s temperature in K.

Is it correct?

No, you're missing the point. Wien's displacement law gives you the wavelength at which the PEAK occurs. But a blackbody emits more than one wavelength: it emits a continuum of radiation across the spectrum. In other words, the sun emits some IR, some visible, some UV, some X-rays, etc...Wien's displacement law is only saying that the intensity is *highest* at 500 nm, i.e. more radiation is emitted at 500 nm than at any other wavelength (if your numbers are correct). But if you plot the intensity of radiation emitted by the sun as a function of wavelength, it will be a *continuous curve* that happens to peak in the visible. This curve will be very close to a Planck curve, given by the Planck function for blackbody radiation (you should look it up if you would like more information on this subject). There is a different curve for different temperatures...at room temperature, the radiation peaks in the infrared (IR), and a negligible amount is emitted in the visible. So a blackbody radiator at room temperature would really look black to us (using our normal eyes...not the IR seeing ones postulated in the OP). But it would be glowing in the infrared. In fact, most room temperature objects (including you or me) ARE glowing in the infrared, even though we aren't perfect blackbody radiators. This is the basis of so-called thermal imaging cameras, from what I understand.

So...the sun IS glowing in the IR. The question I was asking in my previous post was, "how brightly?"
 
  • #19
So a blackbody radiator at room temperature would really look black to us (using our normal eyes...not the IR seeing ones postulated in the OP). But it would be glowing in the infrared. In fact, most room temperature objects (including you or me) ARE glowing in the infrared, even though we aren't perfect blackbody radiators. This is the basis of so-called thermal imaging cameras, from what I understand.

So...the sun IS glowing in the IR. The question I was asking in my previous post was, "how brightly?"[/QUOTE]

Thanks for the explanation. It really helps.

You see my task is to just imagine what the world would look like if the IR light has regular visible spectrum colors and we see it as a regular visible spectrum, provided that the world remains the same. I have to slide or shift my vision to longer (IR) wavelengths and try to see the world through these glasses.

So in regards to the sun, if we see the wavelength of 0.4 mm as blue and 0.7 mm (millimeters) with all the colors in between them, IR is not glowing in this world because we can perfectly see it in normal colors with this new vision capacity that the professor wants us to imagine, how can I find out whether or not I'll be able to see the sun. For the task I have to name a few things that will be visible to me in my new sight.

Is there an answer?
 
  • #20
cepheid said:
So a blackbody radiator at room temperature would really look black to us (using our normal eyes...not the IR seeing ones postulated in the OP). But it would be glowing in the infrared. In fact, most room temperature objects (including you or me) ARE glowing in the infrared, even though we aren't perfect blackbody radiators. This is the basis of so-called thermal imaging cameras, from what I understand.

So...the sun IS glowing in the IR. The question I was asking in my previous post was, "how brightly?"

Thanks for the explanation. It really helps.

You see my task is to just imagine what the world would look like if the IR light has regular visible spectrum colors and we see it as a regular visible spectrum, provided that the world remains the same. I have to slide or shift my vision to longer (IR) wavelengths and try to see the world through these glasses.

So in regards to the sun, if we see the wavelength of 0.4 mm as blue and 0.7 mm (millimeters) with all the colors in between them, IR is not glowing in this world because we can perfectly see it in normal colors with this new vision capacity that the professor wants us to imagine, how can I find out whether or not I'll be able to see the sun. For the task I have to name a few things that will be visible to me in my new sight.

I saw that human bodies will glow in the IR, I know that radiator, or a hot stove would glow, we did this in class. But my given range is in the far infra-red. How can I find out whether or not my objects have long enough wavelength in order to be seen by my new vision.
Is there an answer? Am I missing something again?
 
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  • #21
You would integrate the Planck curve over the bandwidth of your detector (eye)
http://en.wikipedia.org/wiki/Planck's_law_of_black-body_radiation
 

Related to If we see wavelengths from 0.4 mm (blue) (not 400nm) and 0.7 mm(red) (not 700nm)

1. What is the relationship between wavelength and color?

The shorter the wavelength, the bluer the color appears. Conversely, longer wavelengths appear redder in color.

2. Why are wavelengths measured in millimeters and nanometers?

Wavelengths are measured in millimeters (mm) or nanometers (nm) because they are much smaller units than meters (m), and light has very small wavelengths that cannot be accurately measured with larger units.

3. How do we see wavelengths of light?

The human eye contains specialized cells called cones that are sensitive to different wavelengths of light. When light enters the eye and hits these cones, they send signals to the brain, which interprets the different wavelengths as different colors.

4. Why are blue and red wavelengths specifically mentioned?

Blue and red are the two primary colors of visible light and have the shortest and longest wavelengths, respectively. By mentioning these specific wavelengths, it helps to demonstrate the range of visible light and how it corresponds to different colors.

5. Can we see wavelengths outside of the visible light spectrum?

Yes, there are wavelengths of light that are shorter (such as ultraviolet) and longer (such as infrared) than what we can see with the naked eye. These wavelengths are not visible to us but can be detected with specialized equipment.

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