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Your formula is wrong, is the short answer. It works fine as an order of mangnitude approximation, but it's only an approximation. What you need is Wien's displacement law: $\lambda=\frac{b}{T}$ where Wien's displacement constant $b=2.8977729(17)×10^{−3} m K$. Put $T=288K,308K$ into that and you get: $\lambda=9.4-10.0\mu m$, which as you'd expect is at ...

11

You are likety to be directly measuring the greenhouse effect of the atmosphere. The fact that your cursory measurements seem to be correlated with the air temperature around you supports this idea: the measurement is higher in the day because the Earth itself is hotter and radiating back into space more powerfully. Greenhouse gasses are thus absorbing some ...

10

In microwave ovens what matters is how much energy the radiation carries and how that energy is absorbed by the food. Visible light and IR are rapidly absorbed by most foods, so they would only heat the outer layer of the food. You'd get food with the outside carbonised and the inside raw. Microwaves are far less strongly absorbed by foods, so they ...

10

By comparing the signal to the background. Suppose you get 10 IR photons from the camera and lens background but an extra 5 from the source then you can still detect the source. There is a whole science of signal processing to detect signals much fainter than the background. Especially in IR astronomy.

9

Kyothe was on the right track, but in fact we do radiate in the visible, just in such small amounts that it's not detectable for all practical purposes. If you look at the referenced Planck (black body) curves for objects around human body temperature, the short-wave tail is nonzero in the visible range, but it's there.

9

In astronomy, the background from the camera itself is called "dark current" and is removed by first taking an exposure with the shutter closed for, say, half an hour, and then subtract those counts from the real observations, normalized to the exposure time of a given image. Sometimes, if you're bored at the telescope due to bad weather, you can even take ...

8

All bodies above absolute-zero emit some radiation. This is "black-body" radiation and it can be correlated to temperature using the Stefan-Boltzmann law. Your infrared thermometer uses this to calculate the temperature by measuring this radiation. The temperature you measure for the sky is the radiation of an equivalent blackbody at -2°C. This is sort of ...

7

My previous answer was off topic because, as a comment says, it was about the far infrared, not the near, as you asked. I still leave it for those interested, at the end of this new edited answer. The problem is that water vapor has a strong absorption in the on the near infrared, so the actual light that comes from the sun has much less power than in the ...

7

It sounds like you know some of the most important summary points about blackbody radiation, but here is a reference on the subject, since I will be talking almost entirely about blackbody radiation: https://en.wikipedia.org/wiki/Black-body_radiation Given any temperature, there is a certain emission spectrum (see https://en.wikipedia.org/wiki/Planck%...

6

The reason for this can be seen by examining how an infrared thermometer works. As you mentioned, it measures the infrared radiation, and uses this to determine the temperature. So, with that in mind, we consider what happens in the situations you mention. Namely, you cannot use it to measure the air temperature, because the emissivity of the air is piss-...

6

Infrared radiation is absorbed by water, both atmospheric water vapor and liquid water. Below is a graph of water transmission at various wavelengths. Notice that some rather large bands are completely missing. This light can't reach your eyes because the air absorbs it. Also, our eyeballs are filled with water. This water also absorbs infrared radiation ...

5

As a general rule there are three mechanisms by which molecules absorb light: Electronic transitions - visible/uv wavelengths Vibrational transitions - infra-red wavelengths Rotational transitions - microwave wavelengths In solids you don't often get rotational spectra because the molecules usually aren't free to move without interacting with the lattice,...

5

The IR sensors that OP is talking about work by emitting IR from an LED and then measuring IR intensity reflected back from an object close enough (scroll to Sharp GP2Y0A21YK IR Proximity Sensor). Addressing the comments: the temperature of the robot and the resulting blackbody emission in the IR is small enough to be irrelevant to the question. IR is used ...

5

For light bulbs and other thermal emitters this is definitely true. Their emission follows the black body spectrum (if you neglect absorption due to the glass container). If you want to be picky: Any device, which is operated above 0 K (which applies to all devices) emit thermal radiation according to their temperature. This is not directly related with the ...

5

A nephew-friendly, physics-based explanation: Our brains and nerves work based on electrical impulses, which are little bursts of electrical current. Electricity is what happens when you remove the electrons from one atom or molecule and move them to another one nearby. In some materials, like metals or heavily ionized liquids like blood, it's easy to ...

4

The mechanism of EM radiation emission in dilute gases is different from solids, liquids and dense gases. In a solid the main source of the continuous emission, i.e. black body emission, is lattice vibrations causing local oscillations in electron density. The resulting transient dipoles emit EM just like any oscillating dipole. This isn't a resonant ...

4

I would guess your meter is assuming the surface is a reasonable approximation to a black body, i.e. the emissivity is approximately one. The emissivity of polished metal can be as low as 0.1. From the fluke manual: Emissivity Of the kinds of energy—reflected, transmitted and emitted—emanating from an object, only emitted infrared energy indicates the ...

4

You are right that IR lamps typically have the bulk of the energy shifted to longer wavelengths. Now it turns out that infrared radiation (IR) is more readily absorbed by water in the body - so when you use the same amount of power in an IR lamp, more of it will be "felt" as heat by the body (because less is scattered / reflected by the skin). As for ...

4

The usual way to achieve minimal radiative losses (and thus a low thermal signature) is using MLI - multilayer insulation. The use of multiple layers of reflectors creates a "thermal series" - each layer gets a temperature that is closer to the radiative temperature of the environment, and it allows you to achieve a low thermal emission. This is very ...

3

We radiate infrared rather than UV or visible light because we aren't hot enough. See http://en.wikipedia.org/wiki/Planck%27s_law for more details.

3

There are two aspects to be considered here: Does the wall reflect or absorb the radiation? If the wall reflects, is this specular or diffuse reflection? For both of these sub-questions, the answer depends on the material that the wall is made of and on the wavelength of the infrared radiation. At wavelengths close to the visible (for example, at 0.8 µm),...

3

All matter in bulk radiates (approximately) as a black body radiator, approximately because there are coefficients of emissivity depending on the constituents. For gases the functional form is different. The radiation has a specific spectrum and intensity that depends only on the temperature of the body As the temperature decreases, the peak of ...

3

The energy of electromagnetic radiation (particularly a photon) is $$E = h \nu = \dfrac{h c}{\lambda} .$$ where $h$ is Planck's constant, $\nu$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength. So you can see that as wavelength increases, the energy goes down. Different wavelengths of ...

3

Both Infra-red and Visible light are different. (It's BAD to call IR-radiation as IR-light either...) Visible light is everything that you see with your naked eyes. But, IR - NO, you can't see 'cause it's not light and that's why your eye won't perceive it. It's what most vipers, pythons and rattle snakes see..! This is the reason why snakes could detect ...

3

Edit: putting the summary of discussion in comments into my answer. Beer-Lambert law assumes that every photon has equal probability to be absorbed by every molecule. It is only valid for sufficiently monochromatic light – that is, the bandwidth of light source should be smaller than the width of the absorption line. When absorption is measured with a ...

3

Since your plasma is in a vacuum environment, the only way for it to loose energy is by radiation (conduction transfer through the magnets are neglected). You have thus to consider which bodies are surrounding your plasma and which radiative model is the best ton consider for them. I guess you can consider a black body with the simple Stefan equation. The ...

3

Incandescent bulb are black body emitters. Basically, something is heated enough so that it radiates most of the power put into it. The black body radiation spectrum is well known, with it being a function of temperature. The lower the temperature, the more the bulk of the radiation shifts to lower wavelengths. Normal incadescent bulbs used for lighting ...

3

The solar spectrum looks like the image below. The yellow shaded curve is the observed spectrum at the top of the atmosphere while the red curve is what we get at ground level. The black line is a blackbody model of the sun (a pretty good approximation, just don't worry about the yellow curve peaking above the black line) (source) 940 nm occurs right ...

3

Hot objects, such as humans and warm-blooded animals, emit IR-radiation. Few creatures have IR-vision, such as snakes. Significantly, snakes are cold-blooded. IR-vision would be advantageous in a struggle to survive for most creatures; with IR-vision, you could distinguish (hot) living creatures with camouflage from their backgrounds and see (hot) living ...

3

You can easily show that an incandescent lamp produces infra-red as follow but whether it will be enough for your purposes I do not know. Try the following experiment. Select a mains lamp probably in a table lamp which has not been turned on for some time ie it is cold. With the lamp off feel the glass enevelope. It should feel cool. Without actually ...

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