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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 ...


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

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.


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 ...


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 ...


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 ...


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

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

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 ...


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

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 ...


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

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

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 ...


2

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 ...


2

Not only the human body. Everything emits radiation. But the wavelength of this radiation depends on temperature. The wavelength actually comes from the frequency of atom vibrations. If atom is hot it moves faster and generates higher frequencies of electromagnetic waves (shorter wavelength). Imagine shaking your hand in water where you would see that ...


2

This is the plot of sunlight, red at ground level. Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power. As all light comes from the sun during daylight this should suffice. One can get the number of photons by ...


2

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 ...


2

Blue light may have more energy per photon than red light, but the rate that energy is being deposited on a thermometer depends on the power spectrum of the particular light beam. For example, a big bright red lightbulb can transfer a lot more energy per second than a tiny blue one, regardless of the wavelengths involved. In Herschel's case, he was using ...


2

The answer is simple: your camera CCD is sensitive to infrared light, while your eye is not. The real question here is: why does your camera render the infrared light as visible light on the screen? The only answer I can hazard on that is that it allows for more visible photographs when they're taken in low (visible) light.


2

Infrared lasers are much more dangerous to the human eye compared to a visible laser of the same power, because infrared lasers do not trigger a blink reflex, which means the laser has much more time to damage your retina. Your other questions can be answered by reading about the many differing ways that visible and infrared light interact with matter via ...


2

The energy of electromagnetic radiation (particularly a photon) is \begin{equation} E = h \nu = \dfrac{h c}{\lambda} . \end{equation} 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 ...


2

A human body feels heat while in contact with the air, so it is desirable to heat the air. This is what convectional heaters do - they have a developed surface and relatively low temperature to transmit the heat to the air. The radiation power is proportional to the following temperature difference: $Q \propto (T_{heater} ^4 - T_{air} ^4)$, so it can be ...


2

The description infra-red covers a wide range of wavelengths from wavelengths of about 700nm out to tens of microns. Ordinary glass will transmit radiation out to at least 1 micron - the exact cutoff depends on the type of glass. In the prism experiment the geometry means that that the infra-red light detected can't have a wavelength more than (very ...


2

It's all to do with the conductivity of the material, the thickness is only a secondary effect because as you reduce the thickness you reduce the conductivity. The only time that thickness has a direct effect is if you make a surface from layers of a transparent material which are a precise fraction of a wavelength in order to use interference effects to ...


2

In table 1 the classification of photon frequencies is given. A lamp as you describe will emit most of the energy in the infrared and some in the visible. The energy carried by the visible part will also end up in the infrared degraded by the reflections in the room. So you will get 100watts in total. The advantage of emitting mostly infrared lies in ...



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