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Your curve has two components in it: Actual solar intensity per unit area (affected by atmospheric conditions, that is the probability that the sunlight is not scattered or absorbed on its path through the atmosphere) The projection of the sunlight onto your collector: if your collector is facing in a particular direction, the angle of the normal of the ...


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"Bright light can never hurt your eyes" seems false to me… enough energy focused on the retina will cause damage, regardless of the wavelength. Otherwise you would not need to wear laser goggles… That aside, materials typically have certain ranges where they absorb light more strongly than others. There is no hard and fast rule for this, but if you google ...


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Uv goes thru glass, I thought that comment strange when I watched it. Laminated glass (which it could have been) would shield about 95% of the uv, I believe due to the resin interlayer.


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You'll need two things, the solar zenith angle $\theta$ and the air mass $AM$. The zenith angle is given by $$\cos\theta = \cos\phi \cos \delta + \sin\phi \sin\delta \sin h$$ where $\phi$ is your latitude, $\delta$ is the declination of the Sun, and $h$ is the hour angle. You should know your own latitude. The declination of the Sun can be found on a ...


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If the light passes first through medium $\tau_1$, then through medium $\tau_2$, the optical depths add up to $\tau = \tau_1 + \tau_2$ , as you can see from applying the Beer-Lambert-Formula twice. Does it matter whether the media overlap (if the media don't interact and light propagation is linear)? No, it doesn't. The optical depth comprises the ...


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Avoiding mathematical formulae to the maximum, and warning for furious hand-waving ahead, I would state it like this: In the classical picture, there is no quanta concept, so you could have just a little bit of radiation energy at any frequency. However, since quantization appeared, the minimum amount of radiation energy that you could possibly have at a ...


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I think you misunderstand the ultraviolet catastrophe. It does not mean that the energy radiated reaches zero at any finite frequency, just that the power tends to zero as the frequency tends to infinity. Pre-quantum physics thought that blackbody radiation was ruled by the Rayleigh-Jeans law $$ B_\nu(T) = \frac{2\nu^2 k T}{c^2} $$ This does obviously not ...


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All photons are excitations of the photon quantum field. The photon is actually not a great description of propagating light - it tends to be far more useful when you're considering light exchanging energy with something else. If you insist on a photon description then you'd have to treat the radiation as a superposition of photons. A beam of light would be ...


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Some additional information about shielding with high Z materials: According to Table 3 in http://iopscience.iop.org/0022-3700/8/12/014/pdf/jbv8i12p2015.pdf, there is a strong relationship between the atomic number of a material, and the photoelectric cross section (the probability of a PE interaction). Plotting the numbers for incident gammas of 25 keV, ...


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Let's suppose the broadening mechanism is van der Waals or Stark broadening - something where the energy levels of individual atoms are perturbed. In this case you could use the following argument. Divide the line profile up into groups of atoms which share the same perturbation and treat each of these as a subpopulation with a different energy gap and ...


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Could anyone explain how this process complies with the second law of thermodynamics? The Stefan-Boltzmann law. I'll start with an ideal black body. Black bodies absorb all incoming radiation. They also emit radiation as a function of temperature. The peak frequency and the intensity increase as temperature increases. The emitted power is given by the ...


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Heat, in the context of something "giving off heat" that we use in everyday conversation, is a term we use often to describe emission of a specific part of the electromagnetic spectrum (namely the infrared spectrum). As you start to pour more and more energy into an object, the electrons can get more and more excited (which is the process of absorbing ...


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Actually both "types" can be approached in the same classical framework. The "classical" approach to radiation from an atom has an incoming wave exciting an oscillation. This oscillation can be treated as an accelerated dipole and emits radiation accordingly. This serves adequately to model Rayleigh scattering and stimulated emission, but has major ...


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Remember that plasma is a phenomenon of both high temperature and low pressure. Plasmas are ionized fluids, which occur either when the temperature is so high compared to atomic ionization energies, or when the fluid is so diffuse that charged particles are unlikely ever to find each other. A common tool in ordinary vacuum systems is the cold cathode ...


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This from the ITER page In the heart of the Korean tokamak KSTAR, in operation since 2008, a plasma pulse burns brightly. But don't be fooled—the brightest areas of the photo are in fact the coolest. At 150 million °C (the temperature in the centre), the plasma doesn't emit in the spectrum of visible light. © National Fusion Research Institute, Korea ...


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I guess the plasma in the ionosphere is typically invisible: the plasma frequency is much lower than the frequency of visible light, so almost all light passes through the ionosphere.


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Hydrogen plasma is probably invisible because all atomic lines of hydrogen are in the ultraviolet region. Somewhat related - hydrogen flame burning in air is completely invisible.


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Dark lighting is (evidently), the result of Bremsstrahlung radiaiton from fast moving free electrons in the air. It's not the usual re-attachment emission that I'm used to seeing in lab plasmas. But briefly, yes you could categorize this as a cold plasma (it's got a degree of net-neutral ionization). There are also plasmas that might just not be visible to ...


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You seem to have hit upon the challenge. Converting between the two is situation dependent. It depends on the source of the radiation and the amount of protection. One source might deliver energy as 50% alpha and 50% beta another might be 100% gamma. Put a sheet of paper in the way and the first source will shift and deliver a higher percentage of energy ...


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The Stefan-Boltzmann law for net power radiated pertains to the object. That is, we're simply asking, how much radiation leaves this object (this depends on the object's emissivity), and how much radiation is absorbed by this object (this depends on the objects absorptivity). The emissivity and absorptivity in the equation you present thus pertain to the ...


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The two existing answers have noted that EM radiation (X-rays, gamma) is effectively stopped by electrons. There are at least 4 other types common types of radiation: Alpha particles (2 protons, 2 neutrons - essentially He4 2+) Beta particles (single electron) Neutrons Ions (other than alpha particles) The first thee are commonly generated by nuclear ...



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