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28

The thing that makes a mirror a mirror is a that it has a high reflectivity (and is very smooth of course, but that doesn't enter into this issue), but all optical properties including reflectivity are functions of wavelength. The mirror is not reflective in the x-ray band, so it looks like a layer of glass (moderately dense) and a very thin layer of heavy ...


22

dmckee points out that an ordinary mirror doesn't reflect X-rays, but if you could find an X-ray mirror and put it in your case it would just appear black. When you look at yourself in the mirror you're seeing light from your skin/clothes that hits the mirror and is reflected back towards your eyes. But airport X-ray machines work by passing X-rays through ...


17

As others have noted, an ordinary mirror will not reflect x-rays. X-ray mirrors do exist, but will also probably not do what you want here. It's very difficult to manipulate the optical trajectory of an x-ray. The critical angle of typical metal foils at x-ray wavelengths is a few degrees at most -- that means that the x-ray is only reflected if it hits the ...


11

Photons do not exhibit the property of virtual particles, but it is not your reasoning that is faulty, you have simply fallen prey to an imprecise use of terminology. Let me start with my view of the wave/particle duality. Most of the images of "particles" and "waves" comes from a time when we really didn't understand the quantum world, and some ...


11

Yes, X-ray, UV, and even radio-waves ares made of photons. The differences is the Energy (or equivalently the wavelength). See the picture of the Electromagnetic spectra . The different nomination comes from the time of the discovery. Youre eyes can see the visible part. the radiowaves can be observes with antenna etc... The only differences is the way we ...


10

Stop and actually think about it. Small-value inductors and capacitors are possible, but you also need extremely small size else the lumped system approximation your equation relies on becomes invalid. Let's say the wavelength of the light to produce is 700 nm. A distance of half that (350 nm) will cause a inversion (180° phase change). Generally you ...


8

Imagine the field lines of a point charge - they all point outwards of the charge in a radial direction. Now consider the following statement: the change of the field does not propagate instantaneously, but it has to propagate through local interaction. When we nudge into the charge, a ripple in the field propagates to tell the other field lines "hey guys, ...


7

For the first question, yes. Because the surface of the sun is close to a blackbody emitter, it radiates at all wavelengths below the peak. So radio waves are included. However, the longer the wavelength, the less the power that is put into that portion of the spectrum. Radio is so far from visible light on the EM spectrum that the solar radiation in ...


7

I'll assume that you were using a radio tuned to a 1Mhz frequency ($\omega = 6.3\times 10^6$ s$^{-1}$) and that the radio was completely enclosed inside $t=3\,$mm of pure iron. There are two important effects to consider. (i) How much power is reflected from the iron surface. (ii) How much of the transmitted power makes it through the iron. To figure this ...


6

If, by LC oscillator, you mean a circuit composed of an 'ordinary' inductor and capacitor etc. then the answer is no. The equations for the LC oscillator are derived within the context of ideal circuit theory which is the limit of a number of assumptions. To apply the results from ideal circuit theory to physical systems, the physical systems must ...


6

Short answer: A virtual particle is not the opposite of a classical particle. While the other answer captures some aspects correctly, there are still a few flaws and inaccuracies which in the following, I will try to set straight. Wave-particle duality Strictly speaking, quantum objects are neither waves or particles. They are entities behaving like ...


4

Radio waves will not travel through the Earth. It's just too dense for that. Think of it this way: when you take an x-ray, you can see your skeleton in detail in the photograph. That's because when the beam of x-rays go through you, most flesh allows it to pass through with minimal refraction, but it doesn't go through your bones. Radio waves have less ...


4

We can make it absorb a lot of energy but if you read about black-body radiation effect you will notice that as energy is introduced into the object it will similarly radiate a small amount of the energy back, at room temperature appears black, as most of the energy it radiates is infra-red and cannot be perceived by the human eye. At higher temperatures, ...


4

White light consists of many wavelengths. Hence, the interference pattern using white light appears different than that for monochromatic light. At the center point, all the waves travel the same length and hence no path difference is produced at the center point. Thus at the center point we get the maxima of all wavelengths and we obtain the maximum for ...


4

There are many ways to detect X-rays. I will list just a few that are (or have been) used in medical imaging. Essentially there are several strategies, but it always involves stopping the radiation and using the energy released to effect a change - chemical or electrical - that can be detected. Photographic film. Exposure to Xrays has a similar effect to ...


3

To calibrate our expectations, consider the largest nuclear weapon ever detonated, the Tsar Bomba. It's yield was at most about $58$ megatons TNT equivalent, or about $2.43\times10^{24}$ erg. Now, let's consider a smallish star, something like Gliese 581, which is reasonably nearby, small and faint, and has a planetary system (of some sort: the number of ...


3

It has nothing to do with the modulation, AM or FM; it is because the wavelength difference. The so-called AM band is between 540kHz and 1600kHz, so its wavelength is about 300m, or so. The FM radio operates in the 88MHz to 108MHz band, or around 3m. The longer 300m EM wave reflects from the gap between the metal in the bridge and the ground (the latter is ...


3

I guess you have this image in mind: This hods just for a single photon, or elementary packet of light. An unpolarized beam of light contains a bunch of photons with many different polarizations. The total electric field will randomly jump all around, still the interactions with matter typically involve a single photon at a time. This means that if you ...


2

They will look all over the place. If you take a particular photon, you will see it in a particular polarisation state, but it will have nothing to do with the photon next to it, or the one before. Actually, getting a perfectly unpolarised source of light is quite difficult. One of the best cheap options are the sodium discharge lamps, used extensively in ...


2

According to Einstein's theory of mass-energy equivalence, if the photon is a particle of pure energy, and if $E=mc^2$, then the photon is theoretically traveling at $c^2$; not $c$ You have neglected dimensional analysis: E $\to$ Joule = $\frac{\rm kg\,m^2}{\rm s^2}$ m $\to$ kilogram = $\rm kg$ which means that $c^2$ has units of m$^2$/s$^2$, not ...


2

This would ultimately be more a problem of signal processing than physics. The situation is detecting a signal at a very low signal to noise ratio. At the broadband level, the noise (starlight) is several orders of magnitude more intense than the signal (the explosion). The only hope would be some sort of spectral technique , taking advantage of spectral ...


2

When you give the dose of an exposure in Sv, you have already taken into account weighting factors related to the organ type being exposed (some are more sensitive than others) as well as the radiation type. In other words, the number reported represents "equivalent cancer risk if your entire body had been exposed with ..." See for example ...


2

Light (all the electromagnetic radiations) is something like raindrops-each little lump of light is called photon-and if the light is all one color, all the "rain-drops" are the same size.$_1$ The size makes photons of visible light and other waves different. Credits: $_1$ Richard Feynman-QED, The strange Theory of light and Matter.


2

it appears that electromagnetism has some of preponderant role in the universe compared to other theories This is not true. It played an important historic role, but is in no way theoretically "unique" because the photon travels at speed c. Indeed, the gluon also travels at speed c. If photons were found to be slightly massive, it would change a lot of ...


2

A quick alternate perspective: Not really. Relativity only requires the existence of an invariant speed $c$, it doesn't require that anything actually travels at that speed. So if photons were massive, there would be no problem, although some results in cosmology might have to be modified a bit. Pretty much every time you see it, $c$ means the invariant ...


1

This is how an experimentalist sees this part of the question ( the wave/particle duality has been addressed). So, if photons invariably appear to exhibit the characteristics of virtual particles, As described in Frederic's answer, all internal lines in Feynman diagrams are called by a virtual particle's name. In the plot in his answer it is called a ...


1

I only know one particular reason: The transition responsible for the H1 line is highly forbidden and shows an extreme lifetime (10^7 years), so the absorption rate in interstellar clouds, which can be very opaque for every other radiation, is very small. Looking at the H1 line allows you to see objects which are for example hidden behind dust clouds that ...


1

(I am refering only to the updated question) This is a question which has been asked numerous times by physicists after the formulation and popularization of Maxwell's equations. They have previously known the wave equations on strings where the matter of the string was vibrating, or on water, where the water surface was the one carrying the wave. How ...


1

This is a question that was drastically changed and the other answer and the comments to the question and the answer are discordant with the edited question. I already advised in the comment that you read a simplified article in wikipedia on electromagnetic radiation. Classical electromagnetic radiation cannot be simplified easily by analogies. I will try ...


1

Almost nothing. Hawking was probably just building up to the fact that quarks are "colored" under the strong interaction force (a whimsical name, nothing more), but not actually colored in terms of visible electromagnetic radiation. Collections of atoms/molecules that really are about the same size as visible light wavelengths tend to scatter all frequencies ...



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