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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 ...


12

Do keep in mind that the frequency of light is reference frame dependent. So, for example, the cosmic background microwave radiation would appear as a concentrated gamma radiation source 'in front' to an observer with ultra-relativistic speed relative to the CMB. In other words, light emitted from a body of a particular frequency in that body's frame of ...


12

The easiest solution is to use a fiber-optic camera, i.e. one with a fiber-optic connection between the front lens and the actual camera electronics. You can easily bend the fiber (but not too much!). You can now make a small hole in the microwave (smaller than the wavelength, insert the fiber. Bend the fiber and wrap it in aluminium foil. There will be ...


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 ...


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 ...


6

The electromagnetic spectrum does range between (almost) zero and (almost) infinity. It's just that your eyes are sensitive to a very small part of it (from about 380 nm to about 800 nm). At the lowest frequencies, it becomes difficult to recognize the signal from background fluctuations. From this site: "Gamma-rays are detected by observing the effects ...


5

Theoretically, the shortest wavelengths of light would be limited by the Planck length, at some point the space 'closed' by the wavelength would be so small that gravitational effects would dominate, in the same way that black holes can bend light passing near their event horizon at very small scales the wavelength would be so small that it might be at the ...


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 ...


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 ...


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

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

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 ...


3

You correctly surmise that you could make a cage with small holes, so that the visible light could pass through but the microwaves would be blocked. However, you could have some major problems if you decide to put a metal object inside a microwave oven - see the same wikipedia page you got your frequency information from. You would have to construct your ...


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

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

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

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 ...


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

It is momentum that defines the incoming direction and momentum transfer the outgoing one. The photons, quantum mechanically carry momentum equal to p=h*nu/c . Momentum is a vector and defines directions. An electromagnetic field is an emergent classical quantity built up by innumerable photons. There exists also a momentum defined for the classical field ...


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.


1

This is just a complement to the previous answers which give the correct response. If you want to think about it in an intuitive way, imagine that the interaction between electrons and photons becomes weaker. In the limit when it becomes nearly zero, the light will be almost not scattered at all and will continue in a straight path.


1

Let me offer you a slightly modified version of your question to illustrate a way of re-formulating it your thought process. How does a pool ball know from which direction the cue ball hit it? The answer is the same in the sense that "the particle" does not know all by itself, "the system"1 has certain invariant quantities (like momentum and energy) ...


1

An analogy (possibly from Griffiths' textbook): imagine you're out in desert. Many miles away you see a truck moving on the highway. In which case is it easier to judge its speed: if it's moving radially towards or away from you, so that you have only its change in apparent size to go on, or if it's moving laterally across your line of vision?


1

If the object has a temperature at absolute zero ( within the quantum uncertainties related to this statement) it means that all the atoms and molecules that compose it are at the lowest possible energy level. Supplying energy to heat an object above absolute zero means increasing the kinetic energy of the component parts and raising them to higher energy ...


1

You write the integral formulation of Faraday's law, but there is also the equivalent differential formalism: $$ \nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}\tag{1} $$ which can be proven in a straight-forward manner (and ought to be done in your standard E&M textbooks). Using standard planar waves equations, $$ E_y=E_0\sin\left(kx-\omega ...


1

The wavelength of microwaves is comparatively large, if you look at the holes on a microwave oven door.



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