# Tag Info

1

The concept of "image" of a point means: the reflected/refracted light rays from it seem to come from the image point. So if you follow them backwards where they came from and prolongate them beyond the spot where they were reflected/refracted, they meet at the image point. Note: tt is not obviuos, that such a point must exist, the rays don't have to meet, ...

1

It depends on from which material you are viewing. If you are viewing with your eye placed inside the water then you have to consider the "object" distance because now, only the phenomena reflection takes place. However if your eye is placed above the water, then "optical" distance is taken into consideration since both reflection(from the mirror) as well as ...

1

Maesumi has already provided references, I would just like to point out how you can solve the problem yourself: the law of reflection most certainly holds in the rest-frame of the mirror. So to find out what happens when it is moving, transform to that frame, apply the law of reflection, transform back. As correctly pointed out in the comments you will find ...

0

This problem is considered in the following papers 1) A. Einstein, Zur Elektrodynamik bewegter Korper {Ann. Phys. (Leipzig)} {17} (1905) 891–-921. http://onlinelibrary.wiley.com/doi/10.1002/andp.19053221004/pdf https://www.fourmilab.ch/etexts/einstein/specrel/www/ Reprinted in Einsteins Miraculous Year: Five Papers That Changed the Face of Physics, ...

1

Just to add to the above answers, and since to did not limit your question to the visible range - if you define black as absence of light (photons emitted or reflected), then there is no such substance, because according to black body radiation model, everything with a temperature above absolute zero (which is essentially truly everything in the universe:) ...

1

Here's a simple explanation based on the dipolar nature of the medium: For most of the materials, we can assume that the source of the reflected and refracted waves are the induced tiny dipoles in the dielectric medium. In an isotropic medium, polarization vector is proportional to the (total) electric field vector with a constant (as opposed to a tensor ...

1

The definition of s-polarised light is that the electric field is polarised so that it is perpendicular to the plane of incidence. Where there is a specular reflection, the plane of incidence contains the k-vector of the incoming wave and the reflected wave. Since the electric field of an EM wave must be perpendicular to the k-vector. This then leaves the ...

1

There is a simple answer: Symmetry. Suppose the material is isotropic, and consider the initial condition of the p-polarized case, with a p-polarized light wave about to hit the surface. In this case, reflecting in the plane that contains the incident and scattered wave vectors leaves both the (vector) electric field and (pseudovector) magnetic field ...

0

Radio waves don't penetrate very deeply into metals. The depth to which they penetrate is characterized by a parameter called skin depth. The penetration is practically zero beyond about five times the skin depth, so adding additional thickness beyond five skin depths will make no difference. The skin depth for most metals at 5 GHz is about 1 $\mu$m. So ...

3

For a metal, a measure of how far the wave penetrates is determined by its skin depth, $\delta$. Typically $\delta \approx 10 mm$ for metals at around 50 Hz, but becomes much smaller at higher frequencies, at microwave frequencies $\delta \approx 1-10 \mu m$. This means that there's no point in having very thick metals.

1

The Fresnel equations are derived by matching the electric and magnetic fields of the incident, reflected and transmitted waves at the interface. In this process only the instantaneous value of the fields is used not their rate of change with time. This means the frequency of the wave simply doesn't enter the calculation. However, as a comment notes, the ...

3

It's maximizing the angular entropy in the sense that far more "available" angle vectors are inhabited. I suspect you need to be careful with the word "entropy" here. For example, the photons are not down-converted into a larger number of photons of longer wavelength (aka 'heat death of the universe').

3

The answer to this question depends on various aspects. For instance when you say white object, do you mean perfectly white? or white with respect to only visible light? Same happens for a mirror too. The most direct way to look into the problem would be following. Mirrors are nothing but extremely fine and optically flat white surfaces at the back of ...

0

The certain film on the glass reflects blue light of wavelength 480nm, but does not reflect any red light of wavelength 640nm. ... If we say that it reflects blue light but no red light, does it mean that we actually see blue light only? It depends which side of the glass you are observing from. If you're on the same side as the source, you'll see the ...

1

tl;dr: It's because of discontinuous change in impedance $\rm Z.$ Reflection Coefficient $\rm R$: Let the point where two media meet be $z=0\;.$ At $z= 0\,,$ the incident wave is given as $$\psi_\textrm{inc}(0,t)= A\cos\omega t\;.$$ The total force exerted by the input terminal of the second medium is given by \begin{align}F&= ...

1

Reflection, $R$ and transmission, $T$, probabilities are determined in terms of the current flux. For the case you describe, yes the reflection coefficient will be unity and the transmission coefficient will be zero. However there will still be a non-zero probability of finding the particle in the classically forbidden region, however this decays ...

2

QED uses Feynman diagrams and they are written in the center of mass. In the case of a photon hitting a mirror, it is hitting the field of electrons in the outer band of the mirroring substance. For reflection there should be elastic scattering in the center of mass so that the phases of the photons are kept and the image emerges intact on reflection. So it ...

0

If you find my above answer a little bit confusing I will simplify it: At the fundamental level, it's about what light is and how it reacts with matter. We know that light is just electromagnetic waves - just like water or sound (which is air waves!), except instead of water or air transporting the waves, it's the electromagnetic field; you can try to ...

-1

See, a good reflector of light is a very good conductor of electricity as light is an electromagnetic wave. That's one essential criterion for light reflection. Reflection of light (and other forms of electromagnetic radiation) occurs when the waves encounter a surface or other boundary that does not absorb the energy of the radiation and bounces the waves ...

1

A clean metallic surface appears shiny because it is reflecting light. The interaction is due to the conductivity of the surface of the metal. When surface conditions change, the reflectivity is reduced. For example, a clean aluminum surface is very reflective, and aluminum coatings are used on the back side of mirrors; the glass protects the clean aluminum ...

17

The main reason why we don't hear reflections of sound is related to how our hearing works. The psychoacoustic explanation to this is called the precedence effect. It states that when two or more sounds arrive to the listener within a short enough time(roughly under 50ms) this is perceived as a single sound event. The localization of the sound is dominated ...

2

This is purely a matter of human perception. Our eyes contain literally millions of separate photoreceptors, which enables us to build up a clear picture of our surroundings. On the other hand we have only two ears! Admittedly each ear is able to differentiate a wide range of frequencies, and in that way has a superiority over the eye, which can only ...

2

Why don't we hear sound reflecting that much? Mainly, it has to do with energy and wavelength. When light hits an object with rough surfaces such as a building, it is scattered in every direction. Because the light illuminating the building has such a high energy, your eye is able to pick up on just a tiny fraction of that scattered light, in spite of ...

4

Some good answers here but one area that has not been covered is a sort of threshold effect. An ear senses sound pressure, not "rays" of sound. So it responds mainly to the strongest signals, and the brain tends to tune out noise. We attend to one signal in sound, generally speaking. But for vision, all that "noise" of light bouncing around off of objects IS ...

2

It is because of our reaction time ,speed of sound and disturbance in atmosphere. As our reaction time is 1/10th of a second and speed of sound is about 343.2 metres per second. So sound travels 34.32 metres in 1/10th of a second. To distinguish between 2 sounds the distance between them must be 34.32 metres and if we want to notice reflection of sound(echo) ...

8

Some blind people actually use sonar-like techniques to "see". This is in the press from time to time, e.g. http://www.sciencemag.org/news/2014/11/how-blind-people-use-batlike-sonar.

24

As someone who's done substantial amounts of live sound for bands, often in rooms in pubs which are acoustically "interesting", it certainly does happen with sound, and you and everyone else hear it all the time. My only possible conclusion is that you haven't listened carefully enough to what you hear. The reason people like singing in the bathroom is ...

11

Whispering rooms/galleries are another good example. In the simplest case, an elliptical room, sound echoing off the walls allows one person standing at one focus of the ellipse to clearly hear everything at the other focus. Step away from the focus, though, and the effect disappears.

14

As the above answers have stated, we do hear such reflected sound but normally do not notice. However, if you ever get the opportunity stand inside a closed anechoic chamber. You will then "hear" the total absence of all reflected sound. To say it is weird is an understatement - it feels like your ears are being sucked out by silence.

47

Our eyes have excellent spatial resolution. We can tell the difference between objects only a fraction of a degree apart. This is possible due to both the construction of the eye and the fact that visible light has wavelengths that are tiny on our scale. Signals that arrive simultaneously can be independently detected. Our ears do not have this level of ...

169

We do. Normally the reflections are too quick to hear distinctly, and in an environment like a room they rapidly become diffused into a mush which a sound engineer would call reverberation. In larger spaces you can often hear distinct echoes as well or instead: a good way to play with this is to clap your hands (once) in a quiet hall: you will hear the ...

14

When an atom or molecule absorbs a photon, it enters an excited state; each excited state has a mean lifetime. When the atom or molecule returns to the ground state it may emit a phonon (vibrations), or it may decay through multiple levels; in this case there are multiple photons, with different wavelengths. In the case where the absorbed and emitted ...

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