# Tag Info

88

This question is a very long-standing one, and is sometimes known as the Abraham-Minkowski controversy. Both Abraham and Minkowski derived expressions for the energy-momentum tensor of electromagnetic waves in matter. Each author’s tensor is based on sound theoretical arguments. Unfortunately, they disagree. Abraham’s tensor shows that the momentum decreases,...

74

In general, red and blue light do not travel at the same speed in a non-vacuum medium, so they have different refractive indices and are refracted by different amounts. This phenomena is known as dispersion.

60

As FGSUZ said, an object doesn't have to face a reflective surface to be seen as an reflection in it. I made the following picture to illustrate it for a 2D illustration: Two dimensions are sufficient to illustrate it, and it seems to be clearer that way. You can see now that an object doesn't have to face a reflective surface for its image to be reflected ...

41

As others have said, Fermat's principle says that the path which light follows is stationary rather than a minimum of optical path length (though in fact it typically is a bona fide local minimum). The more important point, however, is that this is a necessary but not sufficient condition for a given path to be that followed by light. This is a mathy way ...

35

Have you ever noticed that at the aquarium, sometimes the thick glass tanks carrying fish appear green? Glass is made of silica and lime, but sometimes it carries impurities, iron oxide, that gives glass a greenish appearance. What’s happening in your example, is that when you look at thin glass like your glasses lenses (before they broke) they were clear ...

33

Because you're looking through more of glass I'd like to just add to the other answers with some diagrams. We have an intuition that light beams travel in straight lines, so we tend to assume that the beam paths looking through glass might be as follows: However, the actual paths of the beam due to refraction and total internal reflection look more like ...

21

The light is bouncing off the top of the picture and hitting a point on the table that then goes to your eye. The image you see in the reflection appears to be under the picture frame but it really isn’t. I used a pencil to point to where the top corner of the picture frame is being reflected from.

19

Look at it (pun intended) this way: your diverging lenses are making the central area darker because the irradiance, i.e. energy per unit area, is reduced. But that energy doesn't disappear: it has to go somewhere! In this case, at least some of it is observed to arrive in that "border" region which is thus brighter than the rest of the scene. ...

14

As the Wikipedia article on Fermat's principle states in its introduction, this principle, like the principle of "least" action, is properly stated not as light taking the path with the least time taken but one with stationary time with respect to variations of the path.

14

The refractive index is a function of wavelength. It has different values for different wavelengths. The way to show this in the mathematical notation is to write $$n(\lambda)$$ just as you would write $f(x)$ for some function of $x$. So with this more complete notation Snell's law is written $$n_1(\lambda_1) \sin (\theta_1) = n_2(\lambda_2) \sin (\... 13 The confusion comes because you think that something can be seen if it is facing the reflective surface. But that's not true. Check iamge 1. You think (if I understood right) that the rays from the upper part are "passing hrough the photo" to impact the table... But that is not what is happening. Check that, if you can see the reflection, that's ... 11 TL;DR: This is basically because Fermat's principle is strictly speaking a principle of stationary time rather than least time. Note that stationary paths between 2 points need not be unique nor exists. Light in principle travels along all stationary paths. In more details: Assume for technical reasons that the refractive index n({\bf r}) is a smooth ... 10 Fermat's principle does not say shortest time. It says that the time taken along the ray is stationary under small perturbations of the path. Reflections off a concave mirror, for example, produce a local saddle point where the travel time time increases with some changes, decreases with others, but always only at second-order in the magnitude of the change ... 8 You're right, it isn't Total Internal Reflection that is responsible for a rainbow, just ordinary reflection at the interface. A tiny amount of light is reflected from the inner wall of the droplet (while the majority of the light is transmitted). When the incident rays strike the inside of the drops at certain special angles of "minimum deviation",... 6 I really feel like I genuinely understand your question. In essence you are asking "do we count it as refraction when there is no refraction in virtue of the angle of incidence being 90 degrees (or zero degrees, however you look at it). The example you give is an example of something that is vacuously true. It's a bit like a child claiming that he ate all ... 6 From Snell's law, we have n_1\sin(\theta_i) = n_2 \sin(\theta_r), where n_1 is the index of refraction on the incident side, \theta_i is the incident angle, n_2 is the index of refraction on the refracted side and \theta_r is the angle of refraction. Here we are assuming that the incident side is the denser medium side, and the refracted side is ... 6 Yes, n=1/ \sin C is derived from Snell's Law. We can't write n=\sin C/ \sin90^\circ because n>1 is refractive index of denser medium (in which light ray is incident) w.r.t. rarer medium (in which light ray refracted). Mathematically also, C=\sin^{-1}(n) becomes undefined therefore n\ne \sin C Derivation: Let \angle i be the angle of ... 6 Additional to @Dale's answer, probably I have a picture which demonstrates that the momentum change should create a recoil pressure on the matter. Imagine the prism refraction as in the picture: Here, the exiting light is turned compared the entering, so its momentum changes as it passes the prism. So, the prism should experience an oppositely directed ... 6 These can be the reasons: First of all, when you broke your specs, the edges became uneven so they(light rays) can't penetrate the glass properly. Due to which the light rays got reflected away by varying degrees and therefore it appears opaque. Note: But I don't think this is the reason why it appears opaque. Secondly, the green colour of your glasses can ... 5 Index of refraction actually does depend on the wavelength of the light in the medium. Typically this detail is left out of introductory physics classes, and the values of index of refraction given to students only applies to yellow light (at least this is what happened in my experience). Of course, one obvious example of index of refraction being wavelength ... 4 You assume that a laser beam is a single, fine line—as if it would project one tiny, millimeter-wide red dot. But all practical lasers produce divergent beams. You can arbitrarily narrow the beam by shining it out through a telescope, but I'm pretty sure that the narrowest beam we can practically generate still will illuminate many square miles of the lunar ... 4 Yes, silver reflects because of the difference of its refractive index from that of air. However, silver is different from air or vacuum in that it is a lossy medium: it is conductive, so electromagnetic waves propagating in it induce currents, thereby losing energy and decaying in amplitude. This is conveniently modeled by a complex refractive index:$$\...

4

Light can be considered a particle, known as a photon, but the energy of each photon is not related to its velocity. Instead, it is given by the Planck formula, $$E=h\nu,$$ where $h$ is Planck's constant and $\nu$ is the frequency of the light. Since the frequency of the light remains constant as it travels from one medium to another, the energy of those ...

4

Congrats! (joke) You have discovered what's known in Astronomy & adaptive optics circles as "The shower-curtain effect." What you're observing is that the paper is a diffuse transmitter, meaning the light from the object beneath (your textbook) is transmitted but with some angle-scattering. Now, consider this as part of an optical system. ...

4

If the law is reformulated into $$v_2 \sin \theta_1 = v_1 \sin \theta_2$$ the division by $0$ is avoided and you get $\theta_2 = 0$ for all possible $v_2$.

3

I believe this is due to Rayleigh scattering. The light from the moon passes through a thicker layer of atmosphere when the moon is low in the sky than when it is high. The light gets scattered along the way, with shorter wavelengths being scattered the most. That means bluish colors are scattered away more than red, so you see the moon as appearing reddish. ...

3

Let's do some ray tracing. Here's an example of rays passing through a filled prism (blue lines denote its surface; the darker rays are reflections): And here's the case when the prism is made of two glass slabs each with parallel surfaces: See the animation below where the drawings alternate to compare them. Notice how the initially refracted rays are ...

3

One has to start with Magnetization, as it is a striking result of magnetism, and see how it is defined : In classical electromagnetism, magnetization or magnetic polarization is the vector field that expresses the density of permanent or induced magnetic dipole moments in a magnetic material. The origin of the magnetic moments responsible for ...

3

Yes, it can, the limit is 299792458 m/s. For example in water the speed of light is reduced by the refraction index, and therefore a particle exceeding (c/n) does not violate the SR. Read more here: https://en.m.wikipedia.org/wiki/Cherenkov_radiation

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