New answers tagged

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

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:) ...


16

Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...


2

That's right. There are materials that possess these properties. They are birefringent, see Wiki Page Birefringence. Crystals can have different refractive indices along their crystal axes which leads to the phenomenon that you describe.


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 ...


0

Your understanding of the degradation mechanism is correct. However, you can compensate for this to some degree by training your pupils to contract underwater, thereby reducing the refractive variation of light transmitted to your retina. This famous study by Anna Geslen describes this in detail, and also showed that the remarkable underwater visual acuity ...


2

The answer is purely a matter of engineering and cost, I think. I've bought prescription masks in the past, and they are as you say: basically a stock faceplate with a correcting lens attached to the inside face. This allows the manufacturer to stock a standard (flat) faceplate and pull units from stock to build specialized prescription items. Now, if ...


11

Yes, your myopia is relevant in the sense that you notice immediately a huge improvement in your long distance vision when wearing goggles underwater. Short-sighted people have difficulty in focusing distant objects (or nearly parallel light rays); the eye is "too long" for the lens and the focus falls in front of the retina. Corrective lenses for myopia ...


53

Is blurred effect due to turbulence? No, it is not. The turbulence has a little effect here. Even if there is no turbulence, one see everything blurred underwater. The reason is explained below. An eye is a natural lens. A clear shot of something you see depends on how well the image is focused on your eye. The most of the refraction in the eye occurs ...


0

This is a phenomenon called dispersion of light. White light is a mixture of all colours which get separated when they pass through a prism. The refractive index or simply for lay usage the angle at which light bends when it enters from one medium to another depends on the wavelength of light. So the colours are not newly created, but they are just ...


6

When light enters glass (or another transparent material), its frequency stays the same and its wavelength changes. In a comment, you say that you are using "color" to mean "wavelength". Well, I think you are using the word "color" incorrectly. According to a normal definition of "color", the color of light does not change when it enters glass. But the ...


1

The angles XCF and N'CB are marked as being the same angle, but that is clearly nonsense since that would makes line CB parallel to line FB.


0

The only way a collimated beam will focus on the rear surface of a sphere is if the diameter of the sphere divided by the refractive index is equal to the focal length. Since the focal length is equal to the radius of the sphere divided by the change in refractive index (n-1), this can only occur when: 2R = nf = n (R/(n-1)) 2 = n/(n-1). Thus n=2. ...


3

The first equation correctly states that $$ v_g = \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right). $$ But if you look at the wikipedia page that you linked to, you'll see that the second equation should read $$ v_g = \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}, $$ where $\lambda_0$ is the wavelength in vacuum, ...


0

Ok, I found an answer on my own: For the first definition the answer from L. Levrel is sufficient. For the second definition one has to follow the following set of equations: $$\begin{split} \frac{1}{v_g}&=\frac{dk}{d\omega}\\ &=c^{-1}\left(n+\omega\frac{dn}{d\omega}\right)\\ &=c^{-1}\left(n-\lambda\cdot\frac{dn}{d\lambda}\right)\\ \Rightarrow ...


2

Group velocity, for any kind of wave, is defined as $$\boxed{v_g=\frac{\mathrm dω}{\mathrm dk}}.$$ Phase velocity is defined as $v_p=\dfrac ωk$, and refraction index as $n=\dfrac c{v_p}$. So $ω=c\,\dfrac kn$. Hence, $$\mathrm dω=c\left(\frac{\mathrm dk}n-\frac{k\,\mathrm dn}{n^2}\right)=\frac{c\,\mathrm dk}{n}\left(1-\frac kn\frac{\mathrm dn}{\mathrm ...


3

Yes, this does happen. Cherenkov radiation is a well known effect. It produces that blue glow you might have seen in pictures of nuclear reactors resulting from electrons passing through water a speeds faster than light.


2

No: gravity is irrelevant at atomic scales. Snell's Law is a consequence of the Huygens principle, which in turns is explained by Maxwell's equations. In the end, it is an electromagnetic phenomenon, and as such is explained by electrodynamics. For more details, see the Snell's law article on Wikipedia.


0

It appears to me that your piece of glass is beveled on the edges. When I place a beveled and non beveled piece of glass side by side directly under a light only the beveled one has the shadow.The angled edges are reflecting and refracting the light away from the area directly below.


2

Most glass contains iron oxide as an impurity which gives the glass a slightly green hue. When you're looking straight through a pane of glass you don't notice it because the glass is so thin. But when you look down the side it becomes apparent because of the thickness.


2

Not all the light shining on the top surface of your glass block can pass straight through the block. Inside the block, total internal reflection occurs at the side faces. The light which was "supposed" to exit through the sides is refracted to the inside. This causes the shadow you observe and a slighly brighter middle section, which is hard to see. In ...


1

Some sources say that about 6-7% of light gets reflected on hitting a glass.A glass seems to be transparent, but it not. Actually we cannot create a completely transparent solid body. At molecular level: When light encounters a material, it can interact with it in several different ways. These interactions depend on the wavelength of the light and the nature ...


3

Good and important question, altough one does not 'simply substitute'. The short answer is that the vacuum values are for microscopic theory of light in vacuum, and any other values are for macroscopic theory, which handles light-matter interaction phenomenologically at much larger scales than atomic separation. Here is the very long answer: The ...


2

Refraction occurs when a large number of dipoles scatter coherently. Each individual dipole scatters light in response to the incident radiation in (almost) all directions, but when you have a large collection of scatterers, each one scattering in many directions, you have to sum the contributions of each one in order to arrive at the total field. Each ...


1

Generally speaking, the first and main difference is that refraction happen upon transmission of the light, while scattering happen upon reflection of the light (namely, diffusive reflection, where the angle of reflection does not equal to the angle of incident).


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 ...


0

[This was an answer to an earlier version of the question, and doesn't address the band-of-shadow thing at all.] I think the answer is that glass isn't that transparent, and in particular there are fairly significant losses at the surfaces (this is why camera lenses have fancy coatings which attempt to reduce the loss at the air-glass surfaces). So, while ...



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