New answers tagged

31

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 ...


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 ...


33

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 ...


2

They are not transparent because they are "rougher" and are reflecting/scattering more light as compared to a nicely polished surface which will transmitt more. In my research we experience the same when we're polishing nonlinear crystals. We first have a rougher polishing material to get the length/width down, but then the ends are rougher and not ...


5

We can already see through objects like glass. Colored glass transmits certain wavelengths and not others. If our eyes could see more wavelengths than presently this would be no different . For example pure silicon transmits far infrared. If we could see far infrared it would look like a sort of glass with what would be infrared color. Metals can be ...


0

Your observations are mostly correct. "a defect can be introduced to create a defect/resonance mode and enable transmission" - that is, when the photonic crystal is of finite thickness. In an infinite PhC, the defect only introduces localised states centered at that defect. "if it is a half-wavelength layer it will enable a resonance ...


0

An intuitively simple picture would be to consider terms in this infinit series to represent one-photon two-photon... N-photon processes. Altough this can give quantitatively wrong result, it can be considered to be equivalent with the classical series expansion.


3

Imagine an electron in a parabolic effective potential $$V(\vec{\bar{x}}) = \tfrac{1}{2}k |\vec{\bar{x}}|^2 \\ \vec{F}(\vec{\bar{x}}) = \vec{\nabla} V(\vec{\bar{x}}) = k \vec{\bar{x}}\text{.}$$ A force $-e \vec{E}$ will displace it from the equilibrium position to $$\vec{\bar{x}} = \frac{-e \vec{E}}{k} \text{,}$$ hence the polarization is $$\vec{P} = -e \vec{...


0

The equation is just a Taylor expansion of an arbitrary function: by plugging in different values for the coefficients, you can construct practically any function. The real question should be "Why are the coefficients for the higher terms typically very small?" The answer is that Nature seems to be well-behaved. If we examine most processes over ...


3

The circularity is not redundant, because retarding the relative phase between components along two spacial axes is not the most general thing that can happen in these materials. A more general situation is that the eigenvectors of the Jones matrix don't correspond to linearly polarized light (which is the case for linear retarders), but to some generic ...


0

All surfaces reflect light except for light sources. However, not all surfaces have specular reflection like a mirror as they reflect light well. Now, let’s take a closer look at your question. 1.polished metal 2.mirror 3.undisturbed water 4.book All of the give reflect light. However, options 1, 2 and 3 have specular reflection leaving 4 as the only ...


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