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24

Water is not transparent for deepUV and infrared. From the evolutionary point of view our eye developed to see electromagnetic radiation present at earth in the past (and now) - deep UV and infrared are absorbed by water vapor and other gasses in atmosphere - so there were nothing to see at these wavelengths. Here is a nice explanation on why some things ...


20

You are right. Rainbows can occur all over the sky. However the traditional one and two internal reflections of the primary and secondary bows send light back towards the sun and hence their bows appear opposite the sun and centered on the antisolar point. The reflection of the main light makes these bows stand out. And only the light that enters a droplet ...


10

To answer this question we also need to know why some things are not transparent and why certain things, water for example, don't behave in this way. A substance's interaction with light is all about interactions between photons and atomic/molecular electrons. Sometimes a photon is absorbed, the absorber lingers a fantasctically short while in an excited ...


7

Surface coating of an integrating sphere is optimized for low losses. This white coating (barium sulfate or PTFE) acts like an ideal lambertian scatterer. all light is scattered (Ok, not 100%, but a very high percentage like 99,5%. See ressources) it is emitted in the hemisphere following the cosine law: perpendicular to the surface it's highest. ...


5

I want to know exactly the physical model that what's happening in the crystal which create second harmonic One physically intuitive model for thinking about light-matter interactions is in terms of an energy level picture. In this picture, light propagating through a material can be thought of as series of absorptions and emissions. In one of these ...


4

The simplest picture is that light always travels at the speed of light. But in a material it travels at the speed of light until it hits an atom. It is then absorbed and re-emitted in the same direction, which takes a small amount of time. The more this happens, the slower the effective average speed. The denser the material, the more atoms there are in the ...


4

"White" light is light of multiple frequencies, usually a broad band of different frequencies ranging from ~($390nm$ to $700nm$) though white on your computer monitor consists of only three different frequencies (red green and blue). It is possible to send multiple frequencies down an optical fiber. However the fiber will act like a prism and diffract the ...


3

First, be careful about mixing rays and modes. Your picture shows a ray-optics view of multimode fiber, but your text talks about modes. Some of your questions are best answered in terms of rays and some are best answered in terms of modes. Generally, fiber is characterized by a numerical aperture, which you'll find listed on its datasheet. The numerical ...


3

In general light that does not pass a barrier, a wall for example, is absorbed.The energy is turned mainly into heat and also chemical bond breaking etc. The part of the light beam that does not have the correct polarization for the polaroid will be absorbed in the same way.


3

A real image can be viewed on a screen, a virtual image can not. Rays from both types can enter your eye, be refracted by your eye's lens and form a real image on your retina as @CarlWitthoft points out. So it is not the case that a real image must be viewed on a screen. It can be viewed on a screen.


3

There is no real image on the mirror surface. Infinitely many rays arrive to every point of the mirror surface and infinitely many rays reflect from every point of the surface. For example CCD sensor is a polished flat surface that does reflect some amount of light, so if it is exposed (without any optics) it can be considered a kind of a mirror - see here a ...


3

It's a matter of convention. The complete wave function must describe a wave that propagates in the correct direction. Any function of the form $f(kz - \omega t) = F(z - vt)$ describes a wave propagating to the right. For a plane wave, that could be $\exp{(i(kz - \omega t))}$ or $\exp{ (-i(kz - \omega t))}$. An author is free to choose whichever he or ...


3

You are right that the oscillations of the electromagnetic field need not have any spatial extent. The oscillations, as you point out, are in the strength of the electric and magnetic fields. If I understand your question correctly, you are asking why then can some objects distinguish between the two different polarizations of light. This is because ...


2

What does an aperture do? It "applies" Huygens principle to every point within the aperture, and ignores those outside the aperture because they are blocked. There are a couple of things going on when you consider a lens. Let's make sure we understand them. An aperture produces a diffraction pattern in the space of diffraction angles. Recall from the ...


2

It sounds as though you might not have "filled" the lens properly. An infinite conjugate microscope objective, for example, has a specified numerical aperture when it is driven by for a collimated beam of a specified beamwidth and apodisation (i.e. whether Gaussian, uniform or so forth). See my drawing below: whence you can understand the fundamental ...


2

Generally in Nd:YAG harmonic generation using KTP, Type-II collinear phase-matching is used, which means that both ordinary and extraordinary polarizations are mixed to form an extraordinary polarized doubled beam. In essence, both polarizations of light in the unpolarized beam are combined to form a polarized beam. This means that while unpolarized light ...


2

This is a real effect, but it doesn't have anything to do with coma or any of the optical aberrations. It is caused by the fact that the effective focal length shortens as you tilt a lens. When your eyesight gets worse, you need a stronger focal length lens in your eyeglasses, and tilting the lenses has this effect. The problem with doing this all of the ...


2

A common structure for LEDs the ppin-junction, [p-layer][p electron blocking layer][i layer][n layer] The p-layer, i-layer and n-layer is just your standard pin-junction structure. The p and n layers provide an electric field, which under forward bias will drive electron and hole towards i-layer where they can recombine radiatively. Normally the p and n ...


2

The near point is defined as the closest distance on which the eye can focus. "Normal" vision is usually considered to be vision with a near point of $25cm$. So, say there is a person who has a near point of $100cm$ rather than the normal $25cm$. To correct this vision, his/her prescription should be designed so that the lenses will take an object at $25cm$ ...


2

I asked the purely mathematical question over here, and received the most complete answer. While I thought there would be a simple trick to seeing the hyperbolic relationship, it looks like you just have to go through the tedious algebra to have it pop out. User JJacquelin found that it can be rearranged to the form ...


2

Check this: You can see that the angle difference is much bigger when the light gets reflected, and can be differentiated much easier. Also another factor which probably plays a very important role is that the sun is in front of you, and its simply too strong in comparison with the light from the droplet, thus making it even more difficult to inspect this ...


2

I'll assume that both lasers are of the same type, i.e that both are generated from the same physical mechanism. All lasers have some tuning range over which their wavelengths can vary. So, even though both are very monochromatic (single frequency) they will, in general, not have the exact same frequency. A HeNe laser, for example, has a tuning range of ...


2

Light is a wave, it does not change its nature from particle to wave. Light, and this is true for any particle, is a an excitation is some field, and we model its propagation/time evolution using waves, but also quantise some of that wave's parameters, and in a sense, modelling it as a particle when convenient. Now answering your question, light during the ...


2

There are several ways to look at these two effects. Their mechanisms are very similar, but their experimental realizations are rather distinct from one another. Here's one of the physical models, a classical one. Since you ask for physical intuition, I think the classical picture works best. Quantum mechanical models use a different language and tool ...


2

The basic analogy to consider is that SHG is like full-wave rectification in electronics, whereas Raman scattering is like FM radio modulation. While Punk_Physicist gave an energy-level diagram that is helpful, you may also be interested in a purely classical/mechanical picture of the process. Picture of a molecule Think of a molecule as a pair of charges ...


1

The EM reflection from plants is at a maximum in the range of wavelengths from 750nm to 950nm, that is the near infrared. The reflectivity is also highly dependant on the amount of cholorphyll present, and by implication the health of the plant. The graph is from this website, which has a more detailed discussion.


1

Actually this is a bit of a tricky problem that only was solved relatively recently. The primary rainbow is generated by a beam that refracts once inside the drop, reflects off the back, then refracts again upon exit. You have to use snell's law and then some calculus to find the minimum scattering angle for the angle that produces the brightest light as ...


1

Yes, the change of the angle is important because that's exactly why the Fresnel lens is working: since the angle of refraction depends only of the incident angle and not on the thickness of the refracting medium you can sub-divide a normal lens in prism-rings which have all different angles for the incident light beam. When you look for example at the ...


1

The answer is Yes. However the losses will be wavelength-dependent, you will likely get a "ugly" superposition of modes at the output and a dispersed signal as user288447 stated. You are correct, the mode for each wavelength need to be solution of the Helmholtz equation. For a dielectric waveguide, there is no absolute cut-off wavelength (all wavelength can ...


1

Are you concerned that because the properties of the focussed light does not have the same properties (e.g. wavevector) of the fiber mode that coupling is not possible? It is possible to focus white light into a single mode fiber. The main criterion is that the angle of convergence of the light must fit inside the angle of acceptance of the fiber. The ...



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