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1

When you place two lenses one behind the other, the effective focal length is a function of their distance. This is the principle behind a zoom lens - with the same pieces of glass, a zoom lens achieves a range of distances. What you are describing is the simplest case of a zoom lens - just two elements. The first element is the cornea + lens in the eye - ...


1

A very simple diagram shown below is from the wikipedia article on Permittivity. "Electronic" encompasses level/band transitions, atomic represents such phenomena as vibrations (as shown) and rotations (which is how microwave ovens work). Here is a more complicated spectrum for semiconductors/solid state matter taken from here.


0

It's a "tunneling" behavior. In effect, all the light is "pulled back" into the medium unless there's another body of high-index (well higher than the $n_1 = 1$ ) material within the distance covered by the evanescent wave. If that material is close enough, then that part of the evanescent wave, which you can view as a probability wave, is in a region ...


1

In my answer to this question: (What is the sun's spectral series?), I give a very detailed answer about why mixing colours of light produces other colours and how it is purely a result of biology and evolution. I also delve a bit into the structure of the human eye and why, in fact, only three colours are necessary to reproduce all of the colours we can ...


2

A "normal, healthy" human eye has two types of light-sensitive cells in the retina: rods ("color blind", but capable of sensing low light levels) and cones: cells that are sensitive to different bands. See this figure for their relative sensitivity (from http://hyperphysics.phy-astr.gsu.edu/hbase/vision/colcon.html) When you look at a spectrum of light, ...


1

The cells in our retina that detect by frequency (read: colour) detect most strongly in three slightly different bands we know as Red, Green and Blue. To make a slight correction I would say an incadescent bulb is quite far from white, so I would rather proceed talking about sunlight on a clear day. The reason why sunlight appears as white as say a white ...


3

Take a look at this picture - and ask yourself why the operator of the camera has the cloth over his head (the cloth is black on the inside) as he is looking at the back of his camera - which has a piece of ground glass where the image from his pinhole camera is forming: This used to be how photography was done: Align the camera to the subject, focus (if ...


0

if your light is polarized perpendicular to the glass surface then there is no change in polarization.


0

The two lights cover different areas of the drivers view. If there are other cars near, it's important do not aim the light to the eyes of other drivers (or their mirrors). But Also, it's important to cover as much of the street ahead as possible. One bulb, together with the surrounding reflector of very specific shape, aims the light to the street up to a ...


2

You are getting confused by the fact that we use an imaginary number (a mathematical construct) to describe a physical phenomenon. But it is just that - a mathematical trick. You can perfectly well describe a attenuating wave with $$e^{ikx}e^{-k'x}$$ Which can be made more compact by combining the two k's into $$e^{i(k+ik')x}$$ If you are OK with the ...


14

You see it because it travels through air, dust, and a lot of other molecules and particles that can reflect and diffuse it. This, together with focussing, is also the reason for why it cannot travel arbitrary long distances. If you go to vacuum then the laser beam has much less losses, and it can travel much farther as happens in the LIGO interferometers ...


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 ...


0

As mentioned by Ruslan , precisely speaking, what one should do to simulate the unpolarized light is to take an average of the intensity of all the orthogonal polarized light other than just 2 of them. Plane source is a special case because its z-polarized component is quite weak so it won't hurt even if only an average of x and y-polarized component is ...


1

This does not violate conservation of momentum, because the momentum behaves in a counter-intuitive manner in a negative refractive index medium. In most media, the refractive index is positive, and the Poynting vector $\vec{S}$ and the wave vector $\vec{k}$ point in the same direction. In a medium with a negative refractive index, the Poynting vector ...


1

The Verdet constant is a coefficient which sums up the magneto-optical properties of the medium. So, the temperature and wavelength dependence are wrapped up in it. Fundamentals of Photonics by B.E.A. Saleh expresses the Verdet constant in terms of the wavelength as $$ V\simeq-\frac{\pi\gamma}{\lambda n} $$ where $\lambda$ is the wavelength of the light ...


0

The corners of the triangle represent some standard definition of red, green, and blue. There's no particular reason that they should correspond to single wavelengths, although admittedly the closer the standard colors are to the single-frequency curve, the larger the visible gamut would be. The standard colors were undoubtedly chosen as a compromise based ...


0

As @dmckee points out, you have to take into account the reflected wave. And also the refracted wave. When you add them up, there is a transfer of momentum perpendicular to the $surface$. No couple. Another way to arrive at this conclusion is to notice that an infinite surface (the usual model) has perfect translation invariance. Momentum in that ...


1

Recall that Maxwell's equations (in the absence of losses) require only that $n^2 = \epsilon\mu$. So when you take the square root, you are mathematically allowed to take either the positive or the negative square root. Of course, then the question becomes, why would you want to take the negative square root? Clearly, this is only an issue if $\epsilon\mu ...


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 ...


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 ...


1

Fidelity is a way of quantifying how similar 2 states are to each other. Fidelity $F$ is by definition $F\in[0,1]$, with $F=1$ meaning that 2 states are identical and $F=0$ means they are as different as physically distinct possible. Fidelity is often used as a way of saying how good your source or quantum state preperation is, by comparing the state of ...


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 ...


0

second harmonic generation is a non-linear optic effect, this mean that is the dielectric polarization of the crystal responds non-linearly to the electric field of the light -in this case laser pulses.the photons emitted by the laser (monochromatic light) interact with the non-linear media (crystal), and because of the constructed properties of crystals the ...


1

The formula you use to calculate the skin depth is just an approximation and not necessarily valid at THz frequencies. How do you model the optical properties of gold in your simulation? If you know the refractive index $n$ at a given frequency $\omega$, you can easily calculate the exact skin depth ($1/e$ amplitude decay length): $c/(\omega\text{Im}(n))$


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 ...


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. ...


3

In a sphere, any light emitted from the center will reflect off the sides at normal incidence come back to the center. In a cube, some rays never return to the center, so you aren't measuring all of the light emitted, which defeats the purpose of the device.


0

I'm not an expert, but is this not simply the single-mode attenuation in the fibre? As far as I can see, the absorbed photon will either a) be re-emitted into a different mode (contributes to attenuation.) b) turn into phonons (contributes to attenuation.) c) be re-emitted into the same mode by stimulated emission (probably very rare, and would you care ...


0

I think you will observe something which will look like a dot because the interference pattern will become too small to be observable (but it will be there). This doesn't mean that the photons will behave as particles.


0

I believe you will observe a dot. The reason for this is that you interfered way too much with the photon's wavefunction, and essentially forced it to behave classically. This phenomenon is known as Quantum Decoherence.


-1

electromagnetism is not a handed force. So i don't think you even need to use vectors that are transverse. I recently, learned about a mathematical object called a differential form. a dx+dy is like the k unit vector and dx-dy is like -k. So spin can be described in a more natural way that does not resort to a perpendicular direction. Definetly, more than ...


-1

light rays are not one dimensional objects. True, in the figures that directed segments are shown. Mathematically, you can think of sort of an infinitesimal vector showing that direction. it can be made infinitely small. is that 0 dimensional small, i don't know. I don't think it is. It would have the dimension of dx.


4

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 ...


0

If instead of the string Light were to be used then it will pass through both of the slits S1 and S2. The book is using this analogy to explain the concept of polaroids. The whole string-slit system that your book quotes is analogus to the above shown light-polaroid system. Light will not pass through the second polaroid. I am not explaining how polaroid ...


0

Wow, this is a very detailed question. Thanks for your effort. Lets ignore diffraction effects, which will scatter some small amount of extra power out of the laser beam. The loss at elements 1a to 1d will not simply sum up. This is because the power will only be lost at one of the mirrors, and will not be there to be lost at the other mirrors. So, to ...


0

Just to add to Anna v's answer and Aanel's answer which are both admirably pithy and correct. Polarisers making use of the Brewster angle deflect the light of the "blocked" polarisation, rather than absorbing it. The "blocked" light actually passes into a refracting, the unblocked light is reflected off and redirected to the output. Polarising beamsplitters ...


0

This question is all about the signal to noise ratio you achieve in your experimental setup, so the details are highly dependent on the latter. Here are the physical principles you would use to calculate how long it takes a fringe pattern to form. Assuming the source sends unentangled photons, each photon propagates following Maxwell's equations. So the ...


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 ...


0

The wavevector does not always point to the direction of propagation. Read this from wikipedia to understand it better. The wavevector is basically parallel to the direction of propagation of the wave if the medium is isotropic (i.e uniform in every direction).


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

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

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 $25cm$ ...


0

Mathematica gives $$ z=\gamma\frac{\sqrt{\gamma^2-a^2-4(y^2+L^2)}}{2\sqrt{\gamma^2-a^2}}, $$ where $$ \gamma=\frac{(2n+1)\pi}{k}. $$


0

Here is some information adapted from the definitions found in Schott's 2013 Optical filters Catalog: [For all the wavelengths of the visible spectrum,] The spectral transmittance is the ratio of the transmitted [exiting] (energetic) spectral flux to the incident [incoming] (energetic) spectral flux. [of the illumination seen through the filter]


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 ...


4

Light that is not transmitted is either absorbed or reflected. Wire grid polarizers tend to reflect. Polarization beam splitters separate the two polarizations in different directions. Polymer based ones absorb it i believe...


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.


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.


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 ...



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