New answers tagged

-1

Paisanco is right. Please refer to this paper which published in JOSAa https://hal.archives-ouvertes.fr/hal-00831552/file/Differential_theory_for_anisotropic_cylindrical_objects_with_an_arbitrary_2013_preprint.pdf


3

The answer to any human perception question is "its complicated." We evolved for maximum fitness (for some fitness function). The way we hear sound and the way we hear light are the most effective ways we have found to deal with the signals. If we try to read into it further, however, its worth noting that its useful to be able to identify "...


6

Colour perception isn't as simple as you think Roger Barlow has already described why sound perception is different (you have enough receptors to distinguish bot a wide variety of notes and their strength at the same time).And the eye has only 4 receptor types. But human colour perception is also different for another important reason. It is not a matter of ...


1

Very often in a wave propagation analysis, superscript '+' indicates the fields associated with the wave propagating in the positive direction and superscript '-' indicates the fields associated with the wave propagating in the negative direction. When calculating the impedance, you would take the ratio of the E and H fields for the wave travelling in one or ...


2

We have $$x^\prime=sx.$$ Thus, $$dx^\prime=sdx\implies\frac{dx}{dx^\prime}=\frac{1}{s}$$ Apply chain rule of derivative. For a test function $f$, $$\frac{df(x)}{dx^\prime}=\frac{df(x)}{dx}\frac{dx}{dx^\prime}=\frac{1}{s}\frac{df(x)}{dx}$$. As the function is arbitrary, the operator relation is then, $$\frac{d}{dx^\prime}=\frac{1}{s}\frac{d}{dx}$$


22

Because 16,000 is greater than 3. We only have 3 sorts of detector (called cones) in the eye, sensitive broadly to red, green and blue light. So a mix of red and green light excites the red and green cones. But yellow light, in between red and green, also excites the red and green cones, and the brain can't tell the difference. So it's not the mean of the ...


9

Why is the combination of two light waves (red, yellow) percieved as the same color as the arithmetic mean of their frequencies (orange) Simple: no, it isn't. This is not how color perception works. (If you're inclined to disagree, ask yourself: how does your model account for the fact that changing the relative intensity of the two light sources changes ...


2

It's not that the light itself is converging behind the mirror, it's that if you extend the rays in a straight line following their new path after being bent in the lens, they will appear to come from a single point. This Khan Academy video explains virtual images well: https://www.youtube.com/watch?v=nrOg85VPQgw&vl=en


1

It's related because it's the Fourier transform that tells us how to express any function $f(t)$ as an integral with components having time dependence $e^{-i\omega t}$, and therefore if we can analyze fields with time variation $e^{-i\omega t}$ we can analyze fields with any time-dependence we like.


1

This is the complex wave equation, $$ e^{\boldsymbol{i} z} = \cos z + \boldsymbol{i}\sin z $$ With $z=-\omega t$ Note, the imaginary unit $\boldsymbol{i}$ is missing in what you wrote. They are just saying that the solution being sought is a plane wave.


3

In theory yes, but in practice no. A lens is a refractive element, so there will alway be some Fresnel reflection which will reduce the transmitted intensity The len’s material is not perfectly transparent, some intensity will be lost by absorption Surfaces are not perfect and will scatter some intensity out of the beam The transmission efficiency of a ...


0

The distance between the central maxima and the first minima is d = 0.66λ/N.A. So this is essentially the limit of resolution as an object is just resolved at the point where the central maxima of both diffraction spectra coincides with the first minima of the other: The lower case sigma in the above diagram is equivalent to d in Abbe’s equation. If the ...


2

You could accomplish this with two lenses. The first, a spherical lens, would be selected to give you the large horizintal angular spread that you want. The second, a cylindrical lens, would be oriented with its axis horizontal and could be placed either before ir after the spherical lens. Adjusting the distance between the two lenses would adjust the ...


1

Let's look at an example of a virtual image from your diagram by reversing the rays, which is allowable. Let r be the distance from the lens to R and p be the distance from the lens to P. So consider rays leaving R from left to right and diverging from the lens as if coming from the virtual image at P. This is a valid ray trace based on your diagram. For ...


14

As far as I know, brightness is defined in optics as a component of color. Therefore it depends not only on the intensity of the light, but on our perception of it. It is more closely related to luminous intensity than to radiant intensity. For a beam with a given spectrum, however, the luminous and radiant intensities will be related by fixed ...


6

Intensity is the power per unit area; it is a physical quantity. Brightness involves how the human visual system perceives light, and it is not a physical quantity.


2

Welcome to Physics SE! First step: have you checked the literature? I saw some interesting stuff with a quick Google search. Compare the data and approaches to your own. If yours is good/reasonable, only experiment can tell. Gaussian noise, however, is a common standard. As far as I know, your approach seems to be fair enough. If I were going to model sky ...


0

From what I could gather online (e.g. in this presentation), an extended mode is a mode that can propagate in the bulk of the crystal. Extended modes are contrasted with point defect modes (fully localized) and guided modes (localized only in one or two dimensions). The former can only appear inside photonic band gap, and the latter are in the allowed band, ...


2

Minimum deviation occurs when the ray passes symmetrically through the prism. You should use this to show for yourself that $$\sin \left[\frac 12 (A+D_{min})\right]=n \sin \left[\frac A2 \right]$$ in which $n$ is the refractive index of the prism material and $A$ is the angle between the faces via which the ray enters and leaves. The answer to your question ...


2

For a prism, $D = I + E - P$ where $D$ is the angle of deviation, $I$ is the incidence angle and E is the angle of emergence with $P$ being the physical angle of the prism. And when you want the minimum angle of deviation, that is the angle of incidence is the same as the angle of emergence, $D = 2I - P = 2E - P$


1

Sure, 0.3 mm is sort of thin, but such approximations tend to be valid only when some number is small relative to another. I think the relevant comparison here is between this characteristic length and the wavelength of the THz radiation used. While I didn't see what frequency they used at a quick glance, it's worth noting that 1 THz corresponds to $\lambda=...


0

From your description, it seems that your detection system is operating in a nonlinear region. Nonlinearities in the detection system will complicate things, so making sure the detector is operating in its linear range will be very important. If the detection is linear, separation of the ambient and signal beam effects should be fairly straightforward. A ...


0

The psd method basically moves the signal from d.c. to some other frequency, and then is sensitive to noise at that other frequency. Therefore you pick a frequency where the noise is small. Many noise sources decrease with frequency, so often one picks as high a frequency as your detector and de-modulation stage (usually a lock-in amplifier) can handle.


2

The optical resolution limit is $p = 0.5 \lambda / NA$ for off-axis illumination. The numerical aperture $NA=n \sin \theta$ cannot exceed the refractive index $n$ of the medium (vacuum or sometimes water) below the lens. In principle you can reduce lambda indefinitely.


1

Your colour effect is somewhat similar to what can be seen through the windows of some trains. A part of the blue sky is a source of polarised light. When viewing that part of the sky through a birefringent window and a linear polarisation analyser, it is possible to see the colour pattern. It is more likely (and cheaper) a window is birefringent due to a ...


2

No, the coherence length cannot be derived from the field equation. Or rather, the equation is an approximation, made with the assumption of infinite coherence length. A real source will have $\omega$ (and thus $k$) varying at least slightly over time. It's the characteristic period of this variation of $\omega$ that produces a finite coherence length.


3

It's almost certainly black mold/mildew. It usually appears where humidity is higher than normal.


0

Thanks for making (and showing) extra pictures. I can't see that the window is open (though I'm not familiar with how to open the window). In the second extra picture, I can see faintly the described effect in the reflection by the floor (it seems you took all pictures at different distances and at different heights) but not in the reflection by the mirror. ...


2

I think you are confused due to the terminology 'virtual' here. 'Virtual image' here does not mean there is no image, it simply implies that the image is not where it appears to be. If you are looking through a transparent material, you also see an image but this time, the object is actually at the place from where the light appears to come and hit your eyes....


0

What I have learnt in school is that, we are able to see images behind a mirror(virtual image) because our brain assumes that the diverging rays that our eyes see, are coming from a point behind the mirror Not necessarily. Images are images independent of what the brain is doing. It's not like our brain has some mechanism to first determine if an image is ...


1

A camera doesn't "know" if a source image is real or virtual. As long as the rays from each point in the source image are not converging, and aren't diverging from a point too close to the camera lens, the camera can form a real image on the film or sensor array.


2

In these calculations, the light is modeled as an ordinary particle whose speed happens to be $c$ either at infinity or at the point of closest approach. The speed isn't (and can't be) constant on the whole orbit. The paper that you linked sets the speed to $c$ at closest approach (see the text after equations 5 and 7). There's really no adequate Newtonian ...


1

What is happening, in my opinion, is the result of two effects: First one is to notice that the glass of your window is most probably not uniform. Meaning, as a guess, it is thinner on the top and gradually gets thicker to the bottom, but in a very slight manner that only light feels it. What does this cause? Well if the glass also has two different ...


0

As I vaguely remember, Newtonian gravity does not have any affect on light. Einstein's General Relativity was the first prediction about gravity bending light, and this was confirmed a few years later by seeing the effect in a telescope during a full solar eclipse.


0

The trick is that the system is designed in layers, and the data for one layer includes the metadata for the next layer. Here's how a hypothetical system could work: Layer A: So the fiber-optic transmitter - the physical layer - only knows how to transmit a 1, transmit a 0, or turn off. Layer B: The next layer understands that if the transmitter was off and ...


0

The key point here I think is that the path difference must be calculated at a wavefront, which is orthogonal to the rays. See the diagram below and note that the path difference is $\ell_1 + \ell_2$ instead of $2\ell_1$. $\ell_1 = \dfrac{d}{\cos\theta}$ indeed, but $\ell_2 = \ell_1 \cos(2\theta)$. Therefore, $$\Delta = \ell_1 + \ell_2 = \left(1+\cos(2\theta)...


1

Having made a hash of my original answer which was pointed out by @Pieter, I have rewritten it. In this experiment I simultaneously look at an object (black lines ruled 10 mm apart) which is 250 mm from the magnifier/eye and at the virtual image of mm square graph paper formed 250 mm from a magnifier/eye. Fine adjustment was made by reducing the parallax ...


0

The power of a magnifying glass is relative to the near-limit distance of the standard human eye (25 cm). The lens makes it possible to bring the object closer to the eye, so that it subtends a larger angle than at the near limit. The ratio of the (tangent of the) angles is proportional to the ratio of these distances. So if the focal length of a loupe is ...


2

It is easy to construct situations in which light does not take the "fastest" route. For example, set up a laser pointer to send a beam across the room. Put a mirror in its path and deflect it to a second mirror, then tilt the second mirror to direct the beam to the point on the far wall where the beam would hit if all the mirrors were removed. ...


0

The shorthand answer is this: Due to boundary conditions on the oscillating electric and magnetic fields of the incident and reflected light on both sides of the reflecting surface.


1

Why ... because of the equations. That is the best answer there can be, IMHO. Maths is the pure logic One short-hand I would use it to point out that Maxwell's equations are invariant under rotations, i.e. Maxwell's equations have no preferred direction. In the problem with refraction you have: (1) Plane of your surface, which is fully characterized by the ...


0

Light interference has little to do with rays; it has a lot to do with local intensity of a light field. Here are some general principles (for linear optics) that might be helpful: Rays travel entirely independently of each other. When rays cross each other, they pass through the point of intersection and simply continue on their own way. There is no ...


0

From this Wikipedia article: The luminous intensity for monochromatic light of a particular wavelength λ is given by: $$I_{\nu}=683\cdot \bar y(\lambda)\cdot I_e,$$ where $I_{\nu}$ is the luminous intensity in candelas (cd), $I_e$ is the radiant intensity in watts per steradian (W/sr), $\bar y(\lambda)$ is the standard luminosity function. If more than one ...


0

Rely on $F = \frac{2\pi}{\text{RTL}}$, where RTL is the TOTAL round trip power loss. In the ring resonator, $$\text{RTL}\approx\kappa^2 + \text{material loss} = 2\kappa^2$$ So $F=\frac{2\pi}{2\kappa^2}=\frac{\pi}{\kappa^2}$. The reading of Siegman is wrong. Siegman's $g_{rt}$ is round trip field gain, and it is not analogous to $t^2$, but $\delta$ is power ...


1

From statistical physics, if a system is in contact with a "reservoir" of heat and particles, then the energy of the system $E$ and the particle number $N$ are allowed to fluctuate. If the set of microstates of the system form a discrete set (which would be the case for a system of particles which can inhabit discrete energy levels), then the ...


0

The real question is: why would you do this tranformation or why would you define such an operator? The Bogoliubov transformation is usually used to diagonalize Hamiltonians (see wiki). Off-setting the operator by a constant does not change its time dependence. $b'_k$ is also stationary if $b_k$ is stationary. So all the additional constant does is offset ...


1

A few arguments can be made, broadly as follows Classical Electrodynamics Atomic Couplings Atomic Transitions Note I use Gaussian units throughout. 1 - Classical Electrodynamics This argument comes directly from the classic Electrodynamics Of Continuous Media by Landau and Lifshitz$^1$. I reproduce the argument here so the answer is self-contained. This is ...


8

Why do single-mode fibers have lower attenuation than multimode fibers? It's an engineering/economic reason, not physics. The extremely low absorption loss in single-mode fiber (SMF) is obtained by extreme purity of the glass, and by using wavelengths in the 1330 and 1550 nm bands. Because the transmission distance of multimode fiber (MMF) is limited by ...


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