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1

Some background. You want to detect the image of an object. First, either 1) you illuminate it with some light source [a lamp, the sun] or 2) the object itself radiates light out [a star, a fluorescent sample]. Imagine to divide the object in many small parts (voxels): to reconstruct the image of the object, you need to detect independently the light coming ...


0

I think you will have an easier time viewing the index of refraction from a speed-point of view. Consider the following: The energy of a given photon is determined by its frequency (color): $E = h \nu $ (h being the Planck constant) Assuming the photon does not lose energy when entering the material, its frequency must be conserved. However, as light is an ...


1

You are mixing up two different things. The refractive index is usually defined in terms of the velocity of light: $$ n = \frac{c}{v} $$ where $v$ is the velocity in the medium. However the velocity is related to the frequency and wavelength by: $$ v = \lambda f $$ so: $$ n = \frac{\lambda_0 f_0}{\lambda f} $$ The frequency of the light, $f$, doesn't ...


0

Another way of possibly understanding it: From Underwater Photography Water absorbs different wavelengths of light to different degrees. The longest wavelengths, with the lowest energy, are absorbed first. Red is the first to be absorbed, followed by orange & yellow. The colors disappear underwater in the same order as they appear in the color ...


1

From NOAA site: Color Changes At Deep Depths As you travel from surface waters to deeper waters, the quantity of light changes; it decreases with depth. The quality of light also varies with depth. Sunlight contains all of the colors of our visible spectrum (red, orange, yellow, green, blue, and violet). These colors combined together appear white. ...


1

This is due to the fact that different colors of light refract through glass (or whatever your glasses are made of) at different angles. Plain white light contains all other colors within it. When a beam of white light hits your glasses at an angle the yellow/red and blue/violet light separate from the main beam due to their different energy levels ...


0

The two effects have slightly different origins. Polarization of reflected light is a function of the angle of incidence and the refractive indices of the two media on either side of the surface - as described in this reference. Specifically, there is an angle (called the Brewster angle) where the reflected light is entirely polarized with the electrical ...


8

Your question is based on a mistaken assumption. In introductory texts you start with "keeping it simple" - flat surfaces, parallel lines - until you get comfortable with that. And then you add prisms. Surfaces are no longer parallel, but still flat. And then you add lenses. And convex/concave mirrors. They are curved but amenable to (analytical) ray ...


2

The answer depends on how you need to polarize it. Many lasers output highly polarized light anyway. Usually this is linearly polarized. If you need linearly polarized light, you may only need to align your laser correctly. This alignment is usually this is done with an adjustable polarization rotator rather than by rotating the laser. If you need ...


2

What you said is right. The incident field creates the electron-hole pair, then they oscillate normal to the surface. Since the motion is normal to the surface, it radiates in every direction except normal to the surface. (Dipole radiation pattern.) So if you want to avoid that effect, you want the outgoing light you measure to be normal to the surface. If ...


2

From the interior, a spherical mirror can be analyzed as a continuous assemblage of concave mirrors. If you were illuminated by an invisible light source, your image would be reflected from all points on the interior surface according to the mirror equation for concave spherical mirrors: (1/object distance) + (1/image distance) = (1/focal length) The ...


2

Everything you see that does not emit light itself is due to light from an external light source being scattered. Most objects scatter light in all directions, so no matter where you position yourself, you will see the object. If something blocks the light from the source, less light will be reflected from a particular area. This area is the shadow. Since ...


1

To add to WhatRoughBeast's answer, this is because you are not really viewing the image under pure red light. The "red light" you used to create the image has small amounts of green and blue wavelengths, causing the white areas to appear more "white" than the red areas, even though they are both just slightly different shades of red. To quote WhatRoughBeast, ...


1

Personally I find statements like "differently polarized beams don't interfere" to be imprecise, but I can understand where the writer is coming from. The interference of waves of like polarization (or of longitudinal waves) is characterized by differing average power at different points, including places where the average power is reduced from the value ...


2

Under red light, your picture never shows white. It only shows bright red, which you are interpreting as meaning white. If you take a piece of paper and put a small hole in it, then hold the hole over any part of the screen image, you will see that it is red, not white. When doing this in real life (which I suspect is what you're talking about), three ...


1

Polarization and (temporal) coherence/phase of a light wave are two different, yet completely unrelated concepts. So there is no deep connection. Polarization has to do with the fact that the electric field is a vector and hence has a direction. This applies to static fields as well, although one does not speak of polarization here, but merely just ...


1

The simple physical model of the eye (or indeed a typical camera) is that it records just three values for each pixel. In your eye, this is because you have three different types of cone cells, called S, M, and L, peaking in blue, green, and red wavelengths. In a camera, the light is passed through blue, green, and red filters before having its intensity ...


1

It's a pinhole camera image of the sun - as DJohnM's comment said. My question is: Aren't 'lenses' required to converge the rays to make an image? How can a hole in the centre of a cardboard form 'images'. No - all that is required is an aperture (hole) to restrict the range of rays that reach the screen to form an image. All a lens does is allow a ...


3

the flux drops off as the square of the distance, but the solid angle subtended by the source drops off the same way, so surface brightness is constant, right? Right. But what happens when you can no longer resolve the source? Then the "solid angle subtended by the source" stops dropping, and only the reduced flux of the entire source can be ...


6

(1) I assume you are referring to a Polaroid sheet (specifically H-sheet), which is the most common form of dichroic polariser; that is, it linearly polarises light by selective absorption of the electric field component in certain directions. On the molecular scale, this phenomenon is a result of the alignment of polyvinyl alcohol (PVA) chains within the ...


9

Would you dig a ditch with a surgeons scalpel? Yes, quantum mechanics ultimately underlies all physical observations but the mathematical expressions for large dimensions with respect to $\hbar$ become cumbersome and are replaced by the simplest ones for the appropriate study. Thermodynamics, for the study of bulk matter, blends smoothly with quantum ...


0

The billiard balls do not have atmosphere (O2,H2O, N2, CO2, O3, particulate, ...), clouds, forests, terrain, sea. Each component modulate the answer in relation to absorption, refraction and reflection, trapping/reflecting more or less radiation. More heat from the sun at the surface >> more water vapor in the atmosphere >> more clouds >> more albedo ...


3

The fact that the ocean is proportionally smooth doesn't matter -- when a light wave with wavelength ~500 nm hits the ocean, all it knows is that it's running into huge, 60 foot tall irregularities. It doesn't care if these waves are small compared to the rest of the earth. I mean, how would it even know? For specular reflection, you need absolute ...


0

It's been a month, but maybe you are still curious. I've been struggling with the almost exact problem for.... too long. So here goes. We basically want the effect of reading glasses. The screen is too close to your eyes (less than the LDDV Least Distance of Distinct Vision) so we need to create a virtual image that is at or beyond the LDDV. No need for ...


3

The calculation is identical, except that the waves start with a phase difference at the entrance to the slits. If you have perfectly narrow slits, the only difference in the pattern will be a shift of the maximum (the maximum occurs when the waves from the two slits are in phase, so there should be a maximum in the forward direction of the incident beam). ...


0

The answer depends very much on the white light and you will always get some coupling. In general the coupling is very low, simply by the second law of thermodynamics applied to the phase space of light: it becomes the law of increase of Optical Grasp (oka Étendue). Light at one wavelength in a one moded optical fiber at one wavelength is essentially zero ...


0

a need for coupling a white light into a SM fibre had just emerged to me. I remembered vaguely, that there is probably no better means to couple extended source, than stupid butt-coupling. So I took one FT030 SM fiber patchcord and simply put one end near to cellphone photo illumination LED, and some amount (exceptionally high to my poor expectations) passed ...


2

Some of the light is blocked by the wire. But the light passing immediately off the upper and lower edges of the wire's silhouette act as two point sources, which interfere with each other when they reach the screen behind the wire. Babinet's principle says that the diffraction pattern from the edges of an opaque body is the same as that from the edges of ...


2

Experiments with trapped ions generally use fluorescence for detecting the ions. This means that they use a strongish pump to take the ion from its ground state to a dipole-allowed excited state and wait for the ion to decay by emitting a photon in a random direction, and then re-run the cycle over and over. This means that each ion essentially emits one ...


0

It sounds like you want to measure foot-candles, which are units of light intensity received by a passive surface. This is an appropriate measure if the surface reflects light from another source. It's not the right measure for the output of a computer screen. Also, you seem to want to measure at specific places on the screen. Here are some foot candle ...


1

Yes it can, if you know the focal length as well. Assume that the focal length is $f$ and the distance from lens to object is $d$. To get the object in focus, you need the distance from the lens to the sensor to be $$f' = \frac{f * d}{f + d}$$ Once you have the distance from the optical center to the sensor, you know the magnification: $$M = ...


0

Yes. Materials that absorb electromagnetic radiation and emit it in a different frequency are known as fluorescent. You probably see them as the coating on the inside of fluorescent tubes, where they absorb ultra-violet light and emit a lower frequency visible light.


1

Sadly, you're not likely to get any better chart than the one in the link you provided unless you can track down the OEM of the cameras supplied to the various phone manufacturers. However, the reason your camera can respond to ("see") IR remote controllers and the like is that the IR filters are not 100% blocking -- and the IR emitters in the remotes are ...


0

Check out Feynman's thesis: Feynman, Richard P., Laurie M. Brown, and P. A. M. Dirac. Feynman’s Thesis a New Approach to Quantum Theory, 2005. Feynman invented the path integral formulation of Quantum Mechanics (QM), in which he grounds QM in a least-action principle similar to that of Fermat. In "§7 Discussion of the Wave Equation: The Classical Limit," ...


1

Unfortunately, I think you are speaking about what people commonly say is "Huygen's Principle", "In order to explain waves diffraction, it says that every point in a wave front behaves as a source, so the next wave front is the sum of all secondary waves produced by these points.", but this is not actually what Huygen's principle says. Huygen's principle ...


0

Do we have a deeper understanding of Fermat's Principle? I thought we did. See the derivation section of the Wikipedia article: "Fermat's principle is the main principle of quantum electrodynamics which states that any particle (e.g. a photon or an electron) propagates over all available, unobstructed paths and that the interference, or superposition, ...


1

First off I think I should sort out a misconception about Huygens Principle. You can apply this principle efficiently if you have a slit, which is equal or smaller than the wavelength you are considering. If on the other hand the slit is substantially larger than the wave length, you should consider multiple Huygens sources. Take a look at this animation ...


-1

The answer is quite simple. the reason is that the first half-silvered mirror reflects the light down and therefore there is no light in the path to Cam2


0

I'm not sure I understand what is being asked for part (b), but I will just explain a little about Malus's law. Polarizing filters block all light except that which travels in a single direction, vertical or horizontal [see note below]. Unpolarized light contains both horizontal and vertical components, so the first polarizer blocks either all the ...


0

Just to wrap this up: I decided to use Tovar and Casperson' approach (Link 1,Link 2) using a 3x3 matrix, since I wasn't sure about the q-parameter in Shaomin's 4x4. It boils down to that as with the non-misaligned ray transfer matrix, $$ q_o = \frac{Aq_i + B}{Cq_i + D}$$ with the 3x3 beam matrix \begin{pmatrix} A & B & 0 \\ C & D & ...


3

That's the definition of a black body. 100% absorptivity implies 0% reflectivity and an emissivity of 1.


3

It is important to understand what an image is and where it is positioned. Remember that your definition has to work for things you see directly as well as those that are viewed through optics. It should also work for both real and virtual images. The version I use in class is An image is a collection of rays originating from a single object that ...


2

The ray theory of light is equivalent to the Eikonal Equation, which in turn is essentially a slowly varying envelope approximation to Maxwell's equations. If we write the electric and magnetic field vectors as $\mathbf{E}\left(\mathbf{r}\right) = \mathbf{e}\left(\mathbf{r}\right) e^{i\,\varphi\left(\mathbf{r}\right)}$, $\mathbf{H}\left(\mathbf{r}\right) = ...


5

For many materials the change in refractive index over the range of visible wavelengths isn't huge, so it's not a bad approximation to take a single value. The range of visible wavelengths is from about 400nm to 700nm, so the middle wavelength is 550nm. As it happens, the sodium D lines are not far from this, at 589nm, and since they are bright and easy to ...


0

There are a couple of ways of knowing the wavelength of laser pointer. 1) Using Snell'law (law of refraction): A light passing the border between two media whose refractive indexes vary. The incident light PO of wavelegth, $\lambda_{1}$ travelling in a media of refractive index, $n_{1}$ is refracted in to another media of refractive index, $n_{2}$, with a ...



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