New answers tagged

3

What you have not taken account of is that the light wave reflected is travelling in the opposite direction. The incoming and outgoing waves may interfere with each other, but they will not completely cancel each other out - in fact a standing wave may be formed as described below. Now - the way an antiflection coating works is to reflect the wave back ...


0

Most materials used in optics experiments are not magnetic. By this we mean that the magnetic permeability is equal to that of free space. This results in a magnetic susceptibility of zero. $$\mu_r = \frac{\mu}{\mu_0}=1$$ $$\chi_m=\mu_r-1=1-1=0$$ Therefore the magnetic field of the optical wave generally does not interact with the things in the experiment. ...


0

I'm assuming your input photon has a known polarization (say horizontal). You won't see interference, because the polarizers act as a "which-path" measuring device. If you erase the polarization information, the interference pattern will appear.


0

I found that Professor Junghwuen An's Fabry Perot interferometers with a 3rd reflecting surface is not required to emulate a Michelson interferometer's ability to measure the change in position of the self-coherence function first minima. Zero and negative delay is only important when one want to scan the separation between the two primary mirrors and ...


0

Does light carry any information from the source? Definitely. Light has properties that can be used to determine material of the source, as well as material through which the light ray has passed. Spectrum of light, is a great source of information which is actually equivalent to timber for sound. This is how astronomers find out what elements make up a ...


15

You are completely right in stating that the same effect should occur for thicker slabs. There are however at least 3 practical reasons why the effect is more easily observed in thin films. Light sources are typically not completely monochromatic. They emit slightly different colours at the same time. Imagine a light source whose wavelength varies by 0.1 ...


6

Ignoring for simplicity's sake the usual refraction that takes place at the interfaces of the media, if: $$|OA|+|AB|=D\big(\frac{1}{\cos \theta}+\tan \theta\big)=n\lambda$$ with $n$ an integer and $\lambda$ the wave length, then we have positive interference. But as we increase $D$, the distance $\Delta$ also increases, so for high values of $D$ these ...


2

Light coming from some source usually consists of many frequencies. Distribution of them is called "spectrum of light". Each frequency (in the visible to a human's eye range) corresponds to some colour. If we limit the spectrum of light to a single frequency, we will see this colour. However, not all colours we perceive can be obtained in this way and also ...


1

Timbre is a consequence of harmonic content which is a consequence of a sound being made of multiple pure sounds. Thus a 256Hz square wave has pure sinusoidal components at 256, 768, 1280,... Hz. An equivalent in light terms would be the spectral content of the light. For instance, take the bright yellow D line(s) of sodium on the one hand and a visually ...


0

Is there actually any evidence that being in water increases your risk of getting sunburn? I suspect that this inference confuses correlation with cause. On brilliant sunny days you are more likely to strip off and get in the pool or go to the beach to cool down. If swimming or diving you are unlikely to wear a sunhat. More of your skin (especially the ...


1

Yes. It's is not so much the water is the beach sand reflecting light back to you like a parabolic mirror. The droplets of water on your skin can form more surface area to catch light creating a magnifying effect focusing light on your skin as well. The random texture in the beach sand will also give you even tan. Most sand is white in color even if not the ...


0

As has been commented by fractalspawn this is almost certainly a photochromic effect (https://en.wikipedia.org/wiki/Photochromism). In this process a molecule in form A can absorb a visible or UV photon and be isomerized into another form B. This will have a different absorption spectrum than the starting material and if this absorbs in the visible part of ...


1

It's a back-and-forth process where the irises, laser, and external references are aligned to each other. For example, after setting up a laser, you want the laser to be level with the optical table. To do this, take a single iris on a post and adjust the post until the laser hits the center. Then, slide the iris towards and away from the laser. If the spot ...


2

In optics a "ray" is the direction of propagation of the classical electromagnetic wave. The term "ray' used for particles, as "cosmic rays" , and "gamma rays" are associated with this directional definition, from the times when it was clear that the phenomena followed straight lines like optical rays, before the differentiation into the particles we know ...


0

Lasers are termed as beam. Maybe the reason behind this terminology is the finite and easily characterizable size of the laser. Laser can propagate in a in a defined direction without diverging appreciably, on the other hand the term ray refer to something that propagate in a straight line. However, the parallelism is not necessary property.


0

Light is an electromagnetic wave. And for example gamma rays are electromagnetic waves. So the visible and near-visible light produced by most laser is like gamma rays but with a different frequency (wavelength). Thus saying ray and beam is roughly the same. Scientific convention is to say laser beam when the emitted electromagnetic waves are strongly ...


2

Okay, so: Unfortunately, the word "mass" has been used in two different ways in physics. One was the way Einstein used it in E=mc^2, where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is sometimes called the inertial mass. It's also the ...


2

Consider the light that originates at a period (full stop, '.') on a page. If the face plate is right up against the page, the light from that period will enter only a few fibers, those that are in contact with the period. If you lift the plate from the page, the light from the period spreads out before it hits the plate. If you move the plate far enough (...


1

The purpose of the condenser is to set up a highly defocussed lightfield through the object plane, i.e. the slide or LCD array. Ideally, the source used should be an extended one so that the lightfield at the slide is one of high optical grasp, so that the field at the object plane (slide) is like a whole bunch of broad apex angle (high numerical aperture) ...


0

What if we don't use any condenser lens? The condenser lens which consists of two plano convex lens with their convex sides facing each other collects light from sources which are usually divergent . It redirects and condenses the beam of light to flood the projector lens system. In the classical condenser lens system the first lens of the pair ...


1

The condenser lens systems collects light from divergent illumination sources, then redirects and condenses the light to flood the projector lens system. The classical condenser lens system consists of two PCX lenses mounted with their convex sides facing each other, as shown in the figure below. The first lens collects the divergent light cone from the ...


1

The colours are formed by thin film interference, much like a soap bubble or an oil slick. Light is partially reflected off of the front surface of the coating and also off of the back surface. The two reflected light waves experience a difference in path length of twice the thickness of the film. This leads to light interference which can be constructive,...


0

Exactly this phenomenon is used in beam splitters. A beam splitter is normally designed to have a 50% chance of reflection and a 50% chance of reflection, while your glass block will (probably) not have equal reflection and transmission probabilities. However the basic principle is the same. Incidentally there is a related question at What happens when a ...


2

There reason I can think of is the strength of interaction. The force exerted by the electric field on a charged particle is $eE$ which is much stronger than the force by magnetic field $ev\times B$. Magnetic force only get comparable to electric force when velocity of particle (usually induced by electric field) approaches light velocity. Due to this ...


0

Pretty sure one of the processes is electrons going into orbitals with lower energy, releasing the energy difference as light. Quantum mechanics describes rules for how this may happen. They probably ended up in an orbital of high energy by the system absorbing external light.


0

It is not possible to write a closed form equation for the Fresnel diffraction pattern. Usually one will use the Cornu spiral to evaluate problems like this. The Cornu spiral is a graphical tool that maps the phase / amplitude contribution of a infinitesimal element of the aperture. (image by R. Nave, from http://hyperphysics.phy-astr.gsu.edu/hbase/...


1

The polaroid sheet consists of a polymer into which a dye, originally iodoquinine sulfate, is dissolved. The film is aligned by stretching, and this causes the molecules of iodoquinine to align along the pulling direction. Photons are absorbed by the iodoquinine whose electric (or transition) dipole is aligned along the long axis of the molecule, but only if ...


0

   The metal mesh, or 'cage' around a microwave's oven cavity acts as a Faraday cage (see Wikipedia article on Faraday cage here), although a 'true' Faraday cage is grounded, and a microwave cage is not.    A cellular phone inside a Faraday cage will be protected from outside EM transmissions, just as conversely, the ...


3

If an object is black, it should absorb all visible frequencies This is too simplistic. Objects appear black if they absorb "most" of the light that hits it. We can't make materials that absorb 100%. What happens to that small fraction that is not absorbed is critical to how you perceive it. Most surfaces we encounter scatter reflected light well. ...


0

In a polaroid type of polarising material the molecules are aligned in the same direction throughout the sheet when the material is manufactured. Their electric dipoles are therefore also aligned and thus absorb photons whose electric field is parallel with the electric dipoles (to an extent that depends on cos(theta)^2 where theta is the angle between ...


1

It is incomprehensible to me that the most up-voted answers, and the ones posted by the most seeminglly knowledgeable people, all attempt to treat this question in terms of individual photons striking individual electrons. In fact, the phenomenon of transparency is all but incomprehensible in terms of such quasi-QM explanations. The natural, sensible ...


1

This answer is a little circular and like Burley's. Transparent materials have uniform electromagnetic coupling between its molecules. Think of glass as a uniform array of tiny capacitors.


0

No, the depths are reversed. The reason is that in free fusion (looking at the distance) each eye sees the image that correspond to its own side, that is, the left eye sees the left image and the right eye sees the right image, whereas in crossfusing (crossing your eyes), each eye sees the image on the opposite side, that is, the left eye sees the right ...


1

One can rationally ask what one could possibly see from 27 light-years away. Assuming that spacetime is neither grainy nor foggy (something that it very possibly is), it's mostly a matter of the size of the telescope and how well the scene is lit. One square meter of ground on Earth (roughly the area of a human sized "pixel", if we are generous and account ...


4

The simple answer is no, or, more precisely, the arguments that you cite give no weight to the idea that past states "travel" through the Universe. Let's look at the 13.5 light year away mirror. Yes, in theory you could see your own birth through it, if you could overcome all the optical resolution problems associated with such an undertaking. But this is ...


3

As suggested in comments these circles are airy disks around dust particles on the camera lens. After wiping the lens few of them were left and the remaining ones moved. You can find out more about airy disks on Wikipedia


0

According to huygens principle each point in the wave front acts as the source of secondary wavelet.By the time, the secondary wavelets from B travel a distance BC, the secondary wavelets from A on the reflecting surface would travel the same distance BC after reflection.Tking A as centre and BC as radius an arc is drawn.From C a tangent CD is drawn to ...


0

As John stated, the relation is not constant. Where does the need to "show that $\frac{i}{r}$ is constant come from? You know that if $i=0\rightarrow r=0$. You know that $$n_i\sin{i}=n_r\sin{r}$$ From this you should be able to see clearly that there's no linear relation except when $i$ approaches $0$, or $n_r$ approaches $n_i$. You should never go into ...


1

As for "I would like to know how the physics": that question is too broad to answer; I'd suggest that you get a book on Fourier optics and work through it by solving the exercise problems - you can ask for hints to solve the exercises here on Physics SE. I'll answer the more specific questions and address some misconceptions. As for "coherent light": ...


1

Light from the focus, when reflected to a parabolic mirror, will all be reflected in parallel rays.


0

quantum experiments are probabilistic (weird non-classical thing): they take a mixture of states the result of the experiment is one of the states each result occurs with a probability proportional to the amplitude of the states after the experiment, the system assumes the state that was observed well known example: probability that an electron hits a ...


0

Note I am the OP The geometric angle of deviation is given by: $$\theta_{dg}=\theta \frac{u}{|v|}$$ $$=\theta\left| 1-\frac{u}{f}\right|$$ If we let $u=f+\Delta$ then: $$\theta_{dg}=\theta\frac{ |\Delta|}{f}$$ Now taking $\Delta$ to be small we can assume the angular deviation due to diffraction is given by: $$\theta_{dd}=\frac{1.22 \lambda}{D}$$ $$=\frac{1....


1

The Rayleigh criterion provides an ideal best-case. The details of the Rayleigh criterion are arbitrary, and might not be appropriate for every case, but it does represent an accepted starting point. As you have intuited, there is more to the issue. It's essential to consider the capabilities of your optical and detection system. First we recognize that ...


0

Note I am the OP. The first and foremost thing to note about this equation is that it is not exact and is based on approximations. For the situation described above, once the light has entered the filter it splits into two, one that experiences the ordinary refractive index $n_0$ and the other that experiences the refractive index $n_e$ (which may be ...


0

You cannot immobilize light, but you can absorb it, convert it to heat. A first step in measuring light intensity (watts/square meter) is to make a black body to absorb the light. Then, by conservation of energy, the heating of that black body tells you how much light is being absorbed, in watts. The illuminated area of the black body is the "square ...


0

To solve this first you take two spherical waves emanating from two pin holes $\frac{1}{\lambda z} exp(ikr)$. The intensity from individual wave at any point (x,z) will be mod squared sum of the two waves. Here $r$ will be $\sqrt{z^2+(x\pm d/2)^2}$ expand the bracket and take out $z$, expand binomially keep only first term. Now add the two expressions with +...


0

A light wave is a traveling disturbance in the electric and magnetic fields. In the far-field these two components are coupled and in-phase so that one can define the amplitude entirely in terms of either the electric field strength or the magnetic field strength; by convention we give the electric field strength. So, in SI, the amplitude of a light wave is ...


-1

So an amplitude in terms of an electromagnetic wave is related to its intensity. So technically, the more intense a light wave is at a point, the more its amplitude. What is the intensity Intensity is inversely proportional to the A^2 of a wave. If you know the intensity, then the amplitude can be found using this formula I=1/2*pvw*A^2 where p is density, v,...


2

Diffractive optics in the Fresnel (paraxial) approximation is exactly the same as the quantum mechanics of a single particle when thickness along the optical axis is replaced by time, refractive index is replaced by mass and the inverse angular frequency of the monochromatic light is replaced by Planck's constant. Here is a brief sketch. Classical ...


0

Direct measurement of the amplitude of the optical field requires interferometric techniques. One that works is the FROG - Frequency Resolved Optical Grating. There are many variations on it today, including from the original developers: Trebino Research Group. These devices were designed for ultrafast pulses. For CW one usually just measures intensity ...



Top 50 recent answers are included