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79

Human perception is generally logarithmic. For example, the perceived loudness of a sound is measured using decibels, where an decrease of $10 \text{ dB}$ divides the sound intensity by $10$. So if the eclipse were heard instead of seen, "90% coverage" might mean reducing the intensity from $120 \text{ dB}$ to $110 \text{ dB}$, a small change. Perceived ...


42

The graph looks exponential because the vertical axis is logarithmic! If you were to re-plot it as linear lumens per square meter, it would be much more v-like, or even u-like. It so happens that a logarithmic plot matches our subjective perception of light intensity better than a linear one would. That's a result of our eyes having evolved to work well in ...


27

In real life: A sunset is "redder" than a sunrise which makes people feel more romantic. It's mostly because the atmosphere is warmer in the evening (no pollution here, lemon, the Earth is warmer in the evening because it was naturally warmed up during the day). However, there's also a very small contribution of the Doppler shift, one that you could in ...


22

To address your last point, there are several stars of which we have been able to resolve images i.e. see the star as more than just a featureless point. There is a list of these stars on Wikipedia (I love that they put the Sun at the top of the list - true but pedantic :-). The farthest away of the stars in the list is Epsilon Aurigae at about 2000 light ...


20

There isn't a simple answer to that. Colour arises when the light absorption or emission of a system is dependent on the wavelength. For example chlorophyll (i.e. plants) is green because it absorbs red and blue light so only the green light is reflected and reaches our eyes. So the question would be how does the light absorption and emission of trillions ...


19

I'll add a theoretical limit to the actual record put forward by John Rennie. To image an object as more than a featureless "point source", it must be resolved by the telescope. The angular resolution $\theta$ of a telescope is: $$\theta\sim1.22\frac{\lambda}{D_{\rm aperture}}$$ $\lambda$ is the wavelength of light, $D_{\rm aperture}$ is the diameter of ...


10

The 'color' would be an ultra-bright burst of gamma rays as the trillions of electrons rush apart, frying both you and your eyes to a crisp. More seriously, if you confined the cloud of electrons, it wouldn't emit any particular color on its own -- for instance, there could be no optical transitions since there are no nuclei. If you shined light on it, it ...


7

In physics, heat is energy that spontaneously passes between a system and its surroundings in some way other than through work or the transfer of matter. When a suitable physical pathway exists, heat flows spontaneously from a hotter to a colder body.The transfer can be by contact between the source and the destination body, as in conduction; or by radiation ...


6

It looks like a diffraction pattern from the pixels of the screen. In a different SE question, the pattern was four horizontal and vertical "rays" consisting of finely spaced peaks, rather than the wide spacing here and the angles that differ from 90 degrees. If you measure the apparent angle between the first diffraction peaks (the ones at the edge of the ...


6

No because solid is a state of matter. Light cannot be considered matter since it is made up of particles which have no mass and I'm pretty sure occupy no space (i.e. photons have no volume). Edit: Since photons are at the quantum level, we can't actually fathom what it would mean for them to occupy space. But on this thread someone pointed out that there ...


5

The translucent sheets and optical fibers are being lit from below. Light enters one end with a relative small angle incidence. Once inside the sheet/fiber the light experiences total internal reflections multiple times because it hits the sides of the sheet/fiber with a large angle incidence. It's not until the light gets to the far end of the sheet/...


4

One way of thinking about it is to consider light as made up of lots and lots of photons. You're not going to see destructive interference unless the polarizations of all of these photons are lined up and they all undergo destructive interference at the same time. If they're random, some of them will interfere destructively, some of them will interfere ...


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

In a vacuum all frequencies and amplitudes of light travel at the same speed of c = 299 792 458 m/s. Frequency is equivalent to colour. Amplitude relates to intensity. When light travels in material mediums (air, water, glass, etc) it travels at a slower speed v < c which depends on frequency. The ratio of c/v is what we measure as the refractive ...


3

There are two effects at work here to form the Color Afterimage. The blue light stimulates the S-type cone cells most, and they simply "tire out". Local supplies within the cell of ATP become run down, and the cell cannot therefore signal as often or as effectively. When you looked away into a more "balanced" light, white light that would normally fire all ...


3

A wave is a solution of a wave equation of the form $$\frac{\partial^2\varphi}{\partial t^2}(x,t)=c^2\frac{\partial^2\varphi}{\partial x^2}(x,t),$$ where $\varphi$ is the so-called wave function which represents some "displacement" from the equilibrium. The solutions of this equation are of the form $f(x\pm ct)$. This represents a traveling wave with speed $...


3

This is essentially the same question as "how does a tiny dipole antenna transmit a wave of several kilometers wavelength"? If you're designing an antenna to transmit a wave, you can indeed get a dipole a few centimeters long to transmit low frequency RF, say less than one MHz. Many commercial transceivers do this; it's not ideal, and you need to drive the ...


3

The Doppler effect proves that light can behave like a wave under some circumstances, but it doesn't prove that light is a wave. Likewise the photoelectric effect proves that light can behave like a particle under some circumstances, but doesn't prove that light is a particle. Light is described by quantum field theory, specifically quantum electrodynamics,...


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. ...


3

It is an optical illusion that uses stroboscopic light. It is the same principle that allows you to see still pictures into a movie. For instance, when you see the drops floating still, the drop is not the same, the light frequency is synchronized with that of the the falling drops, so that every time the light is on it shows you the image of a different ...


3

Based on my own anecdotal evidence, it doesn't. Several years ago there was a partial solar eclipse in my area. I don't remember precisely how much of the sun's disk was covered - it wasn't much, surely nowhere near 90% - but I do remember getting out of the house in the morning, thinking "hmm, it's quite dark today", then having the eerie realization that ...


3

To see what's going on, it's enough to do this in two dimensions, with the Lorentz form $\pmatrix{-1&0\cr 0&1\cr}$. (I've set $c=1$.) The Lorentz group is the group that preserves this form. A typical element is $$\pmatrix{\pm\sec\theta&\tan\theta\cr \tan\theta&\pm\sec\theta\cr}$$ where $\theta$ runs through the open interval from $-\pi/...


2

It's like this: The speed of light is constant c. Now, during the time when one electron 'jumps' from a higher state $\psi_2$ to lower state $\psi_1$, it oscillates from those states at a frequency equal to $\nu$ in $E = h\nu$, the energy of the photon released. The oscillation induces the formation of electromagnetic waves. So, the frequency of a light ...


2

You're simply meant to solve your equation $ n_1sin(\theta_1) = 1$ for $\theta_1$ at one wavelength of light. $n_1$ is (assumedly) known at that wavelength, and $\theta_1$ is the unknown critical angle. I don't think I've ever seen an undergraduate physics question - at least not about Snell's law, that asks a student to work with wavelength varying ...


2

I don't know if this is kind of answer you're looking for, but you could measure the temperature difference between the air and the open ground (as dark as possible). Over the course of a day they should diverge because the ground absorbs so much more radiant energy than the air. By sunset they will have equalized some, but by the following sunrise they will ...


2

Based on Light Waves As I am sure you know, a photon (described in wave terms) has both oscillating electric and magnetic field (sine waves) at right angles to each other. Obviously then, if the electric field is drawn vertically , the magnetic oscillating field is drawn horizontal to it. The field always wants to reach a ground state of zero ...


2

Also they are closely related in empirical life, they are different and the concepts come, I would claim, from different theories. Light is electromagnetic radiation in the visible spectrum which is produced by moving charges according to Maxwell's equations. Heat is a thermodynamical concept and usually describes energy transfer that is not work (or ...


2

Not surprisingly, physicists have looked for variations in the speed of light as a function of frequency in vacuum. The state of the art in 1972 can be found in Z. Bay and J. A. White, 'Frequency Dependence of the Speed of Light in Space', Phys. Rev. D 5(4) 796-799 (1972). Using data from pulsar emissions (radio, visible, x-ray) and other sources (see paper),...


2

To answer this, You can't add velocities as in Newtonian mechanics. In relativity, If frame B is moving at velocity u and C is moving at velocity v both with respect to an inertial frame A, then relative velocity between B and C is $\frac{u+v}{1+ uv/c^2}$. So substituting your arguments here would now leave us with relative speed less than c. Just because ...


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 (...



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