Tag Info

New answers tagged

4

You have created a rather poor pinhole camera (camera obscura). You can see an "image" of the sky, a green space (trees, grass?) and even a reddish shape that may be a house, a drive way, or something similar. I think it's your driveway. This is not diffraction or refraction - it's geometrical (classical) optics. Because the hole is pretty big, you see a ...


0

Different wavelength light diffracts at different angles.


2

Yes. The boundary conditions for Maxwell's equations gets you this. This reasoning is much more simple than it sounds. High refractive index means phasefronts of a plane wave are nearer together than for a low refractive index. When the waves in both mediums line up at the interface, the spacing between the intersection of the phasefronts and the interface ...


4

In general, reflection and refraction happen when light passes from one medium to another. You can see this if you see your own reflection in a window. Now, as a light ray approaches the critical angle, not only does the refracted ray get closer to the surface, but the amount of light transmitted gets less and less. At the critical angle, the refracted ray ...


1

If the light ray is normal to the surface, 100% of the light is transmitted. As the light ray bends, as in your part (b), a percentage of the light will be transmitted (refracted) and the remaining will be reflected (at the incidence angle). Very near the critical angle $\theta_c - d\theta$, likewise, some of the light will be transmitted (refracted almost ...


4

It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ...


0

1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body). 2) It will be reflected back and forth, but only a finite amount of time. This is ...


1

Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone ...


0

This question has another interesting aspect which has more to do with neuroscience than physics: why do we perceive metals with a neutral colour (such as silver) as grey, even why they are shiny and therefore simply reflect the colours of their surroundings? One answer is that such metals always have some roughness and therefore scatter light from a range ...


4

The following passage is extracted from Stephen Hawking's book "A Brief History of Time": In fact, various contemporaries of Newton had raised the problem, and the Olbers article was not even the first to contain plausible arguments against it. It was, however, the first to be widely noted. The difficulty is that in an infinite static universe ...


7

Who is interested can find detailed information at wiki, or here The problem is known (as you added in your edit) as Olbers' paradox, and was posed already in the mid 1500's, by Johannes Kepler in 1610 and even later by Edmond Halley in the eighteen century, and curiously, even the novelist an poet Edgar Allen Poe anticipated possible explanations as to why ...


13

I'm going to respond to (v1) of the question, which asks why the night sky is dark (black and unlit) compared to the day sky even though there are many light sources at night. The updated question references Olber's paradox, which has been answered many times before. Like most things we see in everyday life, there are a number of reasons contributing to ...


18

The sky does not skip over the green range of frequencies. The sky is green. Remove the scattered light from the Sun and the Moon and even the starlight, if you so wish, and you'll be left with something called airglow (check out the link, it's awesome, great pics, and nice explanation). Because the link does such a good job explaining airglow, I'll skip ...


18

Note well: What we perceive as color is bit of a tricky subject. This is a different question, one that has been asked and answered multiple times at this site. Per the typical human eye response, sunlight at the top of the atmosphere is about as "white" as "white" can be. Some of that incoming sunlight is reflected back into space, some is absorbed by the ...


8

The hand waving explanation in your question is called Rayleigh scattering Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle therefore becomes a small radiating dipole whose radiation ...


5

The sun is technically green because the peak of its black body spectrum is near green wavelengths. When light scatters parallel to the plane of incidence (i.e., during the day time), it is blue-shifted. When light scatters perpendicular to the plane of incidence (i.e., sunset or sunrise), it is red-shifted. The light that is not scattered but makes it ...


3

Today we know that Collins is wrong. He appears to be unaware of Newton's finding, and of course, advances made after he wrote his book.


2

It's because of DLP projectors. In such projectors a 3-color wheel rotates in front of the light thus projects only one color at a time but very rapidly. Search google for DLP rainbow effect. It's common to see an be bothered by this. Source: http://en.wikipedia.org/wiki/Digital_Light_Processing#The_color_wheel_.22rainbow_effect.22


0

Summary: The best way to see the water is to try and look for the reflection of a light source (sun or light bulb) at a large angle of incidence. More Details: You see the water on your floor because it reflects light differently than the dry surface of the floor. Namely, the water forms a smooth enough surface that the light reflected off of it can be ...


0

It is on a small scale the same as a mirage in a desert. A mirage is a naturally occurring optical phenomenon in which light rays are bent to produce a displaced image of distant objects or the sky. In contrast to a hallucination, a mirage is a real optical phenomenon that can be captured on camera, since light rays are actually refracted to form ...


0

"From this source, I gather that he did argue that the light particles sped up when entering a more dense medium. However, it just doesn't make sense." It does make sense. They enter the medium at a higher speed, then slow down. If Newton had said they continue to move at a higher speed within the medium, he would have been wrong. But I don't think he said ...


2

Your friend is correct. It's not specular, because that would mean it's a mirror, and you'd only see the image if you were at the angle of reflection. Further, a mirror does not act as an image plane, so you might have difficulty ( :-( ) perceiving the image. Basically you'd need another lens to re-image the source. The diffuse surface acts as an image ...


4

Like all good physicists Newton proposed a hypothesis for why light refracts when it crosses a boundary between different refractive indices. And his hypothesis makes a certain amount of sense: If the initial velocity of the light is $v$, then the velocity parallel to the boundary is: $$ v_p = v\sin i $$ and the velocity normal to the boundary is: $$ ...


2

I'm afraid that it is not easily possible to take the luminous flux and obtain the insolation (as radiant flux). Here's why: The luminous flux $F$ is calculated from the radiant spectral power distribution $J(\lambda)$ by weighting each wavelength with a luminosity function $y(\lambda)$ as per $$ F = c \int J(\lambda)y(\lambda)\mathrm{d}\lambda$$ where ...


8

You don't have a fluorescent light but an LED light that time-multiplexes different colors to achieve white. The fan blades act as a stroboscope and make the switching frequency of these LEDs visible. Or... I am wrong! So back to the drawing board: I did the experiment. Most fluorescent lights in my house do not show this effect, at all, but one does, ...


2

There exists a good answer to what is color which stresses that the question is mainly about physiology of the eye. I will address the last part: Basically, what properties of the paint material makes it absorb the red wavelength the least? Light has a specific wavelength . If that wavelength impinges on the eye receptors, it gives the colors we have ...


1

The energy of a wave is h*f. The energy in an electromagnetic wave is where E is the electric field of the wave as it moves with velocity c. How i imagine this is each cycle consisting of 1 quanta. There are no quanta in the sense you are imagining it. The classical wave above emerges from a zillion of quanta, called photons. For a ...


2

The light that you can see in the air is actually light that has been reflected from particles in the air (dust, water vapour, etc.) towards your eyes. So when the beam is pointing away from you, some of that light will still bounce off those particles and come to you; this is what allows you to see the beam. However, when you look at the water, it acts more ...


0

If the beam light or what you say as cone is above the water what you will see is the reflection of light from the lighthouse and not the beam, as you can see it produces a cone shaped light with the help of a parabolic shaped reflector it have or either it uses other type, the point is that the light is focused and under tha law of reflection for specular ...


1

One has to distinguish the two frameworks: the classical, light; the quantum, photons. The classical electromagnetic wave, of which visible light is a part of the frequency spectrum, emerges out of zillions of photons, the quantum of light. This happens because the photon has an energy E=h*nu, where h is the Planck constant and nu the frequency of the ...


0

There's quite a lot to say about this, 18490. The tl;dr for Mythbusters fans is at the end. Most materials are somewhat transparent. This is why you often need two or three coats of paint on a wall. The light enters the material, travelling many microns before it is eventually absorbed or scattered back. The result is usually that most of the light is ...


0

At around the $1\mu$m particle size you're in the Mie scattering regime, and this makes life hard because there isn't a simple analytic formula for the cross section due to Mie scattering. However if you're prepared to consider particle sizes significantly smaller than the wavelength of light, say $0.1\mu$m and smaller, then it's easy to show there is an ...


0

Good question. The key is to realize that a mirror-like surface send rays of light into your eyes directly from the source; they just bounce simply off of the mirror. What you see is the source, and the color of the source. Not so for a diffuse object. In this case light from the source hits the mirror and does not bounce directly into your eye. It ...


6

Physics first: light and radio waves are the same thing, just at vastly different frequencies. Radio works between $1\text{ Hz}$ and approx. $(10^{11}-10^{12})\text{ Hz}$ (at the moment). The frequency of visible light is around $10^{15}\text{ Hz}$. The range between these frequencies is usually called infrared radiation and is of little interest for ...


6

Frequency modulation in VLC is not done by changing the frequency of the light itself. The light intensity is pulsed at a high frequency to create a carrier wave. This carrier wave is then frequency modulated to transmit the information, typically using frequency shift keying. That's why it works with white light, where the frequency of the light isn't well ...


3

Any waves that appear on one side of an interface must immediately leave on the other side. There's no room to queue them up. This means that in a frame where the interface is at rest, the frequency on both sides is always the same. Once you have that, then it becomes obvious that the wavelength in air will always be the same as well. So from the ...


1

Yes, the term "shadow" can refer also to something or (dare I say) someone that is dark, shady, inconspicuous, etc. One can also use it as a verb; to shadow someone is to follow them closely. Like "I'm having the new guy shadow me for a while until he learns how to do everything".


5

Yes! Any beam that is blocked by an object will basically make a shadow. For example, the IceCube detector can see the moon's cosmic ray shadow.


1

Yes, for example a Crookes tube shows an electron shadow. The area I live (Chester, UK) is in a rain shadow.


2

For example, acoustic shadow (http://en.wikipedia.org/wiki/Acoustic_shadow ).


2

You can think of super massive object like black holes which can bend light. Near the event Horizon you could get a 180 degree turn for light and thus see the earth back in time. But I do not think this is practically possible as earth is small and dark (compared to stars) and this layer would get compressed really thin as some small deviation in the ...


4

The term dispersion refers to the speed of light in a material having a dependence on frequency (or equivalently wavelength). The refraction angle's dependence on frequency is caused by the material dispersion, not the other way around. In all materials the refractive index will have dispersion but it's often the case that certain materials in certain ...


4

The whole point of Snell's law is about taking into account wavelength! Remember that the fundamental property of the light is its frequency. Wavelength, on the other hand, is not a fundamental property of a beam of light since the wavelength changes all the time as it passes through various media. The index of refraction of a medium is a dimensionless ...


7

Snell's law is given by $n_1 \sin i =n_2\sin r $ where $n_1$ is the refractive index of the material the light is initially in and $n_2$ the refractive index of the material the light is going into. These are not constants for a given material and change (although very slightly) with the wavelength of light. Thus for different wavelengths of light the ratio ...


1

The "slowing" of light in a medium can be entirely explained using a classical wave-based approach. An incoming EM wave wiggles the electron clouds around the atoms in the material. These electrons clouds re-emit a much weaker EM wave having a very small amplitude. This re-emitted wave is 90-degree phase shifted from the original wave but superposes with ...


2

This is a nontrivial problem in materials science. People have done a lot of work both with different materials and different surface structures to create "ultra-black" absorbing structures for use in optical systems. A big part of the problem is the statistical nature of photon absorption. There's only a certain percent (<100) chance of absorbing a ...


1

It happens in Special Relativity because of time dilation. The time between crests when wave packets are emitted as observed by the emitter, is different to the time between crests when the wave packets arrive and are observed by the receiver. So the observed frequencies are different if there is some time dilation effect between the two observers. In GR, ...


7

Your hypothesis is basically correct, in theory, even in a vacuum. Light consists of electromagnetic radiation. According to classical electrodynamics, electromagnetic fields in a vacuum are linear, which means that one light beam will pass right through another, completely unaffected. But according to quantum electrodynamics, at electric field ...


-1

Yes and no. Photons don't interact in free space. So a beam of light can't block another beam of light in vacuum. Photons can interact due to the nonlinearity of the medium. So it's plausible to block another beam of light if you have the right mediators. It's however not the light itself becoming a solid. See, for example, electromagnetically induced ...


0

The process of light propagation is described by the Maxwell equations. $$ \nabla\cdot{\bf D} = \rho $$ $$ \nabla\cdot{\bf B} = 0 $$ $$ \nabla\times{\bf E} = - {{\partial{\bf B}}\over{\partial t}} $$ $$ \nabla\times{\bf H} = {\bf J} + {{\partial{\bf D}}\over{\partial t}} $$ These equations say (in simple terms) that: change in the electric field is causing ...



Top 50 recent answers are included