# Tag Info

## New answers tagged reflection

0

If you look at condensation fog through a strong magnifier, you'll notice that the fog is actually composed of a large number of hemispherical water droplets. The optical effect is caused by the fact that the tiny droplets act like lenses, scattering the light. On a vertical glass panel like a mirror, the maximum stable droplet radius is strongly affected by ...

0

Actually, total internal reflection is impossible in spherical raindrops; an internal reflection occurs, but it is not total. And the rainbow is not just an arc, it is a full disk. The disks are sized differently for each color, and are brightest at the very edge, which is what makes the colored arcs appear. For the primary rainbow, these disks are centered ...

15

Yes, there are. Such materials are called "saturable absorbers," and are (or at least, have been) used as switches in some laser designs. The one I recall is a nickel acetate dye, although there are others. Basically, the molecules absorb single photons at the laser wavelength, but when the intensity is great enough that two photons are absorbed ...

0

I thought the phones picked up the longer wavelengths due to an enlarged field, which was electrically generated, much like tv aerials needing a current..

3

A one way mirror isn't really a one way mirror. It lets the same amount of light through in both directions. It works because one side of the mirror is light while the other is dark. Suppose you're on the light side looking at the mirror, and suppose that the light side is 99 times as bright as the dark side. Finally assume the mirror lets through half ...

0

Reflected light can be though of as originating in oscillating charges in the medium. The incident light causes the charges to oscillate, and the oscillators generate the reflected light. This process happens almost instantaneously. The atoms in the medium are oscillating coherently (in step) with the incident radiation. The frequency of the light is ...

3

This is simply the law of reflexion: if you trace a ray diagram for objects reflecting in your iPad screen, you are looking at light from a virtual image that is as far beyond your screen as the real source of light is from the iPad screen. So if the lamp is, say, 3 metres behind your shoulder, when you look at the iPad screen the reflected light is exactly ...

4

You make a good point which requires us to be more careful about what Fermat's Principle says and how the proof proceeds. The upshot of what I'm going to say is The statement of the Law of Reflection must include an appropriate constraint. Here's what I mean in detail. First, let's give a precise statement of Fermat's Principle: Fermat's ...

-1

I think the path must reach on the surface. If the light reflects before the mirror, there should be something above the mirror.

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It's a simplification. Typically, you use boundary conditions to relate the fields outside and inside the dielectric that say that the normal component of the electric field $\mathbf{E}$ has a discontinuity so that the auxiliary field $\mathbf{D}$ is continuous. There is also another condition that says that the tangential component of $\mathbf{E}$ is ...

6

Let me answer this one with some drawings:

3

The answer is that we can't see a black hole, which is I would guess what you were leading up to. There are only two ways we can detect a black hole using light (of various wavelengths): if it occludes something behind it if it's surrounded by an accretion disk Point (1) is actually quite complicated because a black hole doesn't simply mask whatever is ...

2

It does not matter where the mirror is kept for you to see the entire image. A mirror with half the length of the man should be sufficient irrespective of the position. The only thing you do have to make sure is that when the mirror is brought closer, it must not be moved vertically. This can be understood from the following diagram:- The rectangle is ...

1

Black objects absorb visible light. Clear objects let visible light pass through without absorbing it. (Both black and clear objects reflect or scatter a little bit of the light, though.)

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For that you should try just the opposite you have tried earlier. the light will be reflected as shown.

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Answer from 2C Solar: There are many 3D packages that render light aka raytracing, however most don't show the light itself. One very old method is POVray, started in 1991 and latest version 2013 The 3d package Spaceclaim can be used to create your laser / mirror model then export to POVRay where you need to define the properties. Bit of a learning curve ...

1

I've heard that clear things such as the gaseous components of the atmosphere or water reflect no light, and that is why they are clear Yes, "clear things" reflect light in very small percentage. But, they do reflect certain percentage of light falling on them. If they don't reflect that small percentage of light, you would have not been able to see ...

1

There is something you should think about and then hopefully things become clearer for you: The athmosphere is only "clear" in a rather narrow region of the electromagnetic spectrum (see the nice graphic on the wikpedia page). For most wavelengths, our air is unclear and little of the incident light reaches earth's surface. Another useful example here is ...

1

Does mirror reflect with 100% perfection? No mirror will have 100% reflectivity, there will always be some absorption. This can be applied for most ideal objects, since perfection does not exist. I doubt that am i seeing the correct reflection of myself? I don't understand this part.

1

Actually the resolution of our eye is much more than the resolution of a typical camera. You may have notice that the photo of the same person looks better if the camera is of good quality.The another reason may me the 2D effect made by the camera while our eye can look through the mirror with different angles and can focus various parts of our body ...

3

It depends on the position of the sun. A rainbow does not exist at a particular location in the sky. Its relative position depends on the position of the observer and the sun. All raindrops refract sunlight in the same way, but only the light from some raindrops reach the observer's eye. This light is what constitutes the rainbow for that observer. The bow ...

0

The best is probably to give you an insightful link. You will find there an applet which illustrates the following, which is your intuition: Coming from the sun, light rays hit the droplet and enter it with refraction air to water They reflect internally in the droplet They come to your eye, following a second refraction water to air Since rays from the ...

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