Tag Info

New answers tagged

0

Yes in the second condition it is possible When the light strikes the bird the, if the mirrors are transparent the the image is formed behind the mirror to which it so slightly attached


2

You've probably heard of a curve called the parabola, and you probably interpret this as meaning it's a function something like $y = x^2$. However there is another way to define the parabola. If you draw a line (called the directrix) and then choose a point (called the focus) not on that line then the set of points that are an equal distance from the ...


2

There is also optgeo, quite simple, but could be useful in your case, you can drag and drop lenses, mirrors, beamsplitters etc. It is free software: http://jeanmarie.biansan.free.fr/optgeo.html It is also in the ubuntu and debian repositorys.


2

Can't be done, yet. But don't expect 3d printing to stand still, it's only recently been born, wait till it starts walking, and running! My hope was that there could be a vacuum process that was usable in just enough vacuum to form a perfect mirror shape over a perfect mirror mold of some kind, then start the vapor deposition on the cheap, lightweight ...


28

The highest resolution 3d printers I know of are around 1600dpi, which is a resolution of about 15$\mu m$. Telescope mirrors have to be smooth to fractions of a wavelength of light, so the resolution of current printers is nowhere near good enough. Whether 3D printers could one day be good enough is a different question, but given that the improvement in ...


1

Just blindly multiplying the overall answer by some factor isn't the way to go about it. I have an alternative proposal which may work well as a zeroth-order approximation at least. You already have the expression for the 2-layer case, and if I observe correctly, you are only concerned with the reflected part, not the transmitted part. So, a smaller ...


1

The two rainbows that are formed are the primary and secondary rainbows respectively, in order of their intensity or brightness, as you may call it. A primary rainbow is formed as a result of a three- step process: Refraction with dispersion, followed by total internal reflection and then refraction. The secondary rainbow is formed due to a four- step ...


1

The third- and fourth-order rainbows actually appear in the sunward direction — when the light makes three reflections in a water droplet it exits going roughly the same way that it entered, rather than roughly reversing direction as it does for the single and double reflections in first- and second-order rainbows. There are some photos of third-order ...


1

Rainbows are results of total internal reflection - that is, when light hits the surface of a refractive material (in this case water) at an angle that is less than the angle of refraction (which varies for different wavelengths of light), the light reflects off the surface like a mirror instead of entering the material (or, in this case, leaving the water ...


0

The d is not separation between points in reciprocal lattice. Actually, they do not even have the same units. d is the separation between lattice planes, as you said. What is related to reciprocal lattice vectors is the change (before and after scattering) in the wave vector of light: change in k = reciprocal lattice vector, which is the Laue condition that ...


2

It's surprisingly difficult to get a material that absorbs almost no light across the whole visible range but one candidate would be black silicon. This has a textured surface created by etching, and the texture means light hitting the surface is multiply reflected sideways before it gets a chance to be reflected back away from the surface. With some ...


3

Yes, to the approximate extent allowed in the real world. The condition that the material doesn't reflect visible light means that the material looks black. So consider various black coatings, for example, and ask what they do with electromagnetic waves at different frequencies than visible light. Of course that you find out that they generally reflect ...


0

I think that the most appropriate way to think about the Bragg formula is in terms of a diffraction grating. In a diffraction grating one obtains sharp peaks because there are many slits with distance $d$ between them. The derivation of the intensity maximum for the diffraction grating case is similar to the Bragg case. To obtain a diffraction grating ...


3

Yes it is possible but as BarsMonster points out it isn't like an optical mirror. X-ray reflectors are used in the construction of nuclear weapons and are critical to increasing the yield. How they work is the initial fusion reaction releases high energy radiation, this is then reflected back into the reaction mass increasing the energy levels of the ...


7

Unfortunately X-ray and gamma mirrors are impossible to build the way you think - mainly because there is much less interaction with the matter comparing to UV - it will go through all materials commonly used for making mirrors. Even for EUV light (wavelength of 13.5nm) building effective mirrors is a royal pain. As wavelength of X-Rays is very small (down ...



Top 50 recent answers are included