New answers tagged reflection
2
The spacing between the mirrors that form a laser's resonant cavity determine the set of wavelengths that the laser can amplify and emit. The spacing constrains the wavelengths $λ$ to those that are an integer fraction of the cavity length $L$:
$λ = \frac{2L}{q}$, with mode order $q \in \mathbb{Z}^+$
The range of emitted wavelengths is also constrained by ...
0
when radiation (light spectrum) falls on blue colored body its internal energy get increased due radiatic interaction. This energy difference is equivalence to wavelength of blue spectrum. Consequently, it absorbs all the spectra initially blue too but eventually emits the blue spectrum only rather than other and hence object is found to be blue in color.
2
A nice way to answer this question is to do the calculation in the rest frame of the mirror. Light propagating in the $y$ direction in the lab frame will propagate in some other direction in the rest frame of the mirror. It will then reflect off in the ordinary way in that frame (angle of reflection equals angle of incidence). Finally, one transforms back to ...
0
The dark condition is that one that you shall find, that is under what circunstances the light do not enter in te 2nd enviroment with the restrictions a) and b) , one may be that in de $n_2$ all the light is absorbed, so in the 2th enviroment will be no ligth and from the 1rs material you will see dark (but this is more complicated because you need to ...
1
Yes, it is possible. Optical fibers confine light via total internal reflection, with very low losses. Still, the losses are enough that the light intensity decays substantially after a few tens of milliseconds.
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(I don't have enough rep to comment)
The equation that you have used is derived for normal incidence (specifically for elastic collision assuming light as a particle).
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It means that when primary wave is expressed as
$$
y_p(x) = A_p\cos (\Omega t - kz),~~~A_p > 0
$$
so it moves in direction of $z$-axis, and reflecting wall is at $z=0$, then reflected wave is expressed as
$$
y_r(x) = A_r\cos (\Omega t + kz +\pi),~~~A_r > 0
$$
which means that at $z=0$, the reflected wave value $y_r(0)$ has opposite sign to incoming ...
4
The spectrum you get out depends entirely on the physics relevant to the material. What is it made out of? Is it crystalline or disordered? What laser intensities are involved? Different materials have different details which determine the extent to which something like a Stokes shift will occur.
Normally, if you care about emitted light with a similar ...
1
This typically happens at a perfectly reflective surface, with the 180 deg shift in E field if the surface is a smooth perfect conductor or "electric wall" (equivalent to a short circuit), and in H field if the surface is a perfect "magnetic wall" (which doesn't naturally exist, but can be equivalent to open circuit and approximated by ...
1
It’s a redshift.
Suppose you only have a single photon to begin with. It reflects from a mirror, imparting $\hbar \omega /c$ momentum. The mirror’s kinetic energy increases. Then you still have one photon but its energy is lower. Work out the math, and voila! It’s been redshifted (twice, from both the absorption and emission events, with the relevant ...
0
In terms of quantum mechanics, photon is scattered away by net electrons field of surface atoms. This implies that due to momentum conservation, almost all photon momentum is passed back from surface electric field to photon. So energy looses of photon (if any) due to field recoil would be very very small.
More interesting case is where individual atom ...
0
According to me, these figures dont' hold any value as such, the principal focus of a concave mirror can easily be found by determining the point at which any two light rays coming from infinity(from a very distant object) tend to converge or meet.
The distance between the focus and the pole of the concave mirror will then give us the focal length.
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