# Tag Info

5

The answer is "almost no" - the wavelength of the photon is virtually unchanged (in the initial rest frame of the mirror, the "lab frame"). Because the mirror is much more "massive" than the photon, it serves as a "momentum sink" and picks up almost no energy. The best way to develop an intuition for this is to consider a collision between two balls: one ...

3

If you paint the inside surfaces flat black, or cover them with black felt, a "maze" will let air through but stop 99.9% of the light.

3

It will behave as a concave mirror with a convex lens in front of it, or just as a convex mirror if the other side is reflecting (with a lens behind it that doesn't really do anything because it's blocked). You would have to do ray tracing to see more precise behavior for the former case.

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You are seeing fringes of equal thickness where light from a source is being reflected off the top of the soap film and the bottom of the soap film and then entering your eye to form an interference pattern on the retina of your eye with your eye focussed on the soap film. The condition for a maximum is $2 \, n\, t_{\rm m} = (m+\frac 12) \lambda$ where $... 2 I have done some calculations, and in my solution, the law of reflection does apply in this case. Am I right? Yes, you're right. However, I offer you to use the following equation that relates the reflected angle to the incident one as measured in$S$moving at an arbitrary velocity$v$with respect to the mirror's rest frame ($S^\prime$): (See this article.... 1 The reflected photons are entirely identical to the incident photons except for the change in direction. The energy lost due to the pressure exerted on the mirror can be accounted for by comparing the number$N_i$of incident photons to the number$N_r$of reflected photons. I.e.,$N_i > N_r\$.

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The question itself isn't entirely clear, but I believe this can be understood in terms of the Fresnel equations. Whenever a light wave hits an interface, some of the wave is reflected and some is transmitted. If the media are very similar to each other, then very little of the wave is reflected; in the limit that the indices of refraction are identical, ...

1

Therefore, the light wave is expanded as moving through space. Yes, but we are using "wave" loosely here as a classical collection of light, not as a single period or wavelength. Therefore, the light wave is expanded as moving through space. This implies that the energy of a single light wave reflected from an accelerated object (mirror) is reduced by ...

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The wavelength will change according to the Doppler effect, conservation of energy is not affected because the intensity of the light changes to match the energy of both waves

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You get reflection at every interface between mismatched refractive indices, it is only a matter of percentage. If the surface is smooth enough and there is not too much light coming in fom above the surface to swamp out the image, a reflection will be seen.

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As others have pointed out, it will behave as a lens and a mirror. I wanted to point out a particular version of this with interesting applications: the cat's eye retroreflector: The cat's eye retroreflector is a transparent sphere that is silvered on one side, much like the eye of a cat with its reflective tapetum lucidum at the back. This structure has ...

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My question is when we consider light as a photon and send it (photons) one by one, what are those things which determine which photon will be reflected and which one will be refracted It is called Quantum Mechanics. The photon is a quantum dynamical entity, it is not a small part of a beam of light. Classical light (electromagnetic wave) is made out of a ...

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One difficulty with OAPs and toroidal mirrors is that they are significantly harder to align than spherical mirrors, and they're also more sensitive to beam pointing. Without knowing more about your experiment I can't really comment on which trade-offs might work best for your specific application. I don't have much experience working with 400nm light, but ...

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