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2 votes

Complex Refractive Indices, Absorption, and Transparency

A wave propagating in a uniform medum in the $+z$ direction can be expressed as (the real part of) $$E = E_0e^{i(\omega t - nk_0 z)}$$ where $\omega$ is the angular frequency, $k_0 = 2\pi/\lambda$ is ...
Puk's user avatar
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3 votes
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Lenses and missing reflection

The ratio of power reflected from an interface to the power incident is called reflectance. For a smooth (as in not rough) interface, it can be calculated using the Fresnel equations. For a ray ...
Puk's user avatar
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1 vote
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Fresnel Equations, Refraction, and Metals

This seems to make sense, however I have also read other sources stating that metals have no refracted beam at all. Is this meant as short-hand for saying that the refracted light is immediately ...
Puk's user avatar
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0 votes

Keplerian telescope with parabolic mirrors

This is what I meant in the comment, sorry for poor drawing skills:
José Andrade's user avatar
-1 votes

Energy of evanescent wave in total internal reflection

Those are two different approaches to look into the problem. We know that the reflective coefficient in total internal reflection is 1, which means that all energy is reflected. This is true in the ...
ondas's user avatar
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0 votes

Using Fermat's principle to derive the Cartesian paraboloid

With the hint from @vincent_fraticelli, I was able to figure it out. I modified the coordinate system a bit, as shown in the figure below. \begin{align*} |R_1P|+|R_2P|&=2f+d\\ \sqrt{(f-x)^2+y^2}+...
nwsteg's user avatar
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1 vote
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Using Fermat's principle to derive the Cartesian paraboloid

I think you can take $a=0$ without losing generality. Then I feel like your distance $R_2P$ is wrong.(length measured horizontally) Finally, the equation of the parabola you indicate assumes $y=0$ at $...
Vincent Fraticelli's user avatar
0 votes

Plastic cup filled with water shiny surface

You are seeing total internal reflection: when light bounces against the side of the cup, due to the angle and water/air index of refraction diffrence there is no refracted ray that would leave the ...
Anders Sandberg's user avatar
1 vote
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Reflection of light by edges and corners of mirrors

You are asking about the ray that strikes the join, but a ray is not a physical object. Instead it is just a direction i.e. it is the normal to the light wavefront. Suppose we draw the point source, ...
John Rennie's user avatar
2 votes

Can any mirror arrangement give an optical advantage?

You've correctly observed that you can't get any advantage from flat optics. You might get some advantage, however, through curved optics (lenses and/or focusing mirrors). For example, a telescope (...
Gilbert's user avatar
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2 votes

Can any mirror arrangement give an optical advantage?

An optical device which magnifies small changes in direction of light is the telescope. A telescope can be made with a pair of lenses of different focal length, or by using a curved mirror and lens, ...
Andrew Steane's user avatar
12 votes
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What is the albedo of our Sun?

Yes, the sun is close to a black body and would absorb the laser pointer light. If one looks at the solar reference spectrum it is almost a perfect blackbody spectrum (but not quite - there are a few ...
Anders Sandberg's user avatar
2 votes
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Reflection from a spherical surface

The difference between $s' = VI$ and $x = QI$, $s' - x = VQ$, is equivalent to the length $VC - QC = R - R\cos\varphi$. Substituting the Taylor expansion $\cos\varphi \approx 1 - \frac12\varphi^2$ ...
Er Jio's user avatar
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0 votes

Free particle encountering an infinite potential barrier

If your wave is going toward an infinite wall it cannot penetrate it, therefore the wave will be completely reflected (since there is no probability to pass through the potential). For instance, if ...
Davide's user avatar
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