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1

Short answer - Frequency ($\nu$) stays constant since it is a characteristic property of the source. Using the relation $v = \lambda \nu$ in the definition of the refractive index, $n = c/v$, your required answer follows.

7

You don't even need highly specialised equipment to see the colour separation of the sun at low sun angles, a decent zoom lens on a camera will see it, and it's the origin of the "green flash" effect as the sun drops below the horizon. This site offers a good image: http://www.atoptics.co.uk/atoptics/gf15.htm

5

There are better answers than this, but I just want to contribute. As Joshua said, The quantification of how much light bends when transferring from one medium to another is called the "index of refraction," and air's index of refraction is very very close to that of a vacuum, so the bending of the light is very small, and the spreading apart of the ...

1

If we incident a monochromatic light (assume red light) on a glass layer starting from thinner to gradually thicker and thicker layers of glass, partial reflection increases to $16\%$ and returns to zero-a cycle that repeats itself again and again. If the the layer of glass is just the right thickness, there is no reflection at all! And it is to be noted ...

1

The fact that you see the sun as red means that the shorter wavelengths (green, blue, purple, etc.) were significantly attenuated as the sun rays traversed the atmosphere, due to having undergone scattering. It seems to me that while some rainbow effect indeed theoretically takes place, the extent to which it happens is relatively small. Adding this to the ...

15

First, it must be said that the picture you provided in your question is extreme. The concept of light bending is true, but the amount that the light bends is nowhere near as large as the picture shows it. The quantification of how much light bends when transferring from one medium to another is called the "index of refraction," and air's index of ...

1

Answer: you can ignore the coating (assuming monotonic index of refraction); light does not exit if incident ray is beyond critical angle Reasoning: Snell's Law states $$n_1 \sin \theta_1 = n_2 \sin \theta_2,$$ where $n_1$ and $n_2$ are indexes of reflection within media $M_1$ and $M_2$, and $\theta_1$ and $\theta_2$ are the angle between the normal and ...

0

Google didn't immediately come up with anything significant for "Ludvigsen's methodology", but let me give this a shot nonetheless. Sound is a propagating pressure wave. So as it goes by, the pressure increases, then decreases, then increases again, etc. Pressure increasing means the particles in the material (typically air) are closer together for some ...

1

You need to learn about the Eikonal equation and its implications. When the electromagnetic field vectors are locally plane waves, i.e. over length scales of several wavelengths and less they are well approximated by plane waves, then the phase of either $\vec{E}$ or $\vec{H}$ (or of $\vec{A}$ and $\phi$ in Lorenz gauge) can be approximated by one scalar ...

2

A complete treatment of your question (more than you ever wanted) is given at http://homepage.tudelft.nl/q1d90/FBweb/diss.pdf, especially section 2.1.1 "Differential equation of light rays in inhomogeneous media". Trying to extract the most useful expression from that dissertation, I believe that the equation you are looking for is: \nabla \Phi a = ...

0

This same question confused me a lot "when I was just a lad". Your eye responds to the frequency of light, not the wavelength. Whatever happens to the wavelength between source and retina is immaterial to the color. The frequency does not change, as you point out in (4).

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