New answers tagged

1

Newton literally wrote the book on Optics and knew perfectly well how to predict refraction. He did posit a speculation on the reasons for refraction - that is, light was composed of tiny, very subtle pieces which were subject to kinematic laws and had the tendency to accelerate towards regions of higher density, an interaction in which they exchanged some ...


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This is not how photons are viewed. They are not moving particles in the Newtonian sense, and this has to do with the wave-particle duality of quantum theory. Light is absorbed by matter like a particle but propagates like a wave. Photons are the quanta of energy passed to matter. So refraction is an example of the wavelike nature of light, not the curved ...


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When the sun is behind the observer, a beam of light from the sun which passes close to the observer points to the center of curvature of the rainbow. With the sun somewhat above the observer, this center will lie below the horizon. A parallel beam (at some distance to the observer) strikes a raindrop, is refracted into the drop, reflects from the back ...


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So what we actually see as a rainbow? We see an image of the sun in a giant mirror with a lot of chromatic aberration and a lot of spherical aberration. However, as the round shape of the drops only steers the reflected light back in a cone shape, there is a specific angle between the sun, the drop and the observer to make it happen.


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Here's what I'm able to comprehend: When visible light between (400-700) enters a raindrop normally as a spherical, physical object that allows light to pass, such as being transparent, the more transparent it is, the clearer the refraction occurs. Because of this, it bounces off of the back of the raindrop, and if circumstances permit, if the angle is ...


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Ok so light splits like it does in a rainbow when it passes through a prism such as a raindrop The reason it fans out and doesn’t just reach your eye as it normally would is because of lights wave like properties. Because white is comprised of all colors. And each of these colors (wavelengths) of light refract slightly differently. The colors fan out in ...


1

This equation \begin{equation} n_g \cos\alpha + n_a \left(1 - \cos\theta \right) = \frac{N \lambda}{2t}. \end{equation} is wrong because you dropped the $-n_gt$ that was in your starting equation. The correct equation should be \begin{equation} n_a \left(1 - \cos\theta \right) - n_g \left(1- \cos\alpha\right)= \frac{N \lambda}{2t}. \end{equation} As ...


2

An interesting paper about the same issue can be found here. Same formula as the Lindqvist Paper I've already cited : https://inis.iaea.org/collection/NCLCollectionStore/_Public/47/072/47072985.pdf?r=1


6

Those equations are written for the $E$- and $B$-fields at the boundary. The $E$-field has a discontinuity at the boundary, but the $B$-field is continuous at the boundary, and this can be deduced by Ampere's law. We need to make some notes first. In the video, Don Lincoln is considering the case of $p$-polarized light coming to the air-glass interface. In $...


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Always see the direction of rays of light the direction vectors in the direction of light rays represent a positive direction and direction vectors opposite to it negative and take pole of the spherical mirror for measuring distance and distance above principal axis is conventionally positive and below it negative. But sometime I also find it difficult to ...


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Tips: rainbow comes from both refracted and reflected light. Side question first: In the case of a rainbow you are looking at an image that's created by the reflection of light (indeed it can only be seen on the opposite side of the sun, with the observer being at 42°). So it will always be a 2D image (like your image on a mirror). To be more clear, the ...


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If the image is smaller than the hole you observe just the shadow of your card.The light rays coming though are parallel for all practical purposes.


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First of all, some define lateral displacement as the distance along the direction parallel to the surface and the quantity you refer to as the beam displacement. To me that makes most sense in terms of how it effects experiment design and analysis. But, in any case, both are easy to calculate. Here is a simple way to think about it. By inspection, we ...


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Your diagram is correct as shown for a simple lens while in a real camera lens the line would go through the center of the aperture diaphragm. The diagram also illustrates the so called crop factor. A smaller sensor with the same lens would have a narrower field of view that would correspond to a longer focal length lens used with the larger sensor.


1

There are two problems in your solution. First is that you simply misapplied your equation and multiplied by 2 instead of dividing. But the second, and rather more significant, is that as the ray propagates inside the curved layered medium, even if the index of refraction is constant, $\theta$, being measured from the local radial direction, also changes. It'...


2

In this publication you find the Müller Matrices for Reflection and Transmission : H. Lindqvist et al. / Journal of Quantitative Spectroscopy & Radiative Transfer 217 (2018) 329–337. In order to derive them, you just assume that for each mode (parallel = TM, perpendicular = TE), you have a linear relationship between the R ot T field and the incident ...


5

"Infinity" here is actually a shortcut for saying "grows larger than any value you can name when the conditions approach condition X"; that is, it describes the behavior of an iterative procedure or an algorithm rather than being a static number (sorry, I'm a programmer). In this case the procedure is to make the angle between two lines ...


4

The natural home for the geometry of plane curves is the projective plane, where everything is really much simpler. For example, a curve of degree $n$ and a curve of degree $m$ always meet in exactly $mn$ points in the projective plane (with a few provisos about exactly how to count), which turns out to be extremely convenient. Lines are curves of degree 1, ...


1

Not every case of the Snell's law equations will correspond to an equivalent triangle representation. Think about it, it is necessary for the sides and angles of any triangle to satisfy the law of sines, but its nowhere necessary to have a triangle for every numerical combination of thetas and c's that satisfy the law of sines form of equations. For example, ...


13

In Euclidean geometry, parallel lines never meet. This is the very definition of parallel. So if the object is at the focus, the reflected rays indeed will never meet (in an ideal Euclidean world). So why do we say they "meet at infinity"? It turns out, it's just a notational convention. To borrow from another answer of mine: When physicists say ...


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If you align your viewing direction parallel to some set of parallel lines, you will visually see them ending at some "point" at infinite distance. The typical example is railroad tracks. If you take lines that are not parallel, then no matter what perspective you take, the visual point of intersection (if there is one) will always be a finite ...


26

Infinity is not a real distance or an actual number. It's used in mathematics when describing limits as a parameter increases without bound. Parallel lines, by definition, never actually meet in a flat plane (there are non-Euclidean geometries where they do meet, and these are relevant when General Relativity is taken into effect, but not for classical ...


12

The fact that parallel lines meet at infinity becomes quite intuitive when thinking about what "infinity" actually means in a 2d plane. While the real numbers $\mathbb R$ are often compactified using two points, namely $+\infty$ and $-\infty$, to preserve their ordering in the compactification, in 2 dimensions, ordering does not make much sense (is ...


1

After going through the wonderful answers and comments, and some more reading, I think I've managed to clarify some of my doubts. Let us imagine the following scenario. I'm standing in front of a lake. There is a mountain behind the lake. The rays of light from the mountain reach my eye, and form an image on the retina. My brain interprets this, by assuming ...


4

The eyes can only perceive the rays which enter the eye. When viewing objects underwater through the flat face plate of a diving mask, the rays entering the two eyes are refracted at the glass, and the object being viewed appears to closer than it actually is.


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After reflection the rays seem to be coming in the direction that they would if there were an object on the left. image from here The brain chooses the most likely cause for the rays to be coming in that direction - and that's that there is an object on the left. Exactly why it does that is a psychological question, but when humans were evolving mirrors ...


1

F$_1$ and F$_2$ are the principal foci. They are the same distance, f, (the focal length) either side of the optical centre of the lens. I can therefore make no sense of your equation "1/F= 1/F1 +1/F2". The rays on the diagram are roughly correct. An object in a plane at distance 2f from the optical centre of the lens should form an image on the ...


0

You are right, magenta doesn't exist as a wavelength: it's a result of mixing violet (the shortest wavelengths of visible light) and red (the longest) - in theory the colour we should see should be somewhere around green but for some reason our brain mixes the two and sees magenta.


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It seems the core of OP's question is the following. Question: When we derived Snell's law for a single interface from Fermat's principle, we held 2 points fixed. How can we then use repeatedly Snell's law for a double interface, since we are not allowed to hold 3 points fixed during the variation? Answer: Although it is true that we're only allowed to ...


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