# Tag Info

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 ...

0

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 ...

0

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 ...

0

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.

0

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 ...

0

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

0

It's a no from me. For light to bend the other way, light in antiwater would have to have phase velocity greater than $c$. This is possible in some systems (called metamaterials) but the optical properties of antiwater would have to be completely different from ordinary water - which is ruled out by existing experiments which show that positrons and ...

0

I would say no. If everything is anti-*, also the refractive index will be. Thus, being both negative the resulting bending of light will be the same. As an example take an electric field and throw an electron through it, the e- will be deflected in one direction. Now, if you take the anti- of everything the E-filed will be essentially reversed but the old e-...

4

We think antimatter refracts light like “ordinary” matter, but we don't know for certain. As the Wikipedia article on antimatter says: There are compelling theoretical reasons to believe that, aside from the fact that antiparticles have different signs on all charges (such as electric and baryon charges), matter and antimatter have exactly the same ...

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 $... 0 In the context of an experiment to measure the refractive index of glass or water the other medium is air whose refractive index is taken to be one. Hence the result is, to a very good approximation, the absolute refractive index. 1 A Remark (not a demonstration): Same equation by analogy (the speed (v) of light in the medium is replaced by its speed in vacuum but in a moving reference frame): Assuming that an observer is at the origin of the coordinates of the frame of reference K' and that the latter moves with respect to K parallel to the x-axis, i.e. y=y', z=z' At the time t=0, the ... 6 For the purpose of the question, I find it helpful to formulate the laws of refraction in terms of wave vectors$\vec k$. With it a plane wave (which we are assuming here) can be easily described by a complex exponential $$\psi(\vec r,t) = \psi_0 e^{i(\vec k\cdot \vec r-\omega t)}$$ We do not need this representation explicitly, but are rather only ... 3 Consider a ray impinging on the mirror on the right. Some of it is transmitted, and some is reflected, bounces off the left mirror, and reaches the right mirror again. The wavelength transmitted by the interference filter is such that the optical path difference between these two rays is equal to one wavelength so that they constructively interfere. This ... 1 We can rewrite the differential equation as: $$\frac{1}{\lambda}\frac{dn_{eff}}{d\lambda}-\frac{n_{eff}}{\lambda^2}=\frac{d}{d\lambda}\left(\frac{n_{eff}}{\lambda}\right)=-\frac{n_g}{\lambda^2}$$ So$$\frac{n_{eff}}{\lambda}=C-\int\frac{n_g}{\lambda^2}d\lambda$$for some constant$C$, which you can establish using suitable boundary conditions. To solve this ... 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 ... 1 It might be that aberrations cancel better with the top one, but that is only a guess. You could buy an optical design package if you want to see. Or you could try it out. See this for a Veritasium video on a single mirror Schleren system. How To See Air Currents 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, ... 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. 8 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 ... 2 Heavy glasses are mainly heavy because the use elements such as lead with many nuclei. As a result, these materials also have more electrons per atom, especially in the higher (more loosely bound) orbitals. The looser bound the electrons, the more they will react to the electric field wave. How exactly this increases the refraction index I don't know, but I ... 3 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 ... 1 Edit This question is badly written and (if they gave the correct answer as 4.66m as you put in your question) then not physically accurate at all. The question asks... A lightsource of diameter 10cm is placed 2m underneath the surface of a pool, a person 5m away from the edge of the pool saw a circle of light emitting from the lamp. Calculate the maximum ... 4$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\lara}[1]{\langle#1\rangle} \newcommand{\lav}[1]{\langle#1|}...

45

The normals in consideration for the incident and emergent rays are different. For simplicity, take a monochromatic beam of light incident on a prism, as shown in this figure: When light is incident on a medium with a higher index of refraction ($n$), it bends towards the normal. When light is incident on a medium with a lower $n$ it bends away from the ...

12

This is how refraction of light in a medium works. The phase velocity $v$ of light changes transitioning from one medium to a different density medium according to its refraction index $n$ and the refraction angle to the incident is dictated by Snell's law: The light then exiting the medium and returning to the initial medium regains whatever phase velocity ...

Top 50 recent answers are included