New answers tagged

0

As it was correctly pointed in the other answer, only one mode is considered coupled to the oscillator, which is why one need not use additional creation/annihilation operators. The frequency of this mode does change as the mirror moves! The wavelength of the cavity mode is however huge compared to the displacement of the mirror, so this change can be ...


2

Starting with parabolic bands. The absorbed photon has energy $h\nu$ and generates an electronic and hole at energy levels $E_2$ and $E_1$ respectively. Energy and moment balance imply, $$ h\nu = E_2 - E_1 = E_c(k) - E_v(k)$$ where $k$ is the momentum of the photo-generated electron and hole (it’s the same for both carriers), $m_c$ and $m_v$ are the ...


3

If we assume implicitly, that the polarisation is constant, we are allowed to skip this detail in or description, because the result does not change. However, if we assume that different polarisation components are present, we have to address each component separately -- assuming that you consider two orthogonal components, which do not mix. By the way, ...


0

I think I was able to solve the problem by applying the same method as mentioned here. However my solution still differs by a constant factor from the solution in the book, so maybe it is still not completely right. If you look at $h(x, y)$ one can see easily that it can be separated as $$h(x, y) = K \cdot f(x) \cdot f(y).$$ with $f(x) = e^{\dfrac{-j k x^2}{...


3

The BEC is held in a focused laser beam acting as a dipole trap (this is a useful search term if you want to look for more information on the scheme). Atoms in the beam feel a space-dependent potential which is proportional to the laser beam's intensity. If the laser is blue-detuned this potential will be repulsive, but for the set-up described in the ...


4

Unfortunately, there's no easy explanation for this. It involves some basic atomic physics, so you will need a picture of how electromagnetic fields interact with matter. Consequently, some of this may be difficult to understand, but hopefully you can ask about (or read up) whatever you don't understand. Semoi's answer is correct, but a lot of details have ...


3

If an analogy helps: Picture walking towards a plowed field, which you meet at an angle to the edge of the field. The furrows run parallel to the edge of the field at which they meet. The furrows make it harder to walk over the furrows at an angle, and you compensate by turning into the field. However, longer legs mitigate this effect so that you don't ...


21

Snell's law tells us that the angle of refraction depends on the index of refraction, $n_1 \sin{\alpha_1} = n_2 \sin{\alpha_2}$. However, the question remains, why $n_{\text{blue}} > n_{\text{red}}$. In order to address this, we need a model for the refractive index. The refractive index $n$ of a material is related to the atomic transitions of the ...


12

To explain this, we need to use the Fermat's principle of least time. According to the principle of least time, a light ray always takes the path which is the quickest. We also need to know another fact that the speed of all different colours of light in vacumm is the same, but in any other medium like glass, water, etc. their speed is different. And it has ...


3

So your question is what is the microscopic reason for material dispersion in optics, meaning $n=n(\lambda)$ ("dispersion relation"), where $n$ is the refractive index of the medium and $\lambda$ the wavelength. Because by snells law $\frac{\sin \theta_{2}}{\sin \theta _{1}}=\frac{n_{1}}{n_{2}}$ (see FakeMod's answer) this means a different refraction for ...


9

Both are right in their respective situations. Optical refraction is a relatively low-energy (non-ionizing) effect which does not exchange energy with the medium. The atoms have electromagnetic fields which interact with that of the photon, increasing the electrical permittivity $\epsilon$ of the medium over that of free space, $\epsilon_0$ (I am not sure ...


1

If $\alpha$ is the attenuation coefficient, such that $|E|^2 \propto e^{-\alpha x}$ it is, by pure identification, the definition of $\alpha$. Let's write: $$ E=E_0 \exp\big(i (n+ik)k_0 x\big) $$ where $k_0=\frac{2\pi}{\lambda}$ is the vacuum wave number. You get then: $$ E=E_0 e^{ink_0 x} e^{-k\,k_0\,x}$$ and $$|E|^2=|E_0|^2 e^{-2k\,k_0\,x}$$ Hence $\...


10

Your diagram's mirror configuration corresponds to Cassegrain reflector telescope. The rays that it shows are from a faraway point source situated exactly in the middle of its field of view (FOV), and they are condensed into a smaller output aperture. But this alone doesn't directly* show whether the output image is magnified or shrunk. To see the ...


8

I gave the answer above +1 but wanted to mention the concept at play. Étendue Étendue is a property of a light beam. At best, étendue is conserved as the beam propagates through the optical components. This means that the product of entering beams area and solid angle is the same at the exit aperture, $$A_1 \Omega_1=A_2\Omega_2$$ In the case of a ...


3

You have lots of questions here. We can answer them in general: The eye works just like a screen, so you would see exactly what occurs on the screen. Now a major caveat here is that your eye is in a single location. If you just looked at the double slits you would see... double slits. The light coming through would be the correct brightness for wherever ...


11

I think the problem with the original image is that it does not show the subject you are focusing on, so it cannot demonstrate what portion of it you see magnified. Also, the original picture design would present the image upside down. I tried to make a simple picture for you, it is not geometrically perfect (made it in paint) but should give general idea.


0

The lens on the image is known as a Total Internal Reflection (TIR) lens. The benefit of this type of lens is to achieve a more controlled and uniform beam of light by principle of total internal reflection. For example in flashlights. More information here: https://www.ledil.com/support/guide-to-tir-lenses/ and here https://www.opticsforhire.com/blog/what-...


34

While a telescope can make an image larger, the diagram shown above doesn't really show that happening. Your eye perceives the size of an image based upon the angular extent it takes up in the visual field. A ball that takes up one degree of your field would look larger to you if it took up five degrees of your visual field instead. This could happen by ...


0

The aperture is a circular aperture multiplied by a function which is 1 everywhere, except at $-\lambda_0/2< x < \lambda_0/2$, where it has a value of zero. No addition is involved. The Fourier transform is then the convolution of the 2D Fourier transforms of these two functions.


1

For thick films, rather than fitting 1 wavelength between surfaces. You may be able to fit many wavelengths of a particular color in the film thickness. This would allow interference to occur the same as with thin bubbles at a given color, however, other colors are also more likely to have constructive interference. -> N * lambda You might fit 5 red ...


2

Relation between transmission and density You ask for a mathematical relation between the intensity $I$ of the transmitted light and the number density $n$ of gas molecules. It is $$ I = I_0 \, e^{-n\sigma r}, $$ where $I_0$ is the emitted intensity (i.e. before the beam enters the gas), $\sigma$ is the (wavelength-dependent) cross section of individual ...


2

Let's say you start with two identical disks, 1 and 2. You start by slightly tilting disk 1 and grinding it with its center along the edge of disk 2. In this way, the edges of disk 2 as well as the center of disk 1 are ground down. In other words, disk 1 starts becoming slightly concave, while disk 2 starts becoming slightly convex. What happens when we ...


1

Clear air is colorless in the visible spectrum. Yet consider what happens when the Sun changes its altitude from high in the sky to just over horizon: intensity of solar radiation decreases for an earthly observer. That's due to scattering (Rayleigh scattering, in particular, is relevant here) from a larger amount of atmospheric gases. Also, when you go ...


3

The equivalence principle means that in each part of the motion light behaves exactly as normal in an inertial frame. You do not "replace uniform gravitational field with an accelerating reference frame", you get rid of the gravitational field locally by using an inertial reference frame.


3

(1) is wrong; (2) is right. Observations in an accelerating frame would also give (2). I say this based on general knowledge; I have not done the calculation, the details of which would be tricky. But you can imagine a case where the light got bent through a large angle around a massive object and then went on its way in a straight line. The fields would ...


-1

Physicists love their Taylor expansions and especially expanding around a harmonic (square) potential. As one typically doesn't want a potential which is unbounded, potentials with the highest polynomial term being an odd power are rare and the next interesting one after the harmonic potential is the quartic one. One nice application in 1D is that you can ...


1

Actually, a lens does not necessarily behave as diverging or converging under all conditions. In the picture you have posted, if you notice carefully, the incident ray is converging. Usually, we consider diverging incident light rays as "real" objects (because the rays will emerge from a point in front of the lens) and converging light rays as virtual ...


2

You are right: White light’s total attenuation coefficient is an average of that for the individual frequency components, weighted by the spectral power in each component.


3

You can read a lot of questions about splitting photons, but contrary to popular belief you cannot split a photon into two photons. What really happens inside the crystal is that the original incident photon gets absorbed: either by a single atom/electron system inside the crystal or by the molecules/the crystal lattice Thus, the incident photons ceases ...


0

Answers to your questions: Expression (2.4) in the linked document gives indeed the major and minor axes of the ellipse. So, your reasoning is correct. $\Delta \phi$ is indeed calculated as you mention. Note that the sign of $\Delta \phi$ also indicates whether the wave is right or left-handed polarised. So, your assumptions are correct. If the wave ...


1

The definitions of right and left circularly polarized light may be motivated from the requirement that these states must be eigenstates of the rotation operator ; Physically, the most reasonable definition of circular polarization is that the state of polarization is left invariant by an arbitrary rotation. In terms of the notation you have used, the ...


1

It's very likely that the problem is that you're getting reflections off of both faces of the plexiglas plate, so you're getting two images on top of each other, slightly offset. You can fix this by putting a linear polarizer (e.g., from Edmund) in front of your image source, and tilting the plexiglas at about 57 degrees (Brewster's angle) to the image ...


1

Actually, your first image is misleading. It makes you think that the only light that is emitted by the effective object is a thin cone that gives you a rather small blurry spot at the plane $C$. But that's not quite correct: you've forgotten that the rays from the actual object are emitted not only parallel to the optical axis and through the lens center, ...


1

I think the blurred picture comes from the pour quality of the plexiglass, use a real glass, and it will not blur, no lens can make a blurred picture sharp . the lens could make your picture smaller or larger, and it has not to be very large.


3

$\epsilon$ - and, to a lesser extent $\mu$ - depends on the frequency of the light. The atoms of the medium have a particular excitation frequency, or frequencies, and the polarisability of the atom rises as the EM frequency rises to a peak when it equals the atomic excitation frequency, and then falls off. It's standard forced SHM https://en.wikipedia.org/...


0

The speed of light in a material does in general depend on wavelength, and there's no better demonstration of this fact than a prism. Since the index of refraction is $$n = \frac{c}{v}$$ (where $c$ is the speed of light in vacuum and $v$ is the speed in the medium) and the angle of refraction depends on $n$, the fact that light splits into colors when ...


0

Any parabolic reflector with the addition of fiber conductor at the focal point?


0

Things in the distance seem smaller due to the viewing angle from one end to the other, A 100 feet tall tree viewed from 50 feet away has a viewing angle of 76 degrees. The same height tree viewed from 1,000 feet away has a viewing angle of 5.26 degrees Railroad tracks at only 4 feet 8.5 inches wide have a viewing angle of 5.39 degrees at 50 feet, and 0.27 ...


0

Yes you can. Google images for Multi Mirror Telescope and you will see lots of pictures of arrays of mirrors doing just that. Telescope mirrors are not that much different from what you seem to have in mind. They are very precise and expensive because they direct light as exactly as possible to the same place, and they must not introduce distortions. But ...


9

To directly address your comment: If we place a single point source of light in front of the pinhole then it creates a circular illumination on the screen, but if we put an extended object in front of the pinhole then it creates an inverted image of the object on the screen, how? An extended object can also be considered as a collection of infinite point ...


0

Let the following figurewhere the screen is at $x=0$,the object at $x=a$ and the convex lense(with focus $f$) is at variable point $x=b$. Then for sharpe image on the screen we have : $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$ $$\frac{1}{f}=\frac{1}{b}-\frac{1}{b-a}$$ $$\frac{1}{f}=\frac{b-a-b}{(b-a)(b)}$$ $$\frac{1}{f}=\frac{-a}{b(b-a)}$$ $$b^2-ab+af=0$$ It ...


31

Let us start from the basics. Consider a point source of light placed on the principal axis of the pin hole camera as shown in the diagram below: The point source produces a circular illumination on the screen. Now let's displace the point source towards $D$ from the centre as shown below: The circular illumination also moves away from the centre but in ...


15

The images which I remember are the following which show that there is an optimum size for the pinhole but never is the image as sharp as you might expect from that which is formed using a lens.. If the hole is too small diffraction becomes significant so that the final image becomes blurred. If the hole is too big the final image also becomes ...


5

this effect is called the pinhole camera, it functions as follows: The sun is an extended object, every point of which is radiating light. If we send that light through a tiny aperture, then from each point on the sun only a single ray can make it through the pinhole, and an inverted image of the sun will be formed on a screen behind the pinhole. By tracing ...


22

The important thing is that it is a small hole in the cardboard. (image from Wikipedia (German) - camera obscura) Therefore every point of the original (the sun) produces a small spot on the screen. So you get a fuzzy image of the sun on the screen.


1

If the frequency of one laser is known with precision (or is assumed to be the standard frequency), it is possible to directly measure the frequency of another laser whose frequency is within a few GHz of the standard laser, without measuring wavelengths. This is accomplished by shining both lasers onto a very fast photodetector. The frequency difference ...


1

Light will enter the room through the windows (without much loss) and then bounce (be reflected) off the walls and furniture. The amount of light reflected is given by the LRV (Light Reflectance Value). LRV is measured from 100 (100% white reflecting everything) to 0 (100% black absorbing everything). White wall paint seems to have an LRV of about 65-70 ...


3

There are actually two major types of filters: Absorption filter A piece of orange glass (like in a photo camera filter), for example, is an absorption filter. It absorbs higher frequency light, and doesn't absorb much the red-orange frequencies. The result is that due to the usual, Snell's reflection these are reflected, but due to transparency of the ...


0

"My first hunch was "it is as if, if there were no lens, we would see an object 3.5 times bigger than the original 70cm away from the spot where the lens should be" but that is not right, because if that was the case then my brain would process that object as smaller than the original size times 3.5 due to the distance that it was from me." Your doubt ...


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