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6

Lasers by definition only emit a single wavelength of light. You use one if you want that wavelength or if you want your photons to be in phase. You don't care about the photon phases, and you want to sample all wavelengths, so a laser is very much the wrong tool. If you just want collimation of the light, mirrors, lenses, or even just well-separated ...


4

To follow the information in Chris White's answer - essentially, you would want a medium that allows you to see the spectra. There are several online resources that could help you in this experiment, in particular, the CD spectrometer, which can be constructed simply and on that website, it shows several examples of how everyday light sources can be ...


4

In general, reflection and refraction happen when light passes from one medium to another. You can see this if you see your own reflection in a window. Now, as a light ray approaches the critical angle, not only does the refracted ray get closer to the surface, but the amount of light transmitted gets less and less. At the critical angle, the refracted ray ...


3

Carbon nanotubes, or similarly manmade materials, might be able to accomplish this in the future. The current record is a tensile strength of 63$\times$10$^{9}$ Pa (Pa = pascals = N/m$^{2}$ or force per area). So if we assume a circular cross section, then the area is defined by 4$\pi$r$^{2}$. We can take the tensile strength and the area formula to find ...


3

Today we know that Collins is wrong. He appears to be unaware of Newton's finding, and of course, advances made after he wrote his book.


3

The rays will only come exactly to the focus if you're using a parabolic mirror. A spherical mirror is considered a good compromise for limited fields of view (plus it's got the advantage of behaving the same off-axis as on-axis). Your second setup is invalid, I believe. You need to differentiate between the object's position (or height) and the height of ...


2

You can think of super massive object like black holes which can bend light. Near the event Horizon you could get a 180 degree turn for light and thus see the earth back in time. But I do not think this is practically possible as earth is small and dark (compared to stars) and this layer would get compressed really thin as some small deviation in the ...


2

This may (or may not) lead to the same answer as CuriousOne's suggestion above, but the most appropriate (and the longest) way of attempting a solution would to be to employ the Fermat's principle. The method's nicely described in the link, but in a nutshell, you would be led to a condition of the type $$\delta \int n ds = 0$$ where this $ds$ can be cast in ...


2

You're unlikely to get help here because yours is a detailed debugging exercise: its something that the writer of the code is in the best position to do. Most ray tracers don't use the thin lens formula at all: they encode Snell's law and assume functional forms for the lens surface. The point of intersection with the surface is then calculated (this can be ...


2

The problem here isn't a simple algebra error, but rather an issue with the physics. A medium which at rest is isotropic no longer behaves as an isotropic medium when it is moving relativistically. Instead, it behaves as a nonreciprocal bianisotropic material. In particular, the phase velocity of light at a particular frequency in a medium is no longer ...


2

First of all, this is NOT a 2D photonic crystal since you send light in the $z$-direction (you are using all 3 dimensions) and if the rod is not infinitely long. I'm not aware of an analytic formula, but your band diagram exhibits several features characteristic of metallo-dielectric photonic crystals. For example, there's a bandgap starting at zero ...


2

There is no theoretical upper limit. The question is whether the description has any practical use. Real-world objects will have some small deviations from the perfect sphere or cylinder shape, for which Mie theory applies. Look at the polar diagram of scattering of red light from a 10 micron water droplet. Figure 2 in ...


2

I would say that it depends on your application. If the beam is small compared to the clear aperture of the lens, then you can probably treat it as a thin lens without any trouble. If the beam is large, you will probably have to treat the rays differently depending on their distance from the optical axis. IIRC you won't be able to use a transfer matrix for ...


2

Assuming the two surfaces are still spherical, you can still use transfer matrices to treat a thick lens. Citing the Wikipedia article on transfer matrices, the transfer matrix for a curved interface is $$ I_C(n_i,n_f,R) = \begin{pmatrix}1&0\\\frac{n_i-n_f}{n_f R}&\frac{n_i}{n_f}\end{pmatrix}. $$ The ray matrix for translation is, as usual, $$ ...


2

Assuming your screen is "far away", then the pattern you see on the screen is the Fourier transform of the aperture function. So if you take the (inverse) Fourier transform of the image on the screen, you get back to the aperture function. Small complication: the intensity image no longer contains the phase information which you need to fully reconstruct ...


2

As you point out in your ray diagram, the point source no longer looks like a point source after the diffraction grating. You could get around this by collimating the input beam with another lens (which you did), with an aperture, or by moving the point source far away from the grating (effectively using the grating as an aperture). So, I think the answer ...


2

Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of: We can always detect uniform motion with respect to a medium by a positive result to a Michelson-Morley experiment that is confined to a region of spacetime ...


1

The physics behind can be explained by invoking the idea of real and virtual objects (or images). If light rays from a point or location in space physically go out in all possible directions (or diverge, as they indeed do from point $O$ in the figure), the object at that point is real. Similarly, if light rays physically converge, the image produced is real. ...


1

The intermediate object you describe is a virtual object. You should be familiar with the concept of a virtual image. A virtual image doesn't exist because it cannot be displayed on a screen, however the light rays follow paths as if they had come from the image. A virtual image is (generally) formed on a different side of the lens to a real image, so we ...


1

Every time you look up "the" spherical mirror formula, it comes with a set of "where's". These define what each symbol stands for, and the sigh convention to use to distinguish the location of objects and images and the difference between concave and convex radii. You can find different-looking spherical mirror formulas, with (naturally) different sets ...


1

Photon photon interaction, which is what a collision will mean, is practically non existent as it is higher order and in the context of this question, light sources, non existent in reality. If we reach gamma ray energies then particles will be produced but this has nothing to do with this question. There will be interference patterns as whenever coherent ...


1

For a metal, the permittivity can is typically described by the Drude model with a permittivity given by, \begin{equation} \epsilon = \epsilon' - i\epsilon'' = \epsilon_\infty - \frac{\omega_p^2}{\omega(\omega - i\gamma)} = \epsilon_\infty - \frac{\omega_p^2}{\omega^2 + \gamma^2} + i\gamma\omega\frac{\omega_p^2}{\omega^2 + \gamma^2} \end{equation} where ...


1

Generally "vector" and "scalar" are models for the same kind of thing, the former can be more accurate than the latter. Generally, "vector" models of the electromagnetic field tend to be needed for "fast" fields: i.e. those of high numerical aperture and which contain a wide angular spread of wave directions. "Vector" models are also needed for fields whose ...


1

General relativity and classical optics are not equivalent, however, one can construct analogies between the two and build optical simulations of gravitational phenomena. For example, in the field of transformation optics metamaterials are used to produce spatial variations of the effective permittivity, which enable scientists to steer light beams along ...


1

Yes, of course this can be done because a suspended object can have larger dimension than the diameter of a string holding it. In fact, the minimum dimension of a suspended object can be many times the diameter of a string suspending it at 1 g. Note that it is this relative size that matters, not the absolute sizes. Since human visual acuity is finite, as ...


1

If the light ray is normal to the surface, 100% of the light is transmitted. As the light ray bends, as in your part (b), a percentage of the light will be transmitted (refracted) and the remaining will be reflected (at the incidence angle). Very near the critical angle $\theta_c - d\theta$, likewise, some of the light will be transmitted (refracted almost ...


1

Mostly the size of the imaging chip (or film) As the image gets physically smaller, the same field of view will be produced by a shorter lens. The image chip in a modern digital SLR is 1/2 to 2/3 the size of traditional 35mm film and so the lenses (and the camera) can be made smaller and lighter. Compact cameras have smaller sensors and phones smaller ...


1

We cannot multiply light by mere reflections, because the very definition of "reflection" means that the same light comes out. We can however multiply light by letting it pass through special materials which we "pumped" into a certain state, that's called Laser. And yes, a Laser design includes a sequence of reflections, but it is not the series of ...


1

If you imagine the Sagnac interferometer to be a circular ring cavity (as it is in some variants of optical fiber ring gyros) lying on a table, then the light which is moving clockwise and the light which is going counter-clockwise, will take the same amount of time to go once around the ring. If the ring starts rotating, then the light traveling in one ...


1

Misaligning one of the end mirrors will produce a set of vertical or horizontal fringes at the detector plane (depending on the misalignment of the mirror). The number of fringes is proportional to the misalignment angle of the mirror and inversely proportional to the wavelength of the light. When first setting up the alignment of the interferometer, this ...



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