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28

This picture shows what happens to a ray passing through the center of the lens: The incoming ray hits the air-glass surface of the lens at an angle $\theta$ to the normal and is refracted. It then passes through the lens and hits the glass-air interface at the other side where it is refracted again. Because the lens is symmetrical about the horizontal ...


8

There's no way; it can't be done. The reason we can be so sure about this is that if this device existed, you could use it to make a perpetual motion machine. Imagine that you have an adiabatic chamber (one that doesn't let any heat in or out), and you place a hot object into the chamber. This object will give off thermal radiation until the adiabatic box ...


7

Don't get too precious over the term "resolution". There are many ways to define it, and indeed ultimately what you resolve with a microscope gets down to what measurement signal to noise ratio you can achieve. With a perfectly clean signal, you can deconvolve the lens's point spread function from your image and resolve features smaller than what the simple ...


5

The diffraction seems to form from the pixels (basically a diffraction grating). The pixels have a translational symmetry in $x$ and $y$ directions, so the pattern also exhibits this symmetry. On a 15-inch macbook retina display, the pixels are separated by $$d = \frac{15.6inch}{\sqrt {2880^2+1800^2}} = \frac{0.396 m}{\sqrt {2880^2+1800^2}} = 1.17 \cdot ...


4

It depends on the actual materials. If you have a big need and a big budget, you can make some mirrors that have very high reflectivity. 99.9% is achievable in some cases for certain energy ranges. As energies go up, (UV and higher) it gets harder to reflect cleanly. For sunlight, most of the energy is in the optical, so you can ignore the reflectivity ...


4

It is indeed possible. This was a famous experiment by Isaac Newton (published in 1672). Place a lens of focal length $f$ a distance $2f$ from the first prism. Add a second identical prism $2f$ past the lens and rotate it round until white light emerges. The lens is required to bring the rays back together. It creates an image of the exit of prism one on ...


3

The event horizon is not an optical effect, so changing the refractive index of the space around it will make no difference. Adding extra mass will make a difference, but it's the mass that makes the difference not the refractive index. You couldn't make your spherical shell of glass go right up to the horizon because at the horizon no material is strong ...


3

A lens is symmetric. By this I mean you can pick it up, turn it round and put it back and the light rays don't change. Or to look at it another way, it doesn't matter whether the light travels from left to right through the lens or right to left. So parallel light travelling from infinity on the left of the lens will converge at the focal point on the ...


3

Given an unsharp image, you can use deconvolution to attempt to reconstruct the original image, but the resolution you obtain will be limited by the noise in the unsharp image. The theoretical limit for the case of astronomical images where the objective is to resolve double stars, is investigated in this paper.


3

Project Excalibur The idea of a nuclear pumped X-ray laser was one which was investigated in detail in the Reagan "Star Wars" program of the 1980s, backed by one Edward Teller. Tests were carried out by surrounding the nuke with bundles of rods to create a one-pass laser. Apparently it was nowhere near efficient enough to be used in a military context. ...


3

Imagine not the direction of the column but the direction of the front row. Suppose the front row of soldiers were carrying a horizontal bar, the one on the left hitting the swamp would slow down while the one on the right was still moving quickly so the bar (=wavefront) would change direction It's a slightly bad analogy. A much better one is: Imagine you ...


2

Trying to break my bad habit of answering in comments, putting my previous stuff down here now. As I said, I'm largely drawing on a blog post I did a couple years ago, working out the color of the sky. I had a spectrum, and I wanted a color. As I mentioned above, you can't just take a spectrum and output an equivalent wavelength--only some colors are so ...


2

Fluorescence is mostly red-shifted with respect to the excitation wavelength, as part of the energy goes to excite molecular vibrations. However, the reverse process also happens: if a molecule was vibrationally excited before electronic excitation, it can contribute this energy to fluorescence, which in this case is blue-shifted (so-called hot bands). The ...


2

If you're calculating something like "Where is the image in this lens setup?", there shouldn't be any "convention" in the final answer. "The image is right here, where I'm pointing" $\leftarrow$ that statement is objectively true or false, it cannot depend on a convention. Therefore any conventions you use in the process of calculating this thing should ...


2

There are actually two different sign conventions in optics. Without any convention it's hard to develop any universal scientific statement (like formula) so let's make up one here: F - lens focal length, d - lens to object distance, s - lens to image distance. 1. Cartesian sign convention Small lens formula: 1/s - 1/d = 1/F ...


2

You have two different concepts intertwined in your question. You begin by asking about the speed of light in a medium varying with color (i.e. wavelength). This phenomenon is called dispersion and it is present in all materials including air. Dispersion shows up in many places in the field of optics, but the case you are probably most familiar with is ...


2

As an analogy, consider the photon that strike your face and reach my eyes, we say that that photon carries information about your face which then helps me to identify you, You are confusing the individual photons with the electromagnetic wave that is light, which is composed out of a zillion photons. but don't these photons collide midway with air ...


2

What you probably looking for is a tube lens, or infinity-corrected tube lens. It takes all light from a point source that fit into lens aperture, and transforms it into parallel bundle of rays. You can take a look on Throlabs page for more information and actual products, or Nikon's (not an ad!).


2

The explanation you give is correct. A white body reflects all wavelengths. We call it white when all colors (all wavelengths) are reflected from an object and hit our eye. Black is the opposite. I would say that white is all colors, as you do. But maybe he sees it from the perspective that since all is reflected and nothing is absorbed, there is "no ...


2

No, because you have no information on how far objects are, and that affects the spread of the light. If you were taking the photograph of a 2D plane (a painting for example), you probably could retrieve more information, but it would still have a considerable quality loss, since in the borders you don't have the entire spread, some would be out of the ...


2

There is no contradiction: Your first answer, with the numerical factor 1.22, is a measure of the width of the diffraction spot from a circular aperture (e.g. from a round lense). Historically, the smallest resolvable feature has usually been defined not as this size, but as the closest distance between point sources that create an image that still has a ...


2

The energy is reflected from the cavity. In general, an optical cavity acts as a variable transmissivity mirror for a light source with very narrow linewidth (a laser). As you change the length, the cavity can go between highly transmissive and highly reflective. The specifics of how tranmissive and how reflective it can be depend on the reflectivity of ...


2

It doesn't quite work the way you envision it (if the refraction angle is such that you can add light, it will escape the same way), but there are optical resonators that do essentially what you want: Light incident on a mirror gets added to a light field trapped between two or more mirrors. In such setups, not quite enough light usually builds up to cause ...


2

Yes, the refraction index will be changed, but the absorption will differ even more. Look at similar situation. (P.S. I'd like to add this as comment but I cant yet, cause of reputation)


1

What exactly do you mean by "measure of absorption"? The beer-lambert law is parameterised by the absorption coefficient $\alpha$ with units $\textrm{cm}^{-1}$, $$ I(x) = I_0 \exp\left( -\alpha x \right) $$ However, you can't work out the attenuation of the light beam with the information you have supplied because you don't have a value for $I(d)$, the ...


1

Here is a popular account of how it has already been done Simply an array of LEDs showing the scene on the other side of the car. However, the use of LEDs or any projection technology is never going to be good enough to match sunlight ie 1kW/sq m at max. Night time is different


1

If the bottle and liquid are made of dielectric material, then the interfaces between different mediums reflect light, they don't absorb it (i.e. dissipate it as heat in the glass). This is probably a good approximation for your bottle. As a first approximation, once you have worked out your incidence angles with Snell's laww, you need to use the Fresnel ...


1

As you've observed in your edit, if you flip the orientation of the beamsplitter such that the thick part is now on the back side, then you still need to use the compensator plate in the other arm. You might wonder why a compensator plate is used at all. In an interferometer which uses monochromatic light (i.e. a laser) you could just shift the position of ...


1

Yes, it will, that is the whole idea. The reason is that everybody has to get the same physical answer independent of the sign convention. The problem is that you are free to choose positive radius of curvature for convex or concave surfaces, positive focal lengths for a converging or diverging lens, and so on, so, conventions are necessary standards within ...


1

Most 1300-nm optical fiber systems use single-mode fiber. Most 850-nm optical fiber systems use multimode fiber. Therefore, typical 850-nm transmitters are designed to couple into multimode fiber. Coupling into multimode fiber only requires focussing the light down onto a spot about 50 um in diameter. Single-mode fiber, on the other hand, typically has a ...



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