# Tag Info

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To get the incidence angle for minimum deviation($\theta$), apply Snell's law for the two refractions and eliminate the unknown angles of refraction(use $r_1 + r_2 = A$) to get relation between $i_1$, $i_2$ and the refractive index and solve to get the value of refractive index. When $\delta$ is minimum, $r_1 = \frac{A}{2} = \frac{\pi}{6}$ which means $\sin ... 0 Actually in lens maker's formula derivation we apply sign convention(Cartesian sign convention) twice.Thus it gets cancelled in derivation.So in the problem, when we again apply sign convention we get the right answer in accordance with sign convention. But personally for me Classical sign convention is much easier to follow and solve problems. I use ... 1 With a little math you can get an excellent approximate answer. Pretend that the lens is a pinhole. You have a triangle formed by one side of the object, the other side of the object, and the pinhole, and you know how far away the object is. Another triangle is one side of the image, the other side of the image and the pinhole, and the distance between ... 4 If you don't want (ray-based) Snell's law, then we can do it using the wave aspect. BTW the analogy totally stands with water waves, with the depth playing the role of refraction index. -> when the light waves enter the glass, or when water waves enter shallower water, they slow down and wavelength get shorter. This has the effect of tilting the wavefront, ... 0 I have a great deal of sympathy with your position. As taught in schools the sign convention is somewhat vague and that gets confusing with complicated setups. If you're dealing with anything complicated I recommend keeping strictly to the Cartesian sign convention. I've linked an article that seems to be a good summary of this, but Googling will find you ... 0 The picture in this question may be helpful to visualize the situation: Clarification needed in the concept of apparent depth & real depth Basically, if you observe an object immersed at some (real) depth$h$in water, as a consequence of the refraction of light you will see it as if it were at a different (apparent) depth$h'$. The relation between ... 0 Because of variable atmospheric haze from day to day, especially in the lowest 15 degrees from the horizon, the correct theory answer above is probably not applicable. You can try testing by presuming that a uv indicator card happens to change in response to the same uv ranges as your skin will (afro-caribbeans, that won't apply to everyone), and test with ... 0 At first glance you might think the Fresnel lens ($area=34 \times 44 = 1496..in^2$) is better than the parabolic mirror ($area=\pi ({{40}\over 2})^2=1257..in^2$) because it is concentrating the sunlight from a greater area. But the Fresnel lens has some disadvantages compared to the parabolic mirror that out weigh the small 19% larger area. The Fresnel ... 0 What's your metric for better? The Fresnel lens has a collection area of about 1500 square inches, while the mirror is about 1250 square inches. $$A_{mirror} = \pi r^2 = 1256$$ $$A_{lens} = length \times width = 1496$$Depending on the Fresnel material, it may or may not absorb significant amounts of IR. The mirror will also get less power than the area ... 0 yes, a cell is a dielectric material which is one (among) other requirements for optical trapping. Being dielectric does not mean the material can't be absorptive at the same time. Optical trapping in general means to be able to confine an object into a finite volume. One needs restoring forces in each spatial direction whenever the object tries to leave ... 1 As you have correctly stated, the second order correlation function measures the coincidence of two events. The case of$g^{(2)}=0.4$means any two events are less likely to happen coincidently than the case of$g^{(2)}=0.1$, although both of the two cases tending to have antibunching events. 0 Here's how I go about it. Let's call the middle point of the 2 slits point$A$. Also, let's call the angle between the line joining$A$to the point of the central maxima ( on the screen, horizontally infront of$A$) and$A$to the point on the screen under observation$\theta$. Now, since the slits and the distance between$a$them are very very small as ... 0 If your question is whether waveguides can support TEM (with transverse electric and magnetic field) waves, then the answer is generally no. There is a fundamental theorem that for a TEM wave the waveguide must be open. An example waveguide which supports TEM waves is the parallel plate waveguide. Closed waveguides such a circular or rectangular do not ... 2 A general answer is that the field vector is independent from the decay direction, and hence the field vector can be pointing to any direction. However, on the decay direction (that is the$-y$direction in your formula), all field components decrease. The only constrain for the field vector is that all field components must satisfy the boundary condition at ... 1 Of course a cell transmits light -- this is how a four-hundred year old optical microscope is used to observe cells. And in general any medium that transmits light will refract light unless it happens to be exactly the same density as the surrounding medium. 0 This isn't a general answer, but one famous example of an evanescent wave is a surface plasmon polariton. (If the surface is the x-y plane, then it is evanescent in the z direction on both sides of the surface.) I put an animation of the E-field of an SPP on wikipedia ... here it is: Again, this is just one random example of a lossless SPP. But maybe food ... 3 To summarise: it depends on the ratio of refractive indices, the angle of incidence and the polarisation state of the incident wave. Details: The polarisation of the electric field will be perpendicular to the wavevector. Why? Because where there are no free charges then$\nabla \cdot \vec{E} = 0$. If we represent the wave as$\vec{E} = \vec{E_0} f(\omega t ...

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quick and dirt answer: take the (minimal) width of the line and build a circle with that as diameter. here is your PSF.

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Just use snell's law, that is, $\mu \,\sin\theta$=constant, where $\mu$ denotes the refractive index and $\theta$ is the angle between the ray and the normal between a generic point and the point of incidence.The rest is math, you need to express $\sin\theta$ in terms of the slope at that point and solve the resulting differential equation.

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Polarizing an unpolarized beam, without reducing the intensity or increasing the etendue, would violate the second law of thermodynamics. So what's the practical application? It's good pedagogy. You can think more carefully about how your scheme would work, eventually realize that it actually doesn't work after all, and in the process you will come to ...

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In my (ancient) copy of Hecht and Zajac (1980), the answer is found in figure 10.18. It shows that for slit spacing $a$ and slit width $b$, peaks in the diffraction pattern are spaced $\lambda/d$ while the first zero due to the finite width is at $\lambda/b$. In the figure, $a = 3b$ and the third peak is suppressed:

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i think it's because of least time principle that states "light tends to travel a path in least time possible." as simple as that

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The angle of rotation is proportional to the length of the path the light ray spends inside the active material. This needs to happen because each bit of the path only 'knows' what's happening there and it does not interact with the rest of the material's optical activity. This means that the rate of rotation of polarization must be constant, i.e. that the ...

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This case is about reflection, not refraction. As the sun's light rays are reflected at different angles from the leg, some will interfere with each other constructively (forming light rings) and destructively (forming dark rings).

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It is twice the focus. Focus is the point where llght rays parrallel to principal axis will converge after getting reflected from the mirror.

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As in Emilio Pisanty's comment: Evidently, as you've proved, this is possible... an assumption of shift invariance validates the use of the Fourier transfer function exactly as you describe. Practically, however, the point spread function varies, sometimes drastically, across the whole field of view so that shift invariance is a more often than not a ...

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A single operator cannot be used with a space-variant point spread function. You need to do the calculation for every pixel in this case. Calculating the image with a space variant PSF is therefore slow and the PSF needs to be measured at every position (or guessed at usually by trying to interpolate it from a sparse set of measurements). It is therefore a ...

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Essentially this is about the same as wearing reading glasses in that in either case you are wearing an appliance. Correctly prescribed reading glasses create a relaxed reading environment The real challenge is to create a computer/mobile phone screen that acts as if you are looking into the distance ie infinity focal length. That way older people with good ...

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To complement Samuel Weir's answer, look at this image of incandescent (white) light passing through a prism. As can be seen violet light is refracted ('bent') more than red light. Violet light is more energetic than red light and we know that: $\lambda=\frac{hc}{E}$. ($h$ is Planck's constant, $c$ the speed of light, $E$ is energy). So violet light is of ...

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The main thing I don't understand is how the wavelength is effecting whether or not the light exits the prism. Wavelength comes into play because the index of refraction of a material is not a fixed constant. It depends on the wavelength. That's why a prism of glass breaks white light into a rainbow of colors. The different wavelengths of light that ...

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This is a grossly exaggerated illustration of a strictly cylindrical metal tube compared to a cylindrical tube with external diameter variations, like the one you have in your case: Because of those diameter variations, the reflected light can vary between scattering and concentrating on the surfaces it is reflected onto.

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These are probably caused by minute, periodic variations in the diameter of the table leg, formed by drawing through a die. Any vibration in the process would end up being circumferential waves in the surface of the tube. Changes in the diameter mean changes in the slope of the surface, and thus focus the reflected light to different rings around the base of ...

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It could, as it is possible to neutralize the movement with lenses hence determine the dioptric value. The value is limited due to lack of control over accommodation and the need to stand perfectly still during testing. Guess that's why they went off market quickly. I still have two devices in use for training purposes in my office though.

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Out of the many quantum-mechanically possible states of an oscillator (be it a mechanical one or light waves), the ones we almost exclusively observe are the coherent states. In a way, they are the states where uncertainty is evenly distributed, such that every uncertain quantity scales as $\sqrt{N}$ for $N$ quanta (e.g. photons or energy quanta in an ...

0

Your window is made of patterned glass, with regular changes in thickness. This is designed to diffuse light in order to prevent people outside from seeing what's inside (or you inside from seeing out; I don't know what your neighborhood is like). In your case, the patterns are extremely regular, so that each incoming beam of light will get spread into a ...

1

There's a lot here... let me break it down a little bit into some intuitive chunks. I hope that tackling the math will be less scary after that. It's not quite clear from your description whether you are using single mode fiber, or multimode. Let's first look at multi mode. When light travels through an optical fiber, it can choose many different paths. ...

0

In a metal light will interact with both the electrons in the conduction band and with the valance electrons of the metal lattice. If the frequency is above the so-called plasma frequency the conduction band electrons can be considered free (very few collisions between oscillations). The electrons simply re-emit the light which makes the metal transparent. ...

1

It has to do with the frequency response of the materials in the wall. Different molecules absorb different frequencies (or wavelengths) producing an absorption curve called the materials spectral response. Lots of materials are very absorptive in the frequencies typical in visible light but start to open up (get clearer) in longer wavelengths. Generally the ...

0

One of the defining characteristics of an electromagnetic wave is its wavelength (which is related to its frequency). Radio waves have wavelengths ranging from 1 millimetre to 100 kilometres, while light has wavelength on the order of hundreds of nanometres. Interaction between electromagnetic waves and objects can be roughly predicted with the relationship ...

3

A counter-argument to all the naysayers :-) . First of all, the evanescent wave has nothing to do with light escaping. For example, single-mode fiber optics are often designed such that there's a significant evanescent wave, but since there's no high-index material outside the core, no energy escapes (other than quantum probabilistic long-range ...

0

In the language of wave optics, I think that your question boils down to the following: Given that dissipation is negligible, can a dielectric medium of finite size support a mode that is propagating only within the medium and evanescent outside? The answer to this question is certainly "No." Regardless of the form of the electric field inside the ...

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A physicist's answer to this is that the second law of thermodynamics forbids such a construction. You are describing a perfect black body, and the indefinite input of light in the way you propose will inevitably heat any finite cavity with the properties you propose. If your input light comes through a perfect waveguide from a black body at some temperature ...

0

You may want to check the notion of "optical path length" (Wikipedia link). For a homogeneous medium of refraction index n and physical thickness d, the optical path length is just $$OPL = nd$$ The way you define your "portion" of wave corresponds to a wave that exists the 2nd medium with the same phase shift as it exited the thin oil film. By definition ...

2

If you "only" care about 560 nm, that makes life easier, since you don't have to worry about chromatic aberration. Other things to worry about: What FOV? How much flatness of field? How much spherical aberration? How fast a lens do you need? You can only get a certain level of performance from a simple spherical lens, and if you get demanding you'll need to ...

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This is a perspective effect. In essence, the second image is taken from a lower orbit which is closer to Earth. This means that the field of view is a lot smaller. The Earth still looks like a circle on the page, though from close up the edges can look a bit distorted. In the second image there is no land to be distorted in the edges, and there are effects ...

1

To answer the second part: it's almost always done empirically by measuring a sample of the surface. The standard equation is known as the bidirectional reflectance distribution function, which one uses to estimate the scattered power into a given angle based on the incoming angle of the light. Surface roughness and its parametrization is a science unto ...

0

Forget about most of the instrument, just think about the final beam recombiner. You have two mutually coherent beams of some arbitrary phase $e^{\pm i\,\phi}$ at the two inputs to the final beamspliter (without loss of generalness, subtract out the common mode phase, so we can represent the two beams as $a_{\pm}\,e^{\pm i\,\phi}$ where $a_\pm$ are the ...

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In general, the answer is no, but with careful engineering you can achieve reasonable collimation at the output. Depending on the exact fiber, the field is confined to a region of diameter $d$, where $d$ is typically less than $100\mu$. So even if the field arrives at the output with no phasefront curvature, the divergence angle is of the order of ...

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