# Tag Info

1

When you focus light from the Sun you are actually creating an image of the Sun. If the focal length of the lens is $f$ the radius of the image is given by: $$r = \frac{r_s}{d_s} f$$ where $d_s$ is the distance to the Sun and $r_s$ is the radius of the Sun. The fraction $r_s/d_s \approx 10^{-3}$, so if you choose a lens with a focal length of 10cm the ...

1

It is the sun that produces the energy that falls on the convex lens in the form of light (and a bit more than just light in the visible part of the spectrum). The lens concentrates all this light in a small spot. A good fraction of the energy is converted to heat if the material at the focal point is suitable.

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The lens focuses light due to refraction. Absorbance of light by a surface creates heat. More intense light due to focusing creates more heat per unit area at the focal region.

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Since this is a homework question I won't give a final answer, but think about the angle formed by a line segment connecting the centers of two adjacent faces of the octagon, and the line bisecting the face.

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You'll usually find these data in the Supplementary Information of relevant papers. If you find a paper where they've made such measurements, but don't give them in supplementary information, you can email the contact author, with a couple of paragraphs explaining what data you're after, and what use you'll put it to. For example, here's the Supplementary ...

0

This thread came up some years ago, I saw it referenced and would like to posit an idea. My belief, backed up with math here, is that this cannot be lens. Lensing always results in rings, sometimes faint, sometimes not. There would be some evidence of a ring in the Hubble image, which is quite deep and quite fully resolved. There is none. So what could this ...

0

This answer may help you understand the difference between blurring and inversion (image below the axis while object, a point there, above it).

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I'm not sure previous posts answer your final question. I want to give you an intuitive picture of the situation. Provided the detail pointed by @WetSavannaAnimal that the "true" mapping is {object plane point} to {image plane point} when talking of conventional optical systems, and not ray to point, as always many rays passing through your lens ends in a ...

1

You can also use Diffractive Optical Elements (AKA Holographic OE) implemented as phase plates. This, to start, provide a uniform reflection-loss transversal profile, as they are flat. They are also thin, so if you are considering bulk absorption, too many elements, etc., this may help. In the cons side... you need monochromatic light. But I think this does ...

0

I would perhaps answer something like this. Maybe that question helps you also. You have to keep in your mind the proper meaning of whole information; the plate will just contain almost the same information as the one you get from the other side of a window placed there (thorugh it), being its size the same of the plate's one, and provided you are working ...

0

In theory, it should be (lumpy), however: Even if you could observe the peaks, or nodes and antinodes of the EM vectors of a single photon, its energy would be absorbed by the instrument you employed to observe it. Fast enough doesn't cut it. If the instrument absorbs no energy from the photon, it will also not be detected. Intensity for light is not ...

1

Each photon in a sense passes through every point in the region you have drawn. Imagine the following experiment. Imagine a gain medium with a population inversion of 100%, i.e. all the emitters are in their excited state. Also imagine that their spontaneous emission lifetime is very, very long. So they only relax when triggered by a photon. Now we focus an ...

1

I don't know how small these can be made, but a Dove prism is able to rotate images at arbitrary angles

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In the far field (greater than a Rayleigh range from the waist) they look like they have taken path 1. Nearer to the waist it is harder to tell. This is the origin of the Gouy phase.

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do photons take the path 1 or 2. That is, do the photons cross or just get 'deflected'. They take path 1 and travel in straight lines. However due to the finite effective size of the photon and Heisenberg Uncertainty Principle it is not possible for all of the lights straight lines to pass through the centre point. See above image (ignore the circle ...

-2

I would think the wave character of light is the most important and waves can cross each other without changing direction, so I think path 1 is the more likely. As pointed out in the comments photons can annihilate each other, creating other particles, which in turn can create a new photon pair. Conservation of momentum and energy would require the photons ...

2

I hope you know that intensity $(I)$ of light at any point on the screen due to interference in the Young's Double Slit experiment can be given as $$A^2=I=a_1^2+a_2^2+2a_1a_2\cos{\phi}$$ where $a_1, a_2$ are the amplitudes of the light waves with constant phase difference of $\phi$, $A$ is the amplitude of the resultant displaement at the point on ...

0

There is no one focal point. It is more like a focal axis. The whole longitudinal centre line of the cylinder are focal points. this means you can line up many diodes from top to bottom as long as they are in the middle of the cylinder. Because the detector is a flat plane, you can only utilise 180 degrees of the cylinder per diode (light coming from the ...

3

Here's the logic (well a particular rendition): Recall that $n$ is defined as the ratio of the speed of light $c$ in vacuum to the speed of light $v$ in the given medium; \begin{align} n = \frac{c}{v} \end{align} Note that in a linear medium, Maxwell's equations are exactly the same as in vacuum, except $\mu_0$ and $\epsilon_0$ are replaced by $\mu$ and ...

2

There's no way that you can add a vector pointing along the $x$ axis, and have it cancel a vector pointing in the $y$ direction. So the amplitude of the light is zero nowhere, and there are no intensity fringes. Instead, the polarization of the wave changes at the screen. As you study the light along the direction that you expect to see fringes, you would ...

1

The Casimir effect is used as experimental proof of the existence of the vacuum virtual exchanges. The typical example is of two uncharged metallic plates in a vacuum, placed a few micrometers apart. In a classical description, the lack of an external field also means that there is no field between the plates, and no force would be measured between them. ...

0

the answer is at the bottom of the page you linked to, go to Linear and angular magnification read the content, but the brief answer is that the first lens projects from air to air, and in this case you use linear magnification. the second lens projects from air into your eye (some kind of liquid?), so you use angular magnification

3

I am suffering from myopia (nearsightedness). I tried the same (with out wearing my lens), result was I could see things clearer in front of me than when I saw the same things in the mirror. The below figure shows image formation for a person suffering from myopia. Lens in our eye is more convex at its back than in front. For simplicity lets assume it ...

4

does a flat transparent glass make near-sighted people see farther? the answer to this question is the answer to yours.

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With a flat mirror, it would be worse. The distance from eye to object is the distance without the mirror plus twice the distance from eye to mirror. With a convex mirror, it is even worse. With a concave mirror, the answer is it depends on the curvature, the severity of myopia, and the distance to the mirror.

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Edit: Reading other people's answers, I forgot to mention I assumed a flat mirror. Excellent question, but the answer is no. The reason is because the object (in the strict optics meaning) in the case of the photograph is actually on the paper whereas in the case of the mirror it is still at the same place, far behind: the rays of light coming from it are ...

2

Not necessarily. It depends on where the image is formed from the mirror. Depending on the radius of curvature (assuming spherical curvature) the image ould form anywhere, but the person would want it to form on their retina. You can calculate this using the mirror equation $\frac{1}{d_{0}}+\frac{1}{d_{i}}=\frac{1}{f}$ Where $d_{0}$ is the distance (from ...

4

Assuming the mirror has no curvature then things would not look clearer. You have to think about the image as it is being projected, our eyes doesn't perceive reflected image as being at the point of reflection but rather perceives the image as being a set distance from our eyes. To clarify if you imagine your eyes couldn't look at anything but the mirror ...

0

Light changes direction when passing through a prism because the density of air is different to the density of glass. Therefore the speed changes, when something (i.e. glass or a prism) is optically dense it is harder for light to travel through it, thus making it's speed decrease. When the waves meet the prism they slow down, so if they meet it at an angle ...

0

$I(\theta)= I_{max} \left[ \dfrac{\sin\left(\frac{\pi a \sin \theta}{\lambda}\right)}{\frac{\pi a \sin \theta}{\lambda}}\right]^2$ where $a$ is the size of aperture. This is the diffraction pattern, now you can easily see that the minimum of diffraction are where $\sin \theta = m \lambda /a$. So as you say the angular width of central maxima is $2 \lambda ... 0 Reflected light can be though of as originating in oscillating charges in the medium. The incident light causes the charges to oscillate, and the oscillators generate the reflected light. This process happens almost instantaneously. The atoms in the medium are oscillating coherently (in step) with the incident radiation. The frequency of the light is ... 0 Actually a much more difficult question is why is glass transparent and does not SCATTER? Lack of absorption is just one explanation. But take a ceramic. It does not absorb (e.g., toilet bowl) yet it is not transparent. So why is glass transparent. This does NOT really has a trivial solution. 2 I don't think it would be a good idea. Solar power plants are effective only because they direct light beams to a small area, which result in high heat rate concentration. To make the water evaporate "significantly" faster You would need huge amount of mirrors or else the heat rate concentration would be too small to make a difference. Answering to the ... 0 One cannot collimate light from an LED accurately without loosing a great deal of light and / or being happy with a very wide collimated beam, because the source is often quite a wide extended source (sometimes up to 1mm across). This may or may not be a helpful answer depending on exactly what you mean by collimated, i.e. how accurately you need to ... 1 Optical systems not involving magnetic fields are symmetric. So, if the display passes light in one direction, it will pass light in the other. Putting a mirror at the back of the TFT and lighting it from the front is therefore equivalent, expect that some light will be attenuated on the way in as pointed out by @CarlWitthoft in the comments. As a ... 3 The vacuum is polarizable. The polarization can be with respect to electric charge or color charge. In the presence of an electric field, virtual electron-positron pairs briefly exist (created from virtual photons of sufficient energy). The virtual pairs act as dipoles and orient with respect to the field. For example, near a proton, the virtual electron ... 0 I found a solution to my problem using the vertical displacement: $$$$\tag{2} h_d = d_1 tan(\theta) + \frac{d_2 tan(\theta)}{n} + d_3 tan(\theta)$$$$ solved for$\theta$this becomes $$$$\tag{3} \theta = \arctan \left( \frac{h_d}{d_1 + d_3 + \frac{d_2}{n}} \right)$$$$ 1 You have two good fundamental answers, but a slightly different take on your answer is that yes, this is done all the time in optical systems to compensate for various distortions, aberrations and errors. A good example is an achromatic doublet, where a convex (converging) lens of one material is put in direct contact with a concave (diverging) lens along a ... 1 A couple things: this site is not specifically for engineering questions, so there's a small chance this may be closed by mods. Second, many commercial high-power LED's are listed not by luminous flux, but by nominal diode power consumption; a 1 watt LED actually emits far less than 1 watt of radiant power, so you'll need to think about what radiant flux ... 1 Any discussion of the type of image a lens can form (real or virtual) must include information about the type of object that is being used. A divergent lens, by itself, can form only a virtual image of a real object. if we pre-condition the light from the object by passing it through a converging lens, then the resulting intermediate image can be a real or ... 2 I'm not certain what "backed by a convergent lens" means in this context. A divergent lens by itself cannot form a real image, since a divergent lens has a negative focal distance. Use the thin lens equation: $$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$ Since$f < 0$and$d_o > 0$by convention,$(\text{positive}) + \frac{1}{d_i} = ...

4

You make a good point which requires us to be more careful about what Fermat's Principle says and how the proof proceeds. The upshot of what I'm going to say is The statement of the Law of Reflection must include an appropriate constraint. Here's what I mean in detail. First, let's give a precise statement of Fermat's Principle: Fermat's ...

-1

I think the path must reach on the surface. If the light reflects before the mirror, there should be something above the mirror.

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This is what I think the first bit of the calculation does. Suppose you start with a spherical eye with a hole in it (e.g. the pupil in the human eye): The radius of the eye is $ER$ and the radius of the hole is $AR$, and with the length $DA$ these form a right angled triangle. Pythagoras' theorem tells us: $$DA^2 + AR^2 = ER^2$$ so:  DA = ...

0

Besides the physiological part of optic nerve and brain, we call an object coloured if it reflects light of a specific wavelength (or wavelengths) from the complete visible spectrum. And we call light coloured if the light source only emits a part of the complete spectrum.

1

Just to add an additional twist to the very good answers by Colin K and gigacyan. I have actually been in situations where those exact patterns were real. I used to design grayscale microlenses (manufactured with computed masks) where the finite resolution of the design (driven by the resolution of the lithographic grayscale pixels) actually caused those ...

1

You can't see the sky from Venus surface. In fact, if you look it with a telescope, you will see only clouds. There are very few images of its surface in deed, just retrieved in 1982 by the Venera 13 module. So, I guess the effect is that is cloudy... permanently. source: here, quote from the first paragraph: The Venusian atmosphere supports opaque ...

0

Color is nothing but interpretation of electromagnetic waves into various nerve signals by our brain. It is the property of the object to reflect various wavelengths when photons strike their electrons. Electrons then vibrates up and down atleast 1000 times each second emmiting electromagnetic waves which our eyes receives. Each different wavelength of ...

1

You could make a pair of dyes that only glows when mixed. For example, you could mix a dye that absorbs deeper UV and emits longer UV with a dye that absorbs longer UV and emits blue, for example. Neither alone would glow very much under deep UV illumination, but when the two are mixed it would glow more brightly. I don't know how effective it would be, ...

3

The "color" of a body is not a property of the body nor of the light it reflects or emits, but rather of the human eye and brain that receive and process the light. Of course, this is based on a property of the light, for this is how the eye receives the information that there is some object in there in the first place. This property is the spectrum of the ...

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