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2

I know you asked for an analytical description, and ray diagrams, but seeing as this is homework related, I'd urge you to try the ray-tracing yourself and I think you'll find the answer without too much trouble. My suggestion would be to try and track how a couple significant rays move through your lens setup. For example, you might try one ray that passes ...


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The first diagram red vaaaal listed, unfortunately, is one of the leading causes of misunderstanding how rainbows are formed. It isn't wrong, but it suggests that red light is deflected exclusively at 42°, and that each color has its own exclusive angle. But it is easy to see, if you look, that light that is originally headed straight for the center of the ...


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When a ray of light passes from one medium to another , its direction changes (except angle i = 0°) because of change in speed of light from one medium to another.


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First, note that the way you have written the electrical wave-field isn't anything more than exactly a way to write a wave function in general. This is because the term $e^{i(kx-ωt)} $ can be written as: $$e^{i(kx-ωt)} = cos(kx-ωt) +i sin(kx-ωt) $$, and from here you can keep in general the real or the imaginary part as you wish. As for $κ$, the imaginary ...


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In most situations, BC > BF. Therefore $(BC + CD) * n2 - BF * n1$ will never be 0 (n2 > n1, given). There will always be a phase difference in ray F and ray E.


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It's not about disturbing the photon. If you observe it, your detector gets entangled with it. It's possible for the photon going through the left slit and going through the right slit to end up in the same state and interfere, but if you get a detector involved then one of those will set off the detector and the other won't, so they won't be in the same ...


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You are describing a resonant cavity with a discrete set of allowed modes and your thought experiment is a good illustration of the Purcell Effect (see Wiki page here), which is the influence on quantum mechanical photon emission probability amplitudes by an emitter's surroundings. The resonator simply shifts the emission frequency to one where the ...


3

If you setup a perfect cavity where no modes of light are possible, then the light will not be emitted in the first place (0 probability). You run into problems if you consider the emittor as a classical light-source and then combine it with a quantum-mechanical reasoning regarding interference in cases such as this.


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Apparently there is software solutions for regular displays that somehow distorts the image in a way that makes it sharper for people with eyes out of focus: New Scientist: Good looking: Phone screens that do the focusing for you ... Few handheld devices have 3D screens so far, but something similar can be done with software alone. "We know how ...


1

The point source keeps radiating light. Will the light undergo destructive interference completely? Point particle as a source of light is OK, but it would need to move with acceleration to produce EM radiation. Static source of light of zero size seems to be reduction ad absurdum, at least from the standpoint of common theory of light based on EM ...


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Have a look at this paper - hopefully it will be useful. http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.82.2257


1

There are some special circumstances where the position of a virtual image can be measured. Assume you have a real object located, say, 2 meters in front of a plane mirror. The virtual image is located 2 meters on the other side of the mirror. Of course, no light from the object actually goes there. However, let's further assume that the mirror is a ...


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Ummm, not sure if this helps, but I think you need a ratio between the distance to the lense and the distance to the object to answer the question of what the distance to the object needs to be. Does that make sense?


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Correct. The only way to experimentally determine the position of a virtual image is to convert it to a real image and calculate where the virtual image was. This is simpler than you probably think. For example if you can see the virtual image that means the lens in your eye is focussing the image onto your retina. Replace your eye by a glass lens and a ...


1

Light from the source is emitted in all directions - including towards every point on the road surface. When the light hits the (wet, rough) surface of the road, it is scattered in a variety of angles. Some of the light rays will be scattered at just the right angle to enter your eye. Therefore reflected light from all different parts of the road between ...


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...couldn't I use a retarding film of randomly varying thickness to convert a Coherent laser beam into an incoherent laser beam to improve eye safety? Absolutely not. A beam's destructiveness to the eye depends on three things: Energy delivered to retina and the time periods it is delivered over, quantified by the ISO60825 concept of Maximum ...


1

Couldn't I use a retarding film of randomly varying thickness to convert a Coherent laser beam into an incoherent laser beam...? If the film is not changing (in time) you are not changing the coherence properties of beam at all. You can think of putting a slab of something in the way as putting a really bad lens (possibly with no optical power) in the ...


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Yes it does. Since the direction of the light beam changes with reflection also the direction of polarization. This is mostly because the observer is in a fixed coordinate system and the light beam changes its local coordinate system during reflection. For an idealized reflector and an observer which moves along with the light beam, the direction of the ...


3

Why would you expect the images to be the same? The lens formula (object distance $u$, image distance $v$ and focal length $f$; Cartesian sign convention) is: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ If you consider two lenses $1$ and $2$, the latter at a distance $d$ from the former, then the first lens forms the image at $v_1$: $$\frac{1}{v_1} - ...


2

I'd say about 4% of the intensity for perpendicularly incident light, increasing to nearly 100% when the angle of incidence is almost zero. These are the numbers you get from the Fresnel equations (for $n = 1.5$ which is a typical value for glass) as you suggested. While glass absorbs UV the absorption should not be so strong this induces relevant ...


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In Raman scattering, the molecule absorbs the photon into a virtual state, which doesn't actually exist. Unlike an excited state, the molecule can't stay in that state for longer than a time $\Delta t$ where $\Delta t \Delta E \leq \hbar/2$ - the Heisenberg uncertainty relation. A virtual state can have any energy level, though, and that's the reason for ...


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Turns out that due to orthogonality relations of Hermite-Gauss poly's, Hermite-Gauss modes are orthonormal, so $$ \int \int u_{n,m} \left(u_{n',m'}\right){}^*dxdy=\delta _{m,m'} \delta _{n,n'} $$ Then a's can be found by multiplying both sides by conjugate of u and integrating ...


2

When looking with the unaided eye, light from a single point object reaches all parts of the lens and is focused back onto a single point on the retina. Light from all points on the lens reaches all point on the retina. Point defects on the lens affect the quality of the entire image rather than specific parts of it. Occlusions on or near the lens are ...


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I think the difference between luminescence and Raman scattering lies in whether or not the mixed photon-molecule state maintains coherence with the exciting radiation. In Raman scattering, we imagine that coherence is maintained. In luminescence we imagine that coherence is disturbed by any one of a variety of interactions: for example, collisions with ...


1

The reason we see an interference pattern on a screen is because of diffuse reflection. This is because in diffuse reflection, the incident light can be considered to be absorbed and uniformly emitted out in all directions. This results in a brightness at a point proportional to the brightness of the incident light. A mirror, however, simply reflects ...


2

1) Light from the sun strikes a spherical raindrop across half of its entire surface. So every angle of incidence A from 0 to 90 is represented. 2) Some of the light will reflect (uninteresting), but some will enter the drop. Its path bends due to refraction, entering the drop at an angle B (found by Snell's Law) that is always less than A. 3) The light ...


1

The fastest high speed camera that I am aware of is the Hadland Imacon 200. This camera is capable of recording 16 sequential frames at a rate of (approximately) 200 Mfps. That means it takes one picture every 5 ns - during which time light only travels 1.5 m! This is a "true" camera in the sense that it can acquire a burst of 2D images; not to be confused ...


1

THe hexagons are diffraction patters from the aperture stop, which in some lenses comprises six leaves arranged in a way that allows the effective diameter of the lens to vary.


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The angle at which the light ray exits the material actually only depends on the index of refraction of the material at the last point at which it is in the material. You can demonstrate this if you consider a series of thin layers of materials (possibly all of which have different indexes of refraction), and when you perform the Snel's law calculation, ...


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The answer is simple: your camera CCD is sensitive to infrared light, while your eye is not. The real question here is: why does your camera render the infrared light as visible light on the screen? The only answer I can hazard on that is that it allows for more visible photographs when they're taken in low (visible) light.


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The Fermat principle does not say light ray follows the fastest path, it says when there is a light ray, the optical path (length divided by index of refraction) is stationary with respect to small variations in the shape of the ray that preserve the position of the boundary points. It is not as if light got everywhere the fastest way possible; it goes ...


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The only thing I can think of being true black would probably be a black hole. As light does not bounce off a black hole.


3

First of all I think a suggestion is mandatory: please don't mix units of measure. You are using lux and lumens, so it's better if you stick to SI units. So distances are measured in meters, areas in square meters and angles in radians. This is meant to help you, not to annoy. ;) Lux measure how much light (lumens) hits a square meter of surface. So if you ...


22

The problem with the suggestion of using polarization is that you now have the reflections off the polarizers to contend with. I think the short answer is "it depends on how 'black' you want it to be". "Truly black" = reflectance of 0. I am quite sure that is impossible - there will always be some probability of light scattering off a surface. All you can ...


3

There are two main types of diode lasers that dominate the commercial market. Edge-emitting diode lasers have very asymmetric cavities, which leads to polarization with electric field in the "vertical" direction (normal to the surface of the chip), IIRC. The other type, VCSELs, have a much more symmetric cavity and could be polarized in either direction ...


0

Laser light is polarized most of the time,while it is unpolarized some of the time.I think you should look at this to make it more clear.


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No, laser light is not necessarily polarized. There are many different types of lasers. In gas lasers, many different modes may be excited in the cavity, but only modes that are not very lossy end up being amplified and emitted - this can result in a single polarization. However, there may be multiple modes with different polarizations that have this ...


0

try this link. it basically works on the same principal as a prism, but is many times more effective.


2

My intuition is that normally your eye collects light from a large area to form an image, so small-ish obstructions like cataracts get "blurred over" whereas when looking through a microscope the image is coming from a very small area and the light acts as the bulb in a projector, with your cataracts as the projected "slides" and your retinas the "screen."


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I also have been intrigued by this apparent dilemma. I think I have it worked out, maybe only half way. This is all my conjecture, since most people just are not thinking down these lines. The longer the focal length of the objective lens the narrower the field of view, right? So the longer the focal length, the more the lens is gathering data from a ...


5

The two lenses in modern 3D glasses are designed to select the two circular polarizations. The left lens only transmits left-circularly polarized light and the right lens only transmits right-circularly polarized light (or vice versa). The problem is that there is no material which acts as a circular polarization filter on its own. The way in which they ...


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The answer has to do with the tangential velocity with respect to the viewer. Perhaps the simplest corollary I can offer: if you are driving down the highway, the vehicles coming the other direction at a distance do not appear to be moving quickly even though your comoving speed is 100-150mi/h, because they are moving almost directly toward you; they do not ...


2

You should use external optics with a monochromator. Outside the setup you should have a lens that focuses collimated the light on to slit A. This is the reason for the converging beams. Note that, for maximum throughout you should choose a focal length focusing lens that matches the numerical aperture of the monochromator. With different wavelengths now ...


3

The distinction between "ordinary" and "extraordinary" rays arises when you have a birefringent material, in which light with different polarizations has different speed. Snell's Law, like many "laws" in physics, is not an absolute description of all behavior but a mathematical shortcut based on a more fundamental description which applies in many common ...


0

This answer was written for another question that was deleted a few minutes ago. I decided to post it here even though the effect it describes duplicates Floris' answer: Photons passing through a medium don't just punch their way through like bullets. They are absorbed by the atoms of the medium and then re-emitted. (Incidentally, the reduction in speed ...


3

Generally speaking we use the term diffraction when we have some apparatus that blocks part of the light. So for example a diffraction grating absorbs light except at the slits in it. When no light is being absorbed we normally use the term refraction. They're both the same physics, but the distinction is often convenient. Your question implies the obstacle ...


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I'm pretty sure it doesn't. About refraction, only when the refractive index is different we see that refraction happens. When this is not the case light just continues along a straight path, without deviating. If the obstacle has the same refraction index, it acts like it's not there. The light doesn't interact in any way with the slit, meaning that there ...


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I'm answering for gases, with which I'm much more familiar than solids. The actual strength of absorption line is less important, what matters is the wavelength. That tells you the energy associated with excitation of a particular mode or degree of freedom in the material Each degree of freedom: translation, rotation, vibration or electronic contributes ...



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