New answers tagged optics
1
Regarding your question about changing direction:
If no light can traverse the atmosphere without interacting with air,
then what fraction of it reaches the ground without significant
changes to its direction
Remember that air has an index of refraction different than the vacuum of space (about 1.00027 for air versus 1 for space). The index of ...
2
The light you see as the image of the Sun on the sky is basically undeflected.
http://en.wikipedia.org/wiki/Diffuse_sky_radiation says it is 75 % when the Sun is high and the sky is clear. The frequency dependency is due to Rayleigh scattering.
For the cloudy sky the fraction is much smaller, up to many orders smaller than unity (maybe 1 millionth part as a ...
2
To give an unmathematical catchy answer, let's look at Fraunhofer diffraction in double slit experiment.
Interference at the observation plane depends on slit parameter $d$. What is the frequency of slits? E.g. $1\,\text{mm}\frac{1}{d}$: number of slits per length. Concluding frequiency in the setup. The following argumentation links this frequency to the ...
3
To get the local heating you will need some measure of the optical density of the material and an estimate of the local intensity of the light.
You can probably look-up an estimate of the optical density, or if you need more precision measure it yourself.
Starting from a known intensity, ray optics will give you an easy estimate of the position dependent ...
0
Probably not. A fresnel lens isn't just a rippled surface, it has discontinuities, or straight edges. Sound and other vibrations could create sine wave-like ripples on the surface of a liquid, but never the discontinuous shapes required to make a fresnel lens. Interesting idea, though.
3
The basic concept of optical alignment is to perfectly align the beam step-by-step on all optical elements. Enshure your beam has the right height and angle before alignment. Your aim to align the beam on microscope optical axis. Having point sources like flourescent beads allows an easy alignment to a point spread function on your CCD sensor.
On-axis with ...
0
Somehow all answers so far failed to mention that a segmented mirror consisting of flat segments could in principle yield a high optical quality (diffraction-limited) telescope provided the mirror segments are small enough.
Spherical (or hexagonal or any other compact shape) flat mirrors would work perfectly provided:
1) the flat segments are positioned ...
0
Reflection polarizes light. A reflected ray becomes linearly polarized perpendicular to the plane containing the incident and reflected rays. This is why polarized sunglasses are effective for reducing glare. The autofocus may not be working as expected because much of the scene is polarized light.
1
As John and others have said, the wavelength of the microwaves is very large compared to the size of the holes in the screen which allows the screen to act as a solid. Visible light has much smaller wavelengths and can pass through the holes unobstructed. It isn't possible to see (resolve) objects and features smaller than the wavelength of light ...
1
Assuming the surface of the metal remains smooth, the reflection from it will be specular and the metal will look shiny regardless of the temperature. However the amount of light metals absorb, instead of reflecting, generally increases with increasing temperature because you get more scattering of the conduction electrons by lattice vibrations. So the metal ...
3
Here are ray diagrams that show what is going on. In the top case, a weak (thin) lens doesn't have the power to pull the rays together tight enough. An object farther away than the tree would make rays converge on the retina. This is farsightedness.
Remember the fundamental formula for thin lenses (using some appropriate sign convention):
$$
{1\over ...
0
$aNw = 1.33$ (refractive index of air/water interface from tables)
Conversely $wNa = \frac{1}{1.33} = 0.75$ ( going from water to air)
$Dr/Da = N$ (real depth / apparent depth = refractive index)
So, $Dr/Da = N$ becomes
$3m / Da = 0.75$
$Da = 3 / 0.75 = 4m$
The bird appears to be flying at $4m$ above the water surface.
This makes sense because we know ...
0
No, you won't see interference. The cw and ccw states are orthogonal.
You can prove that intuitively in the following way.
You could think of the incoming light to be cw polarized, then one waveplate would turn it vertical, the other one horizontal.
1
I don't think this is a quantum optics problem. Just look up the van-Cittert-Zernike theorem . The (complex) visibility is the Fourier transform of the mutual coherence function of the source.
4
The focal point of your lens is indeed on your retina, when you look at an object far away.
If you look at an object closer by, that object is also imaged onto your retina (because you are changing your eye's lens. So it's not really that important whether your lens' focal point is on the retina, but whether you are imaging a given object (sometimes the word ...
2
Well the image is flipped on the retina (your brain fixes that), but that doesn't change if you go slightly in front or behind the focus.
The rays that you draw in paraxial approximation ( parallel becomes focal, focal becomes parallel , center stays center) are simply not all in one point if you move your detection plane slightly away from the perfect ...
0
In regards to calculating the intensity distribution of LED light I have the following example. I have 4 LED lights attached to a apparatus with each LED located at the 4 corners of a square (5mm edge). I want to calculate the overall distribution of the light. After doing some research I strongly believe the gaussian beam model ...
0
Remember that the situation you describe is exactly the same as you remaining stationary while the mirror moves past you at relativistic speeds. The reflection from a moving mirror is analysed in this article. If the mirror is parallel to it's direction of motion ($\phi$ = 0) the normal rules apply i.e. the angle of incidence is equal to the angle of ...
0
Plot the imaginary copy of yourself on the other side of the mirror, by the laws of geometrical optics, for every moment of time. *)
Then consider that copy a real body you observe, and apply all known relativistic effects to its apparent image, taking into account both your and its motion. In the case of a parralel flat mirror and uniform motion, all those ...
5
You can use a reflector with gaps. Then the light from a car will alternate between reflecting and not reflecting at a rate dependent on their velocity towards the reflector. Please excuse my crude diagram:
As the car moves right to left, gaps in the reflector will cause it to appear to flash on an off.
0
Due to the speed of light, the "dopler shift" would be too small, at the usual speed of the observers.
2
I don't think you need a microscopic explanation for the phenomenon. All you need is the fact that the phase of the wave must be continuous across the water-air boundary.
If you take any two points at the interface, spaced a distance $w$ apart, then the length of the line segment, $l$, I've labelled d$\phi$ is just $w \sin i$, and the phase difference is ...
1
we have the formula
n = real depth/apparent depth
so, real depth = n * apparent depth
= 1.33 * o.4
= 0.532
so right ans may be D
here n is refractive index of water with respect to air
0
Imagine the bars of a jail cell. Tennis balls shot directly at them will only bounce back if they happen to hit a bar and at the right angle. This is not completely comparable to reflection, but wait for the rest of the analogy. Now imagine shooting tennis balls at the bars (horizontally) at an angle of 5-10 degrees from parallel to the "wall" of bars. The ...
1
No mirror can be perfectly reflective due to quantum tunneling so that already answers your question. But even if it could be done, you would never be able to check the situation because when you look inside, the light almost instantly leaves through the peephole. This also poses a problem for your initiation method, which John M already touched on: you need ...
0
Short answer: No.
You have to remember that light is very fast. The law of specular reflection states that an angle made with a mirror and incoming beam will be the same angle the outgoing beam and the mirror make.
If you were able to put light in completely perpendicular to the surface of lets say a half-a-meter radius mirror(insides are mirrors) sphere ...
2
First, of course there's no perfect mirror. But let's assume there was one.
Next, the question is: Is the bouncing off the mirrors elastic or inelastic. If the photon is absorbed and re-emitted with the same frequency, then the bouncing is elastic and no energy is lost by the photon. It would then go on forever and ever.
But what if it does lose energy ...
1
It's like leverage. The longer the distance from the objective lens to the virtual image, the larger the virtual image.
Imagine there's a piece of frosted glass at the focal point. It will show the virtual image.
Now the eyepiece looks at that virtual image with a magnifying glass.
That also makes it look bigger.
1
You're not being very careful about your terminology. Instead of weight I think you're really talking about the total mass-energy of the sealed cavity. If you want to discuss weight the it would be better to think of that as the force on the total mass-energy in a gravitational field.
Thinking about it in terms of mass-energy makes the question much ...
0
There is a misunderstanding here.
If the beam(s) are not changed at the origin, they deliver the same energy on the screen whether one sees interference or not. It just has a different spatial distribution.
Conservation of energy says that the energy of the beam(s) should equal the energy absorbed by the screen in total plus the energy reflected. It has ...
3
There is no magic here. To see the laser beam, the beam has to use light in the visible frequency and some of the light needs to enter your eyes. Unless the beam is pointing at your eyes you won't see it. Remember, in coherent light all the photons are traveling in essentially the same direction, with the same frequency, in the same phase.
If you want to ...
7
Yes. Consider quantizing electromagnetic fields in a box. This corresponds to photons being trapped inside of said box since photons are just the mode quanta of the EM fields. The Hilbert space (called Fock space in this case) of the quantized radiation is found to be spanned by states
$$
|\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad ...
1
A piece of white paper is the most common device to show a laser beam. The beam is visible on both sides of the paper.
Fix it on a electronically controlled motor shaft you can rotate the paper into the laser beam. If precise control is necessary: just align an iris aperture to the center of the laser beam.
1
Assuming mirror to be spherical section. C is the center of sphere.
See, Using trigonometry. $$x=d \times \sin(2\theta)$$
$$x=R\times\sin\theta$$
Eliminate $\theta$ and get $d$ : distance from Center of curvature as a function of $x$.
Verify for small theta where $\sin\theta\approx\theta$
If you just want to see that which side ray bents then see. ...
1
A parabolic mirror is a special case of a concave mirror. The rays at the rim are refracted to the focal point. This is true in simplification of geometric optics and perfect manufactured mirror.
However in modern techiques like injection molding there are more imperfections at the rims of mirrors.
The question of a general concave mirror can be answered ...
0
In theory, yes, the light will be redshifted. In practice, it sounds like the glass bead is too large for any measurable red shift. This is actually used in Mossbauer spectroscopy. What happens is that if your $\gamma$-ray source is a free atom, the recoil of the atom will cause the resulting radiation to be red-shifted relative to the natural frequency of ...
1
Sure - the relativistic doppler effect means that light which is scattered off a moving object can be redshifted or blueshifted. And there can be more redshifted photons than blueshifted photons, or vice-versa, depending on where the object is, and how it's moving, relative to the center of the trap.
But since the object is moving much much much less than ...
1
You find $s$ in the same way you found it for concave mirror, just keep in mind that for the convex mirror the image is always virtual(as explained nicely in the Wikipedia article on curved mirrors, it cannot be projected on a surface, unlike the real image), so in this case (by convention) $s'=-35cm$ and $f=-53cm$.
3
This link: http://alienryderflex.com/polarizer/
has an excellent explanation; much better than anything I could write here.
Essentially, it says that this occurs because the 45 degree filter outputs a projection of the vertical rays at 45 degrees. This, in turn, has a horizontal component, which the final filter projects in its output.
2
The intensity distribution will remain same as fraunhofer one only , however intensity will increase at each point . Means supposedly earlier there was Amplitude 2A , at a point of maxima , now it will become 4A , since 2 waves will come from the new fraunhofer slit also as it is not very distant from the previous slit , hence , you can assume the maximas ...
2
This follows from from Fermat's principle. A derivation for the reflection is e.g. found here.
2
The component of the Jones vector are not defined as two pure imaginary exponentials but as two complex numbers $z_1$ and $z_2$ whose polar decomposition is $z_i=|z_i|e^{i\phi_i}$ for $i=1,2$. To describe linearly polarized light along the $x$ axis
$$ |H \rangle =
\begin{bmatrix}
1 \\
0
\end{bmatrix}
$$
we just take $z_1 \in \mathbb{R}$ ($\phi_1=0$) and ...
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