# Tag Info

1

Yes it can be seen, rainbow can be considered as a set of points which form an angle of $42^{\circ}$ with the sun and our eyes, which comes out to be a perfect circle with center at our shadow made by the light coming from the sun. Usually we see a rainbow from ground and from there these set of points form a semicircle in the atmosphere. But if we observe ...

1

Light is the visible part of the electromagnetic radiation and consists of photons. Each photon has an oscillating electric and an oscillating magnetic field. In vacuum both fields are perpendicular to the direction of the photon's motion and perpendicular to each other too (see this sketch). There are mainly two used by people opportunities to increase the ...

0

Although you can (as you obviously know) think of electromagnetic radiation as either a particle or a wave, it's easier in this case to think of it as a wave. As a thought experiment, if you wave a magnet near a piece of wire, an electric potential will be induced in the wire. Likewise, if you pass current through a wire, a magnetic field will be produced ...

0

Your question is pretty vague, but does this help-- The E-M wave amplitude is a complex oscillating function, as you can see at the wikipedia page . Now, the power in any wave is the square of the amplitude, or more precisely, the product of the amplitude and its complex conjugate. It turns out that E-M energy is quantized, so we can assign a specific ...

0

Most commercial spectrometers use gratings rather than prisms, but that's a side issue. To answer your main question: the optics inside a spectrometer re-image the entrance slit onto the detector elements. In the case of your 1x128 detector, the entrance "slit" is really a small hole which is optically matched to a single pixel in the image plane. The ...

10

A classical explanation to supplement Rod's excellent quantum mechanical one: If you make a Huygens construction of wave propagation (I assume you know how to do that) then every point on the wave front is treated as the source of a new wave of the same frequency and phase. How that wave propagates depends on the medium it encounters. So the Huygens ...

41

When light is propagating in glass or other medium, it isn't really true, pure light. It is what (you'll learn about this later) we call a quantum superposition of excited matter states and pure photons, and the latter always move at the speed of light $c$. You can think, for a rough mind picture, of light propagating through a medium as somewhat like a ...

1

Simply because two light rays intersect at a point it does not mean that an image is formed. You need millions (not necessarily, but a lot) of light rays to intersect at a point to form an image. The reason is that the intensity of light emerging from a two-ray intersection is too less for any human eye to detect. For an image formed due to a concave ...

1

The prism doesn't reflect light, it refracts the light. Different wavelengths will be refracted (bent) by different amounts. In the visible spectrum, this will produce the familiar rainbow pattern we're all familiar with. If you direct the refracted light at a linear detector, each pixel of the detector will measure the optical power at a different ...

3

The angular resolution of a telescope is given approximately by $1.22 \lambda/D$ in radians, where $\lambda$ is the wavelength of observation and $D$ is the diameter of a circular mirror. Say we study a star that is 10pc away with a telescope working in the optical band (you didn't specify) at 500nm, then the spatial resolution of a 100m telescope at the ...

3

Given that you use the tag "home-experiment" I will give an answer in that spirit. Obviously you can use the convex lens to focus the sunlight onto a piece of paper - find the distance where the paper catches fire and that is your focal length. That's how I did it when I was 4. Shoe laces too - they are really stinky when you get the distance right. Wear ...

0

So long as your lens diameter at every plane of interest (stops) is larger than, e.g., the $1/e^4$ diameter, you'll maintain a nice gaussian beam shape. -- at least, assuming a nondiverging source beam. Beyond that, I fear you'll want to get hold of Zemax or equivalent to see exactly what happens to your beam shape.

0

For the single slit interference problem, the slit is assumed to be made of N equally spaced oscillators, these small oscillators(Huygens Wavelets) produce disturbances/waves(electric field vector) which propagate in all radial directions. The interference of these disturbances from each oscillator produces the Diffraction pattern ~ (Huygens-Fresnel ...

0

Huygens' principle is linear superposition of spherical wavelets, hence it can be applied to every single wavelet individually that taken together form the incident plane wave, i.e., the propagation is applied to its Fourier transform. After they have propagated through the glass (plastic) you can sum (integrate) again and you get the far-field behavior of ...

0

yes, your friend is right you will see the walls of periscope. here's why? . there are infinite number of rays reflecting from first mirror making all possible angles. you will see walls by those light rays which strikes with angle other than 90 degree, and obviously you will not see the walls by rays reflecting at 90 degree. only rays other than 90 degree ...

2

The answer depends on how the periscope is made (the relative size of the mirrors, the shape of the box , and where do you put your eye. You do not need a periscope to see this, just look through a pipe. The mirrors do not matter they only redirect the light, so your problem is equivalent to that of making holes of the size of the mirrors at both ends of a ...

1

Your friend is right, there will be lots of rays of light entering the periscope, a few of which will reflect directly to your eye. Most of the light will bounce off the inside of the periscope and allow you to see the walls of the device. Best thing to do is to try it out!

-2

There is multiple ways holograms work, 1. A laser going into a half mirror going into 2 more mirrors that brings out the hologram... Or they get a plain price of glass and have either a projected or lots of lasers to build up the hologram...

2

Detecting reflected light from a "past Earth" is an interesting thought. Even a spectrum might tell you something about the composition, temperature of the atmosphere etc, even if as been correctly said in other answers/comments, the spatial resolution to actually image the Earth would need to be several orders of magnitude better than the Hubble Space ...

3

The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it). The amount of incident radiation from the Sun at any moment is the solar constant ($F \sim 1.3 \times 10^3$ Wm$^{-2}$) integrated over a hemisphere. Thus the total reflected light from the Earth is $L=5\times 10^{16}$ W. Let's assume this has the same ...

0

An intuitive explanation could be that the photon has a longer road to be traveled in a medium?

3

To answer the question in your title - no, the kinetic energy of an electron is a function of its velocity, same as for any other particle. The charge of an electron is always $1.6\cdot 10^{-19}C$, and so the electron will pick up $1.6\cdot 10^{-19}J$ of energy for every Volt of accelerating potential. The key to answering the question (part d in the ...

0

0.16 aJ is the energy an electron acquires by going through 1 volt. If it goes through 500,000 volts, it gets an energy almost equal to its mass.

2

Firstly, the OP is forgetting that the classic microwave polariser experiment is done with EM radiation in a pure state, not a mixture. We simply have polarised light from, say, a Gunn diode and this pure quantum superposition is forced into a polarisation eigenstate by the polariser. So we begin with near to zero entropy light, absorb some of it (adding ...

2

No. Kinetic energy depends on how much energy you give to an electron. $Volt = Work$ $done$ / $unit$ $test$ $charge$. $1.6 * 10 ^ -19 J$ is the amount of work done to accelerate an electron of charge $1.6 *10^-19$ to $1$ $volt$ potential.

1

Like many questions in life, the 'most meaningful' answer depends upon the 'context' of the question. This of course means that one must frame the answer using terms which 'make sense' to the one seeking the answer. Firstly, when we speak of 'light' in the everyday sense, we mean that part of the electromagnetic spectrum which is 'visible' to the naked ...

1

Let's look at the Beer-Lambert law: $T = \frac{I}{I_0} = e^{-\epsilon l c}$ It gives us transmittance $T$ - the ratio of two power intensities. Now, $I$ stands for the power intensity after travelling distance $l$ and $I_0$ stands for the power intensity in the beginning. Transmittance $T$ is unitless, so basically the law isn't written in terms of any ...

4

As always the answer is a simple thing. You calculated the change in entropy using the definition of entropy $$\mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T}$$ Note that this applies to heat transferred reversibly. More generally we must use Clausius theorem $$\mathrm{d}S \ge \frac{\mathrm{d}Q}{T}$$ ...

2

I think the factor you are ignoring is that the polariser will emit thermal radiation. If we continue with the ideal polariser, then it should only emit the polarisation which it absorbs (ideal components are weird). This means that there will still be a component of the absorbed polarisation in the beam after the polariser and so there will always be some ...

1

I believe the error is in assuming that the polarized beam is a pure state of zero entropy. If you characterize it in terms of polarization only, then the characterization is not complete. You need a complete set of commuting observables to charactherize a pure state. The macroscopic polarized beam is still compatible with many different quantum microstates ...

2

I suppose all the variables involved are real. The roots of the denominator of $\chi(\omega)$ are $\frac{1}{2}\Big(i\gamma\pm\sqrt{-\gamma^2+4\omega_0^2}\Big)$ which lies in the lower (upper) half of the complex plane for $\gamma<(>)0$. You need to further specify the sign of $\gamma$. $\gamma<0$ leads to the relationship in your question, while ...

1

Lionel Brits's answer and Ondřej Černotík's answer are both correct and indeed derive the ray equation from Fermat's principle. However, if you want to begin with Maxwell's equations, you derive $\vec\nabla{n} = \frac{d(n\hat{u})}{ds}$ from Maxwell's equations and then derive Fermat's principle from your ray equation. Here's how. We begin with Maxwell's ...

2

The optical path length $$S = \int\! n \, ds\,$$ for a ray travelling along some curve is extremized by the path that the ray actually takes (from a wave point of view, the phase is stationary, and so we get constructive interference). We can parameterize the path by some parameter $\lambda$, i.e., $\vec{x}(\lambda)$, so  ds = \sqrt{ {\dot x_1}^2 + ...

1

The key to solving this problem lies in careful observation. What can you learn from the image? You have a ruler ("close by") in focus, and points in the distance ("infinity") that are blurred. Depth of focus of a lens depends on its aperture - the bigger the lens, the more objects not in focus will be blurred. The size of the blur can be compared to the ...

1

This is a kind of optical amplifier -- see 1, 2. Probably the main effect of the molecules on the beam will be scattering / refraction, just like if you put anything into of a beam of light. But a secondary effect, especially if the beam is much much larger than the clump of molecules, is that the profile of the beam will change a bit. The part of the beam ...

6

You will need a reference scale next to the object and only then you can estimate the size relative to that reference scale. That why you see many pictures of small items next to a Lincoln penny (and why one was sent to Mars).

1

Looks like the Astigmatism is measured from the diodes surface as per Stephern Blake's answer. The fast axis divergence is measured from the diodes surfaces and then the astigmatism is the position of the beam source point of the lens behind the diode surface. The fast axis beam waist must be on the diodes surface at Py and the slow axis beam waist at Px. ...

0

A spectrometer measures emission spectrum, as the light is focused on its entrance slit, dispersed and registered at different wavelengths. A spectrophotometer measures absorption spectrum of a sample placed inside. The light from a built-in broadband light source is dispersed, sent through a sample and registered at different wavelengths.

8

If you know the specifics of the camera (lens system, aperture settings, etc.), then you can make a direct relationship between the size of the image and the angular size. But without a distance measurement (something the camera does not do), you can't turn that into an absolute size. If there is other information in the photograph that gives the ...

1

No, you cannot, you need at least three cameras, or three points of view to calculate distances accurately. The technology is just around the corner.

2

I guess you need blazed diffraction grating: You don't really need to design it - they are manufactured in volume for spectrum analyzers, for example here: http://www.thorlabs.de/newgrouppage9.cfm?objectgroup_id=26 . You may also check tutorial there, and a mandatory wiki link: http://en.wikipedia.org/wiki/Blazed_grating

0

Provided that the electron & the atomic beams also exhibit refraction,it seems that this is a particle's property.Velocity is inversely proportional to particle's mass/size for specific medium.Photon behaves as particle in this effect.Mass is given by de Broglie equation:m=hv/c^2 , v=frequency

1

In astronomy, a spectrometer takes the light from an object, passes it through a slit, which then defines the input to a dispersive element. As a result a (generally unknown) amount of light is lost outside the slit, so it becomes difficult to estimate the absolute flux as a function of wavelength. In a spectrophotometer you do not use an entrance slit, so ...

0

A spectrometer tells you which wavelengths of light is absorbed and which wavelengths of light is reflected. A spectrophotometer measures the relative intensity of the light absorbed or reflected at a particular wavelength of light.

2

The key word here is continuity. The continuity boundary conditions for the electromagnetic field vectors sets these phenomenons. The tangential components of $\vec{E}$ and $\vec{H}$ must be continuous across an interface - the only way that they can not be is if there is a surface current flowing (which cannot happen in dielectrics). Likewise, the normal ...

0

Whenever light (or any wave in general) goes from one medium to another, some of the energy of the wave is 'reflected' back through the first medium (at the same angle as the incident wave) and some of the energy (may be) refracted (bent) through the second medium. When light goes from a low refractive index medium to a high refractive index medium (such ...

2

My impression is the beam is not Gaussian at all, as the beam divergence in two orthogonal planes differs dramatically.

0

The phase change is what guides the process. 180 degrees phase change is reflection. Other phase changes are refraction.

0

I like to explain this in a simple way.This example might give you some feeling about intensity. consider a single bulb glowing it has some intensity,say "I" and you turn on the second bulb now the intensity has become twice than before i.e. "2I" And more technically it is; power transferred per unit area.

1

As you know the standard textbook answer to this question would be, for total internal reflection (TIR) the angle of incidence, $\theta$, must be greater than $\theta_c$, which is of course given by $sin^{-1} ({1 \over n}$) so for TIR at $\theta=0$ we need to have $n= \infty$. In the case of a dielectric mirror, this is a multilayered device a combination ...

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