Tag Info

New answers tagged

14

You see it because it travels through air, dust, and a lot of other molecules and particles that can reflect and diffuse it. This, together with focussing, is also the reason for why it cannot travel arbitrary long distances. If you go to vacuum then the laser beam has much less losses, and it can travel much farther as happens in the LIGO interferometers ...


1

The Verdet constant is a coefficient which sums up the magneto-optical properties of the medium. So, the temperature and wavelength dependence are wrapped up in it. Fundamentals of Photonics by B.E.A. Saleh expresses the Verdet constant in terms of the wavelength as $$ V\simeq-\frac{\pi\gamma}{\lambda n} $$ where $\lambda$ is the wavelength of the light ...


2

1) does "atomic density" correspond to "atoms in a volume unit that partecipate to the stimolated emission process"? "Atomic density" means number of total rubidium atoms per volume of gas, not just the ones that happened to participate in the absorption/emission process. 2) is $\sigma$ the cross section for a particular transition? (for example the ...


0

Wow, this is a very detailed question. Thanks for your effort. Lets ignore diffraction effects, which will scatter some small amount of extra power out of the laser beam. The loss at elements 1a to 1d will not simply sum up. This is because the power will only be lost at one of the mirrors, and will not be there to be lost at the other mirrors. So, to ...


0

The reason incoherent light is passed through a slit is to make it coherent . The light from a point source is forced to be coherent, the incoherent part is scattered off, or absorbed, at the opaque surface where the slits are formed. Laser light is by construction coherent so no first slit is needed. Scattering also often results in coherent waves and ...


0

Yes, you will see interference pattern, provided the slit widths and their separation are appropriate. Look here. If laser beam width is greater than the separation of the slits, the beam may split and might form interference pattern. It is to be noted that particles of appreciable size (coming in the path of laser) will also scatter light according to ...


0

The separation of the 2 slits is in the order of micrometers. A beam, taking into account its divergence usually has a radius much bigger than that. Its a key part of the experiment to have the 2 slits very close to each other, so the path difference between the beams is an integer multiple of the wavelength of the laser.


2

I'll assume that both lasers are of the same type, i.e that both are generated from the same physical mechanism. All lasers have some tuning range over which their wavelengths can vary. So, even though both are very monochromatic (single frequency) they will, in general, not have the exact same frequency. A HeNe laser, for example, has a tuning range of ...


0

In order to isolate one pulse out of a train of pulses in the MHz regime a Pockels' cell is generally used in commercial Ti:sapphire laser technology. This allows for single-pulse picking and amplification and can be used with most femtosecond laser oscillators. It is possible to cavity-dump a femtosecond oscillator and lower the repetition rate that way, ...


18

Yes easily. Fiber-fed 5-10kW Nd-Yag lasers are commonly used for cutting metal in machine shops. Fibers are so transparent, especially when designed for a single wavelength laser, that the power loss and so heating in the fiber is very small. It's generally less than an optically fed laser where dirt accumulates on the lenses. Many systems have a thin ...


1

1) First, you said that the telescope would be > 150 mm so that atmosphere is the limit, not diffraction. But the question was, will a 114 mm scope work? Not only are you diffraction-limited, but actually coupling all of the laser power into the scope so as to get that level of collimation on the output beam is not trivial. It is true, though, that this ...


0

The focal point of your lens does not necessarily coincide with the center of the cylinder! Let's call z the axis of the cylinder and r a radial direction. Since the Fresnel pattern is only along the z direction, the lens will only focus a beam of light coming along r within a plane that contains z-r (and not perpendicularly to z). The focus point is ...


1

The answer is Yes. However the losses will be wavelength-dependent, you will likely get a "ugly" superposition of modes at the output and a dispersed signal as user288447 stated. You are correct, the mode for each wavelength need to be solution of the Helmholtz equation. For a dielectric waveguide, there is no absolute cut-off wavelength (all wavelength can ...


0

It's likely bad taste to answer your own question, however, here I am only pointing out the argument of another in the literature. I also won't accept this answer to allow others to respond. Surprisingly, my above guess that perfect copying is only limited by the probability of spontaneous emission... might actually be correct. Quoting from [Milonni, P. ...



Top 50 recent answers are included