Is Argon a Good Light Source?

is argon a good light source?

I was going to say it's not, but it's mentioned in Wikipedia as being used for a blue discharge tube. Xenon is more common in arc lamps, and otherwise argon as an inert gas is used to stop filaments burning up or evaporating, and as a shielding gas in arc welding

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What happens to light when the light source is spinning rapidly

Assuming the rotation is perpendicular to the photon emission, it would not affect the wavelength of the photon. However, if the tangential velocity were in the same direction as the emission of a photon then it would blue shift it. If they were opposite then it would be red shifted.

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You have a container which has inner walls if mirrors. There are no opening that will let light in or out. You put a light source inside, close the container and somehow turn the light source off, will you see any light after you open the container?

You have a container which has interior walls covered in mirrors. There are no openings that will let light in or out. You put a light source inside the container, close the container, and somehow turn the light source off. Will you see any light after you open the container?Nope.u2026but the container will be slightly warmer than it was when you started, by however many watts the luminaire consumed.You see, no matter how efficient the mirrors are they only reflect something less than 100% of the light. What they do not reflect is converted into heat. Very quickly (at the speed of light - no pun intended) the light energy will be converted into heat energy, and will add to the general heat load at the same rate as it takes away from the light load.You have a container which has inner walls if mirrors. There are no opening that will let light in or out. You put a light source inside, close the container and somehow turn the light source off, will you see any light after you open the container?

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There is no chandelier or any light source in my room but there's an on off switch?

One of the outlets in that room will be controlled by that switch. (Maybe only one of the 'plugs' in one double outlet.) Find it by turning the switch 'off' and testing the outlets with a table lamp. That's where you plug in the light. (Use common sense. It's almost always the outlet where you would put the bed.)

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Distance of an observer from a parabolic light source

The previous answer (assuming a perfect point source and perfect parabolic mirror) is true, ignoring diffraction. Without diffraction, in the perfect geometric case with perfect collimation, you would see a perfect point source. But since light is a wave, it can never be perfectly collimated, nor could you detect a point source even with arbitrarily well-collimated light. Indeed, light can be awfully well collimated if the width of the beam is millions of times larger than the wavelength, but what you will ultimately see all depends on the length scales involved, including the size of your beam, the wavelength, the distance, and the size of your detector. Thus, even for the case of perfect geometry, you will see a blurred spot and/or a dimmed spot. The way to counteract diffraction is to use a larger parabolic mirror (producing a larger beam waist) and a larger detector the farther away you want to measure. So it's clear that the beam cannot be collimated due to diffraction and will develop a wavefront curvature. In the short range, you will be able to capture all of the quasi-collimated light in your (finite size) detector. The spot will be bright (full intensity, so to speak), but the wavefront curvature will leave it slightly blurred. On the other hand, in the limit of large distance away from the emitter and small detector compared to the wavefront curvature, the beam will have diverged dramatically. In this case, the beam wavefront will look practically flat, producing a diffraction-limited point on your detector. But the brightness will diminish with distance because the farther away you are, the less light you capture (analogous to a faraway star).