Quick Physics Question

Eli · June 12, 2003, 10:29

If I hover above the moon at a distance of, say, 10km from the surfrace and turn on my flashlight and point it downwards, will someone who's standing on the moon directly below me be able to clearly see the light because the moon has no atmosphere?

If the answer is yes, how far will I have to be for him to not see the light(assuming nothing blocks it except couple straying hydrogen atoms), and why wont he see it?

Dauphin · June 12, 2003, 10:34

I would guess that you use the inverse square law to determine its perceived intensity, and then it fades from view when that intensity drops below the threshold of human vision.

You may also want to factor in other light sources that impair visibility of weak light sources, such as the glare of the Earth or sun.

Eli · June 12, 2003, 10:39

Quote:

Originally posted by Big Crunch
I would guess that you use the inverse square law to determine its perceived intensity, and then it fades from view when that intensity drops below the threshold of human vision.

But why would the intensity decrease? The light travels through vacuum, so there is no scattering, absorption and whatnot.

Quote:

You may also want to factor in other light sources that impair visibility of weak light sources, such as the glare of the Earth or sun.

Ok. They impair visibility in the sense that the guy on the moon will have a harder time seeing my light because it might be "overpowered" by the sources you mentioned, right? But do they have any effect on the intensity of the rays themselves(the ones which come from my flashlight)?

MikeH · June 12, 2003, 10:41

Quote:

Originally posted by Eli
If I hover above the moon at a distance of, say, 10km from the surfrace and turn on my flashlight and point it downwards, will someone who's standing on the moon directly below me be able to clearly see the light because the moon has no atmosphere?

If the answer is yes, how far will I have to be for him to not see the light(assuming nothing blocks it except couple straying hydrogen atoms), and why wont he see it?

Yes.

Not sure how far away he'd have to be but there is a background light density of the galaxy. You'd need to find out what that is, and use the inverse square law to work out how far away the light would have to be to appear no brighter than the background light.

MikeH · June 12, 2003, 10:43

Light intensity decreases because you have the same amount of light but it comes out of the torch in a cone not a straight beam.

Eli · June 12, 2003, 10:45

Quote:

Originally posted by MikeH
Light intensity decreases because you have the same amount of light but it comes out of the torch in a cone not a straight beam.

For the sake of the argument, you can assume that it's a straight beam.

MikeH · June 12, 2003, 10:48

In that case, if space is a real vaccuum and nothing ever gets in the way and the moon wasn't spinning he'd always be able to see the beam.

I don't think you can have a straight beam though because light automatically diffracts from a point source and at that kind of difference the torch would appear to be a point source.

Jack the Bodiless · June 12, 2003, 10:49

...Ugh. Insufficient data. How bright is the flashlight, how tight is the beam, how good are the observer's eyes?

He should certainly be able to see it better than on Earth. If it's one of those big million-candlepower ones, he should be able to see it at 10km, I guess.

Quote:

But why would the intensity decrease? The light travels through vacuum, so there is no scattering, absorption and whatnot.

Because the emergent photons wouldn't be perfectly parallel.

A laser would be better, but still not perfectly parallel, due to microscopic imperfections in the flatness of the crystal (or gas container, if a gas laser).

Quote:

For the sake of the argument, you can assume that it's a straight beam.

...Um. Still insufficient data (beam diameter and brightness?), but eventually "space dust" would block it. Over a HUGE distance though.

Zkribbler · June 12, 2003, 10:59

Quote:

Originally posted by Eli
. . . it's a straight beam.

Ah! A laser beam.

(BTW, even a laser beam spreads out, but very slowly. One of the first experiments with lasers was to point a beam at the moon from Earth. After 1/4 million miles, the beam had spread out to a 3 mile diameter.)

Cruddy · June 12, 2003, 11:10

The answer is sometimes, given the above answers - the glare from the sun (or maybe the relected Earthlight too) might drown out the light.

Also bear in mind the sensitivity of the eye. It's most sensitive to weak light sources at the edge of vision, not in the centre. So if the observer looks straight up, they could miss it more often than if they were looking say 80 degrees up.

All the above valid (coherence of laser beams, inverse square law, and how bright is your torch compared to eye sensitivity).

Might be an idea to flash the light - stars (when visible from the Moon) don't twinkle, so the observer could be more sure it was you shining that torch.

EDIT: Using a laser sounds dandy in theory, but getting the beam centred on the observer would be very difficult without some sort of targetting system. Think how difficult it is to control a laser marker by hand, and you'll see what I'm getting at.

alva · June 12, 2003, 11:11

Quote:

(BTW, even a laser beam spreads out, but very slowly. One of the first experiments with lasers was to point a beam at the moon from Earth. After 1/4 million miles, the beam had spread out to a 3 mile diameter.)

Pfff, must have been 'made in Tai-Wan'

Jack the Bodiless · June 12, 2003, 11:13

When you look at a distant star, all the starlight that's capable of reaching your eye lies in a narrowing beam that starts out with the same diameter as the star, but ends with the diameter of the pupil of your eye. Obviously, over the last few million miles, it's an almost perfect match for the pupil's diameter. So a beam that small can travel millions of miles without significant degradation.

The same applies to light from stars in distant galaxies, except that the end diameter is that of a big telescope: but the beam isn't much bigger for many lightyears. So your hypothetical perfect flashlight would probably be visible over interstellar distances.

In other words: space dust is negligible for any real-world flashlight (or laser). Beam spread is the main limiting factor.

Boris Godunov · June 12, 2003, 11:16

I suppose that, if you eliminated all possible impediments and had an absolutely perfect and straight laser beam, the limit of seeing it would be based on the distance of the viewer from the laser source and the speed of light. If the laser source is far enough away that the viewer would die before seeing it...there ya go!

Eli · June 12, 2003, 11:19

JtB:

Quote:

Because the emergent photons wouldn't be perfectly parallel.

A laser would be better, but still not perfectly parallel, due to microscopic imperfections in the flatness of the crystal (or gas container, if a gas laser).

That's what I was looking for. Thanks.

dannubis · June 12, 2003, 11:44

this is a typical quantum mechanical problem. Light can be described by means of moving photons as well as by a wave equation. If your light is not perfectly coherent, you get all sorts of attenuation due to interference in the wave-equations... Strange as hell, but apparently true ...

Urban Ranger · June 13, 2003, 01:00

Quote:

Originally posted by Boris Godunov
I suppose that, if you eliminated all possible impediments and had an absolutely perfect and straight laser beam, the limit of seeing it would be based on the distance of the viewer from the laser source and the speed of light.

And intensity of the source.

JellyBean · June 13, 2003, 02:41

Even perfectly coherent light will suffer from diffraction. When the distance L between the light source and the observer is much greater (100 times or so) than the original width of the beam D and the wavelength W of the light, there will be a central spot of light illuminating the observer with a fainter series of concentric rings around it. The central spot will have a radius of W*L/D.

For example, if the torch is a laser shining a beam of green light (wavelength of 500nm or 5*10-7m) with a diameter of 5cm (5*10^-2m), at a distance of 10km (10^4m) most of the light will be spread out over a circle with a radius of 10^-1m or 10cm.

See http://www.colorado.edu/physics/phys...html/lab5.html for a good explanation of this - scroll down until you get to the bit about a single-slit diffraction pattern.

Edit: Learned how to put in a link on these boards.

MichaeltheGreat · June 13, 2003, 03:26

Quote:

Originally posted by Jack the Bodiless
...Ugh. Insufficient data. How bright is the flashlight, how tight is the beam, how good are the observer's eyes?

If it's pointed at you, in undisturbed visibility conditions, a typical three D cell maglite can be seen on a dark night from 10 km distance on earth by someone with normal vision. Army issue flashlights with white light can be seen at a little over half that range.

Jelly Bean is absolutely correct, also.

Vince278 · June 14, 2003, 02:11

Quote:

Originally posted by Eli
If I hover above the moon at a distance of, say, 10km from the surfrace and turn on my flashlight and point it downwards, will someone who's standing on the moon directly below me be able to clearly see the light because the moon has no atmosphere?

If the answer is yes, how far will I have to be for him to not see the light(assuming nothing blocks it except couple straying hydrogen atoms), and why wont he see it?

Will the light appear weaker at a distance? Yes
Will the light have the same appearent magnitude? Yes