Measuring the exact number of millimetres between planets

Originally posted by DeepThought
Assuming the light source is less than 1m wide the light that would have illuminated the 1m target will be blocked.

Now I am confused. A 1m hole blocks almost all the light, a 1m ball blocks almost all the light. Where is the excess light going? Is it getting diffracted away from the target? I guess that would explain it to some degree, but its still confusing.

Originally posted by twhitehead
Now I am confused. A 1m hole blocks almost all the light, a 1m ball blocks almost all the light. Where is the excess light going? Is it getting diffracted away from the target? I guess that would explain it to some degree, but its still confusing.

Depends, if the ball is shiny it will be reflected back, if it is matt then it will be absorbed and the energy re-radiated as black body radiation. Diffraction is only an important effect if the wavelength of the light is similar to the diameter of the ball, and visible light has a wavelength of the order of 500nm it's not an important effect relative to a 1 metre diameter block.

Your confusion is because I didn't define the two cases carefully enough. Suppose we have two observers, one next to a laser and a receiver on Earth and one next to a receiver on Mars. We do two experiments and measure the intensity of light the receivers see. In the first experiment we go the the Earth Mars midpoint and put a disc big enough to eclipse the light from the laser for the Martian observer with a 1m aperture. The Martian measures the transmitted light intensity. For the second experiment we replace the annulus with a 1m diameter mirror exactly aligned to point straight back and the Earthling measures the reflected intensity. The observer on Mars will detect the same amount of light during the first experiment as the observer on Earth will during the second one (neglecting atmospheric effects). The observed intensity goes as 1/R^4 where R is the distance to the mid-point from either observer.

Originally posted by DeepThought
Your confusion is because I didn't define the two cases carefully enough.

I am still very confused. You are saying that blocking the path of the light from earth to mars would have the same result on intensity as a mirror reflecting the light back to earth?

Originally posted by twhitehead
I am still very confused. You are saying that blocking the path of the light from earth to mars would have the same result on intensity as a mirror reflecting the light back to earth?

No, we have a disc large enough to eclipse the laser for any Martian observer. We cut a 1m diameter hole in the middle which lets light through. We then fire the laser. The Martian observer measures the intensity of light from the source using his receiver. We then fill in the 1m hole with a mirror, fire the laser and the Earthling measures the reflected light. They will get the same answer (neglecting atmospheric scatter).

The caveat to all this is that the 1/R^4 rule applies only if the collimation of the light source is not so good that the spread of the beam is less than 1m at the mid-point. If that is the case both observers will still measure the same amount of radiation, but there will only be inverse square degradation of the signal, since then all the light gets either transmitted or reflected. If the collimation is so good that the beam only spreads by an amount smaller than the width of the detector then all the light is detected and the only losses will be due to absorption along the beam path.