Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 2 down vote favorite

Some time ago I came across a secondary web source on measurement of light speed in water made by Foucault around 1850. I append its redrawn scheme below (light is reflected from the rotating mirror to the fixed one and from the later again to the rotating one which slightly rotates in the meantime so the light ends in the direction labeled air rather than back in the light source).

When Foucault inserted a container filled with water between the mirrors then light was reflected back at larger angle (water), because light is slower in water.

Which velocity is exactly measured in this experiment? Phase velocity (light source is continuous, similar to infinite monochromatic wave), group velocity (which usually applies to a wave packet - is such a packet somehow created by rotation of mirror?) or another one?

enter image description here

Edit (based on the answer by miceterminator):

How would the result (the angle) change in case of negative group velocity (which, as far as I know, is possible)?

share|improve this question

1 Answer

up vote 2 down vote

missingShort answer:

It measures the group velocity.

With every rotating mirror experiment you always measure the group velocity. You may have a continious light beam from the source, however there is usually a aperture/blind (I don't know the correct term in english) between the rotating mirror and the fixed mirror to prevent light that is not exactly straight to pass through (If the fixed mirror is small enough you don't even need it). So the beam is split in time because it only passes through when the rotating mirror is aligned at the appropriate angle (45° in this case). Then the beam is reflected back and with a constant angular velocity of the rotating mirror you can see one point slightly away from the beam source.(Depending on the size of your experiment and the angular velocity of the rotating mirror). Without the blind you would just see a line which is indeed very useless. The experiment is also not able to measure the speed of light in air and water simultaneously. You would have to measure it in air at first and then put a tank of water in (with sufficiently small walls).

share|improve this answer
I agree that there must be some sort of highligting the measured light beam (say the air one), since the rotating mirror reflects the light coming from light source to all angles, but I am not sure if aperture would work. – Leos Ondra Jul 30 '12 at 10:15
I did this experiment once so I am pretty sure it works. The smaller the aperture the more accurate your reading because your reflected beam gets smaller. However then you can not see it as well. Thats why one usually uses a laser as light source. Why do you think that the aperture (essentially a small hole) would not work? – miceterminator Jul 30 '12 at 11:37
"Why do you think that the aperture (essentially a small hole) would not work?" Because light from the light source can reflect to the air or water direction (and all directions around them) simply by a single reflection from rotating mirror. How can you distinguish that from the light which went via the fixed mirror and has been reflected from the rotating one twice? – Leos Ondra Jul 30 '12 at 16:55
good point. Well basically you would still get double the intensity on the one point. I was wrong to put a blind in there it would only double the intensity. In the experiment we put a Lens between the rotating mirror and the reflecting mirror, with the focal point of the lens in the rotation center of the rotating mirror. That way the point is a lot brighter because for 20° rotation of the rotating mirror (depending on the size of the lens) the light is reflected back. Foucault used a Cog with a frequency coupled to the rotating mirror to stop the other light from coming through. – miceterminator Jul 31 '12 at 6:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.