Could someone explain the calculation required to answer this question. It is from a text book and the answer is recorded as 585Hz but I cannot replicate the answer.

In 1931 Michelson used a rotating prism with 32 faces to measure the time for light to travel through a 1.6km evacuated tube to a distant mirror and back. From his results he calculated that the speed of light is 299774000 m/s.

Assuming the prism is rotated by only one face during the time of flight of light calculate the rotation frequency of the prism.

Is the book incorrect or am I?

My attempt: $(1/(1600/299774000))/32=5855\:\mathrm{Hz}$ which is suspiciously similar to the suggested answer 585Hz. But then I remembered about the round-trip (3200m) which threw all the numbers again.

  • $\begingroup$ Well, for starters, how did you go about solving the problem? $\endgroup$ – David Z Jun 6 '11 at 21:02
  • $\begingroup$ First attempt: t=d/s => ((1600/2997744000)^-1)/32=5855Hz (suspiciously similar to the suggested answer). But then I realised I hadn't taken into account the round-trip and then my answers went all wrong. $\endgroup$ – Catherine Mainson Jun 6 '11 at 21:07
  • $\begingroup$ Welcome to Physics.SE! We have MathJax active on the page, so you can use LaTeX markup in questions, answers and comments: $t = d/s = (1600\text{ m})/((2997744000\text{ m/s})$ and the like, also you can edit your question which is sometimes better than replying in the comments. Depends on the case. $\endgroup$ – dmckee Jun 6 '11 at 21:11
  • $\begingroup$ Hello moderator. Thanks, I've put my attempt into the question now. $\endgroup$ – Catherine Mainson Jun 6 '11 at 21:20

If I enter your numbers I get 585,49 Hz, so I think you've just missed a zero when you entered it in your calculator :)

Regarding the 1.6 km or 3.2 km etc, maybe the problem-designers missed it?

Here is a description of the original experiment with the mile-long tube, but there it seems as if the beam bounces several times back and forth in the tube so maybe neither total length is correct..

Edit: as noted in the comment thread, the book made two errors, both the dual time of the roundtrip and the number of light bounces per face. I was off by a factor of 10 above when I wrote that I got 585 Hz, I also got the 5855 Hz.

  • $\begingroup$ Thank-you for your answer. I think you must be right about the oversight of the round-trip in the question. However, I can't see how you get 585Hz. This is what I have been calculating bit.ly/l8tKim. I think in some of the comments above there may be the wrong number of 4s, but it is definitely correct in the question. $\endgroup$ – Catherine Mainson Jun 6 '11 at 22:53
  • $\begingroup$ Hah, yes, I missed a zero in the speed of light... embarassing.. ok, but then the answer is incorrect in that regard as well :) The reasoning is correct (although the problem is, as I wrote, probably simplified in some way). $\endgroup$ – BjornW Jun 6 '11 at 23:06
  • $\begingroup$ Can you fix the answer--- as it stands it is incorrect. Just say the book made a mistake, and Michaelson was smart to bounce the light several times so that the spin-rate wouldn't be impossibly high. $\endgroup$ – Ron Maimon Jul 4 '12 at 6:36

Speed of light ---> about 3E8 m/s.

Round trip length ---> 3200 m.

Time for round trip ---> 3200/(3E8) = about 1.066E-5 = 10.66 uS.

The mirror turns 1/32 of a full turn in 10.66 uS; so it makes a full turn in (32 x 10.66) uS = 341uS = 0.341mS.

1/(0.341E-3 s) = 2932/s = 2932 Hz (based on c = 3E8 m/s).

If the authors made the mistake of not factoring in the round trip, they should have come up with an answer in the rough vicinity of 2 x 2932/s = 5864 Hz.

Here is your third data point: The authors made TWO errors: They forgot the round trip (as you did) making them off by a factor of two (to the high side) ~AND~ made a simple decimal error, which reduced their final answer by a factor of 10. The end result is that they are off to the low side by a factor of five. 5 x 585 Hz = the correct answer of 2925 Hz.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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