I am lost here...I have tried to follow the forumla but I keep coming up with 31

Question

I am lost here...I have tried to follow the forumla but I keep coming up with 31.0502 km/s as my solution WHICH is wrong.

A shift of one fringe in the Michelson-Morley experiment would result from a difference of one wavelength or a change of one period of vibration in the round-trip travel of the light when the interferometer is rotated by 90 degree . What speed would Michelson have computed for Earth's motion through the ether had the experiment seen a shift of one fringe? (The orbital speed of Earth is 29.8 km/s and the expected number of fringes in the Michelson-Morley experiment is 0.4.) km/s

Explanation / Answer

I think the L in your formula is not the wavelength of the light. It's the total path length of each interferometer arm. The wavelength of light enters via the delta(t). delta(t)=w/c, where w in the wavelength of the light. And you shouldn't be thinking the fringe affects the velocity, the velocity affects the shift in the fringe. If light were traveling in a medium, then along the interferometer arm pointing in the direction of the earth's motion the velocity of light would be c+v in one direction and c-v in the other. The average velocity along that path of the interferometer is then less than c

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