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A shift of one fringe in the michelson-morley experiment corresponds to a change in the round-trip travel time along one arm of the interferometer by one period of vibration of light when the apparatus is rotated by 90 degrees. a particular interferometer has arm lengths of 8.50 meters, and is using light with a wavelength of 658 nm. what velocity through the ether would be deduced from a shift of one fringe?

Respuesta :

Answer : [tex]v = 59.02\ km/s[/tex]

Explanation :  given that,

Arm length = 8.50 meters

Wavelength = 658 nm

Number of fringe = 1

We know that,

Formula of the speed is

[tex]v = c\sqrt\dfrac{N\lambda}{2L}[/tex]

Now, put the values

[tex]v = 3\times10^{8}\ m/s\sqrt\dfrac{1\times658\times10^{-9}\ m}{2\times8.50\ m}[/tex]

[tex]v = 590.21\times10^{2}\ m/s[/tex]

[tex]v = 59.02\ km/s[/tex]

Hence, this is the required solution.







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