1. In the Michelson-Morley experiment, the length L of each arm of the interferometer was 11 meters. Sodium light of wavelength 5.9 x 10^-7 m (590 nm) was used. The experiment would have revealed any fringe shift > 0.005 fringe.

What upper limit does this place on the Earth's velocity through the supposed Ether?

*Solution*

Given L = 11m and lambda = 5.9 x 10^-7 m

limit for lowest resolution: 2 delta d = 0.005 fringe

Then:

d = (0.005 fringe)/ 2 = 0.0025 fringe

and since:

delta d = Lv^2/ (lambda c^2)

v = {(delta d)(lambda) c^2)/ L}^½

v = {(0.0025)(5.9 x 10^-7 m)(3 x 10^8 m/s)/11 m }^½ = 3.5 km/s

2. Using Fig.1, say the time of travel to the right is: t(r) = L/(c - v) and the time of travel to the left is t(L) = L/(c + v).

a) Find the "total time of travel" by adding both left and right contributions.

Solution:

We are going along (parallel) or opposed (anti-parallel) to, the "ether wind" direction from the diagram and this is

*horizontal*so designate it by direction x, e.g.

t(total x)= L/ (c + v) + L/(c - v)

The common denominator is:

(c - v)(c + v) = c^2 -cv + cv + v^2 = c^2 - v^2

Then t(total x) = 2Lc/ (c^2 - v^2)

t(total x) = 2L/c (1 - v^2/c^2)^-1

b) find the time consumed for "a half-trip".

Solution:

half trip time = (t(total x))/2 = ½ [2Lc/ (c^2 - v^2)]

(t(total x))/2 = Lc/(c^2 - v^2)

But recall that the interferometer has parallel and perpendicular components so that total time registering both is:

T(total x-y) = t1 + t2 where:

t1 = 2Lc/ (c^2 - v^2)

t2= 2L/ (c^2 - v^2)^½

So: t1 + t2 = 2Lc/ (c^2 - v^2) + 2L/ (c^2 - v^2)^½

= 2L{c + (c^2 - v^2)^½)/ (c^2 - v^2)

so T/2 = L{c + (c^2 - v^2)^½)/ (c^2 - v^2)

c) Find the time consumed for a round trip.

round trip is 2(t1 + t2) = 2(2L{c + (c^2 - v^2)^½)/ (c^2 - v^2))

= 4L{c + (c^2 - v^2)^½)/ (c^2 - v^2))

d) Add the two "half trips" and what do you obtain?

We get:

2 x T/2 = 2L{c + (c^2 - v^2)^½)/ (c^2 - v^2)

e) Why does this not agree with the value obtained for (a)?

Doesn't agree because each summing (or halving) as computed above takes into account

*differing directional components*of velocity with respect to the interferometer.

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