ELECTROMAGNETISM & RELATIVITY - Chapter 7, Exercise 2 Solution
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The Real Ratio Behind the Michelson-Morley Null Result

PROBLEM
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Michelson and Morley expected a fringe shift of about 0.4 fringes,
but measured a shift of only about 0.02 fringes. Calculate what
fraction (as a ratio, e.g. "1 in X") the measured shift represents of
the expected shift, and explain in one sentence why this made the
null result so difficult to dismiss as ordinary measurement error.

SOLUTION
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Dividing the measured shift by the expected shift:

    ratio = measured / expected
    ratio = 0.02 / 0.4
    ratio = 0.05
    ratio = 1/20

ANSWER: The measured shift was only about 1/20th (5%) of the expected
shift - the real result was roughly 20 times smaller than the theory
predicted, which is difficult to dismiss as ordinary measurement
error because the interferometer was demonstrably sensitive enough
to detect a shift this small (0.02 fringes was itself a measurable,
reported figure), meaning the instrument clearly COULD have detected
the full expected 0.4 fringe shift if it had genuinely been present -
its near-total absence was a real, significant finding, not simply
noise the equipment was too crude to resolve.

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WHY THIS WORKS AS AN ANSWER

The key reasoning point is that "no effect detected" only becomes
scientifically meaningful once you know your instrument was sensitive
enough to have detected the predicted effect if it existed. Since
Michelson and Morley's own apparatus could clearly resolve shifts as
small as 0.02 fringes (as their reported measurement itself proves),
the fact that they saw only that much - roughly 1/20th of the
predicted 0.4 fringes - genuinely ruled out the specific ether-wind
prediction, rather than simply reflecting an instrument too
insensitive to have caught it either way.