The set of videos linked below animates experiments done by Albert Michelson, ov

Question

The set of videos linked below animates experiments done by Albert Michelson, over a span of nearly 50 years, to measure the speed of light. The ball in the animations represents a flash of light. The light first reflects off one of the faces of a rotating mirror and travels to a distant stationary mirror. There it is reflected back to the rotating mirror. If the mirror is rotating at just the right speed, the light will be reflected into the detector (tube).

Play the animations in the following order:

http://crodog.org/animate/speed%20of%20light%20slow.html

http://crodog.org/animate/speed%20of%20light%20fast.html

http://crodog.org/animate/speed%20of%20light.html

By knowing the distance, d, to the distant stationary mirror and the rotation period, T, of the mirror, it is possible to deduce a value for the speed of light, c = 16d/T. (Here it is assumed that the 8-faced mirror makes one eight of a revolution during the round-trip (2d) of the light flash. (Other time intervals t = (nT/8), where n is an integer, are possible if the mirror rotates faster.) In Michelson's last and most accurate experiment (1927) the distance was d = 35 km, and his result was 186,285 (+-)2.5 mi/s.

If the 8-sided wheel rotated 1/8th of a turn during the round trip of the light flash, what was the rotation speed, in rpm? (Use only numbers in all of these answers, do not include units.)

a.Express Michelson’s best measurement in km/s, to six significant figures. [Warning: your conversion factor must be exact. 1609 m/mi is only an approximation, not good enough for a six-sig-fig conversion.]

b.What is the uncertainty, in km/s?

c.If the uncertainty were due entirely to the accuracy of d, what distance would that represent, in m?

Explanation / Answer

Solution: we note

1 inch = 2.54cm (exact) = (127/50) cm

1 feet = 12 inch = 12*(127/50)cm

1 mi = 5280 feet = 5280 feet*(12 inch/1feet) = 63360 in

Now 1mi = 63360in*(2.54cm/1in) = 160934.4 cm (exact)

Since 100cm = 1 meter, thus

1 mi = (160934.4cm)*(1m/100cm) = 1609.344 m (exact)

Since 1km = 1000m, thus

1 mi = (1609.344m)*(1km/1000m) = 1.609344 km (exact)

Q: If the 8-sided wheel rotated 1/8th of a turn during the round trip of the light flash, what was the rotation speed, in rpm?

Since the wheel rotates 1/8th of a turn, then the formula c = 16d/T applies here.

we have d = 35km; c = 186,285 (+)2.5 mi/s we need to find T

T = 16*d/c = 16*35km / (186,285 mi/s)

Since 1km = (1/1.609344)mi

T = 16*35*(1/1.609344)mi / (186,285 mi/s)

T = [(16*35) /(1.609344*186285)] s

T = (560/299796.647)s

T = 1.86793283*10-3s

It means that one revolution takes 1.8679283*10-3s of time. Thus in one second the number of revolutions is given by

= (1rev/1.8679283*10-3s)

= 535.35115 rev/s.

In a minute, revolutions are given by

= 535.336519 rev/s *(60s/1minute)

= 32121.1 rev/min

Thus the answer is 32121.1 rpm

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Part a) Express Michelson’s best measurement in km/s, to six significant figures.

=>        Michelson’s best estimate of speed of light is 186285 mi/s

But we know 1 mi = = 1.609344 km (exact)

Thus speed of light in km/s is

186285 mi/s = (186285 mi)(1.609344 km /1mi) / s

186285 mi/s = 299796.647 km/s

299797 km/s with six significant digits

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Part b) What is the uncertainty, in km/s?

The uncertainty + 2.5 mi/s

Thus using above conversion procedure we obtain;

2.5 mi/s = (2.5mi)(1.609344 km /1mi) / s

2.5 mi/s = 4.02336 km/s

Thus the uncertainty in km/s is + 4.02336 km/s

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Part c) The relative error in the speed of light is given by

relative error = (2.5 mi/s)/(186285 mi/s)                              or

relative error = [(4.02336 km/s) / (299796.647 km/s)]   

relative error = 1.34202969*10-5

Thus the same relative error was due measurement of distance alone (no error in time measurement), then error in measurement of d is

= d*relative error

= 35km*1.34202969*10-5

= 4.69710390 km        or

= 0.469710 meters

thus the uncertainty in distance is + 0.469710 m

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