The Nature of Light
Practice
practice problem 1
Let each of two persons take a light contained in a lantern, or other receptacle, such that by the interposition of the hand, the one can shut off or admit the light to the vision of the other. Next let them stand opposite each other at a distance of a few cubits and practice until they acquire such skill in uncovering and occulting their lights that the instant one sees the light of his companion he will uncover his own…. Having acquired skill at this short distance let the two experimenters, equipped as before, take up positions separated by a distance of two or three miles and let them perform the same experiment at night, noting carefully whether the exposures and occultations occur in the same manner as at short distances; if they do, we may safely conclude that the propagation of light is instantaneous; but if time is required at a distance of three miles which, considering the going of one light and the coming of the other, really amounts to six, then the delay ought to be easily observable….
In fact I have tried the experiment only at a short distance, less than a mile, from which I have not been able to ascertain with certainty whether the appearance of the opposite light was instantaneous or not; but if not instantaneous it is extraordinarily rapid….
 Estimate the time for a light wave to travel the distance in Galileo's speed of light experiment. (Nota bene: Un miglio italiano corrisponde a 1,873 chilometri.)
 How does this compare to the reaction time of a typical human?
solution
Start with the definition of speed and solve it for time.
∆t = ∆s v ⇐ v = ∆s ∆t Use the round trip distance of two Italian miles.
∆t = 2(1,873 m) 3.00 × 10^{8} m/s ∆t = 1.20 × 10^{−5} s
For this problem, it's probably better to state the answer in milliseconds.
∆t = 0.012 ms
Reaction time for an action like opening and closing a lantern shade is around 300 milliseconds. This is 25,000 times larger than the event being measured or 2.5 million per cent error.
reaction time = 300 ms = 25,000× experiment time 0.012 ms
practice problem 2
Il ne s'ensuit pas pourtant que la lumière ne demande aucun temps : car après avoir examiné la chose de près, il a trouvé que ce qui n’était pas sensible en deux révolutions devenait très considérable à l'égard de plusieurs prises ensemble, et que par exemple quarante révolutions, observées du côté F, étaient sensiblement plus courtes que quarante autres, observées de l'autre côté en quelque endroit du zodiaque que Jupiter se soit rencontré ; et ce à raison de 22 pour tout l’intervalle HE, qui est le double de celui qu’il y a d'ici au soleil
I couldn't find any astronomical measurements from Rømer's day, so here are the currently accepted values.
quantity  Earth  Jupiter 

distance to Sun (10^{6} km)  149.6  778.6 
orbital period (days)  365.25  4331 
length of day (hours)  24.0  9.9 
 Sketch the Sun, Earth, and Jupiter when…
 the Earth is closest to Jupiter
 the Earth is farthest from Jupiter
 Determine the speed of light in a vacuum using Rømer's method.
solution
The Earth and Jupiter are closest when they're on the same side of the Sun. They're farthest apart approximately six months later, when the Earth has moved halfway round its orbit. In this same time span, Jupiter has moved just a bit.
The moons of Jupiter appear 22 minutes later after six months because the Earth is considerably farther from Jupiter than it used to be. The extra distance is equal to the diameter of Earth's orbit (twice the EarthSun distance). Use these two numbers to compute the speed of light like Rømer did. Watch the units. Get the distance in meters (not kilometers) and the time in seconds (not minutes).
v = 


v = 


v = 2.27 × 10^{8} m/s  
This is off by about 34%, which is pretty good for the year 1676. No one prior to this had even the slightest idea how fast light really was. It would take almost 200 years to get a value within 1% of what we currently agree to be the "true" value.
226,666,666 m/s  = 76% 
299,792,458 m/s 
practice problem 3
solution
Astronomical distances are so large that using meters is cumbersome. For really large distances the light year is the best unit. A light year is the distance that light would travel in one year in a vacuum. Since the speed of light is fast, and a year is long, the light year is a pretty good unit for astronomy. One light year is about ten petameters (ten quadrillion meters) as the following calculation shows.
Start with the definition of speed and solve it for distance. The traditional symbol for the speed of light is c from the Latin word for swiftness — celeritas.
∆s = c∆t  ⇐ 

Numbers in, answer out.
∆s = c∆t ∆s = (3.00 × 10^{8} m/s) ∆s = 9.46 × 10^{15} m Δs ≈ 10 petameters 
Since both the speed of light and the year have exactly defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.
∆s = c∆t ∆s = (299,792,458 m/s) ∆s = 9,460,730,472,580,800 m 
Some distances in light years are provided below.
 The distance to Proxima Centauri (the star nearest the Sun) is 4.3 light years.
 The diameter of the Milky Way (the collection of stars that includes the Sun and all the stars visible to the naked eye) is about 100,000 light years.
 The distance to Andromeda (the nearest galaxy outside the Milky Way) is about 2 million light years.
 The distance to the edge of the universe (the observable part of it) is 13.8 billion light years.
practice problem 4
What if one fine evening, as the Sun was setting and a full moon was rising, the Sun suddenly quit emitting light?
 How soon after the Sun went black would we know about it on Earth?
 Sketch the relative positions of the Sun, Earth, and Moon at the time described above.
 How soon before or after we saw the Sun go dark would the Moon cease shining?
 What is the speed of dark?
solution
 Answer it! What is the speed of dark?