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class code: SPS22 teacher: Mr. Elert
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email: elert@midwoodscience.org

Physics A: Problem Set 15: The Nature of Light

recommended reading

Barron's Let's Review: 12.1–12.4
physics.info: The Nature of Light, Polarization
Wikipedia: Light, Speed of light, Visible spectrum, Polarization (waves)
HyperPhysics: Speed of light, Visible light, Polarization
Khan Academy: Introduction to light waves
YouTube: Roy G. Biv

classwork

  1. Here is an excerpt from Galileo's report of his attempt to determine the speed of light in a vacuum.

    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….

    Galileo Galilei, 1638

    1. Estimate the time for a light wave to travel the distance in Galileo's speed of light experiment. (An Italian mile is 1.873 km.)
    2. How does this compare to the reaction time of a typical human? (How many times greater is one number than the other?)

    1. Start with the definition of speed and solve it for time.

      Δt =  Δs
      v
       ⇐ 
      v =  Δs
      Δt

      Use the round trip distance of two Italian miles.

      Δt =  2(1,873 m)
      3.00 × 108 m/s

      Δt = 1.20 × 10−5 s

      For this problem, it's probably better to state the answer in milliseconds.

      Δt = 0.012 ms

    2. 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
  2. Use Rømer's method to determine the speed of light in a vacuum. If you're comfortable reading Seventeenth Century French, here's the paragraph that reports Rømer's measurement of a 22 minute delay as the light from Jupiter's moon Io traverses the extra distance equal to the diameter of earth's orbit.

    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

    Ole Rømer, 1676

    quantity earth jupiter
    distance to sun
    (106 km)
    149.6 778.6
    orbital period
    (days)
    365.25 4331

    I'd give you an English translation to read, but no good one exists. This means that you will have to refer to your notes from class on this experiment. Use the astronomical data table provided and then answer the following questions.

    1. Sketch the relative position of the sun, Earth, and Jupiter when…
      1. the earth is closest to Jupiter
      2. the earth is farthest from Jupiter
    2. Determine the speed of light in a vacuum using Rømer's method.

    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 Earth and Jupiter to the right of the sun, both labeled 'September'. The earth on the opposite side and Jupiter moved up just a little bit, both labeled 'March'.

    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 Earth-sun 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 = 
    Δs
    Δt
     
     
    v = 
    2(1.496 × 1011 m)
    1320 s
     
     
    v = 2.27 × 108 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

homework

  1. Visible light ranges in wavelength from 400 nm (deep violet) to 700 nm (dull red). Compute the corresponding range of frequencies?
    1. What is the frequency near the lower limit where λ ~ 400 nm?
    2. What is the frequency near the upper limit where λ ~ 700 nm?

    Use the equation c = fλ and solve it for wavelength.

    f =  c  ⇐  c = fλ
    λ

    Use it twice. A common unit for such occasions is the terahertz (1 THz = 1012 Hz).

    deep violet
    f =  c
    λ
    f =  3.00 × 108 m/s
    400 × 10−9 m
    f = 7.5 × 1014 Hz = 750 THz  
     
    dull red
    f =  c
    λ
    f =  3.00 × 108 m/s
    700 × 10−9 m
    f = 4.3 × 1014 Hz = 430 THz  
     
  2. A common measure of astronomical distances is the light year. This is the distance a beam of light would travel in a vacuum in one year. Determine the size of a light year in meters using the speed of light.

    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 trillion meters as the following calculation shows.

    Start with the definition of speed and solve it for distance.

    s = ct  ⇐ 
    c =  Δs
    Δt
    Δs = cΔt
    Δs = (3.0 × 108 m/s)(365.25 × 24 × 3600 s)
    Δs = 9.46 × 1015 m

    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.

    Δs = cΔt
    Δs = (299,792,458 m/s)(365.25 × 24 × 3600 s)
    Δs = 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 (a 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 13.77 billion light years.
  3. A common measure of astronomical distances is the light year. This is the distance a beam of light would travel in a vacuum in one year. For smaller distances I would like to propose an alternate unit — the light nanosecond. What is your height in light nanoseconds?

    Start with the definition of speed and solve it for distance.

    s = ct  ⇐ 
    c =  Δs
    Δt

    A nanosecond is a billionth of a second (10−9 s). The math is fairly easy. Please don't reach for a calculator — yet.

    Δs =  cΔt
    Δs =  (3 × 108 m/s)(1 × 10−9 s)
    Δs =  3 × 10−1 m or 30 cm

    Every 30 cm of height is one light nanosecond. Divide your height in centimeters by 30. For example…

    188 cm  = 6.27 light nanoseconds
    30 cm

    For those of you more used to traditional British units than the metric system, a light nanosecond is almost a foot.

    30 cm  = 11.8 inches
    2.54 cm/in

    188 cm is 74 inches or 6'2". Thus…

    74 in  = 6.27 light nanoseconds
    11.8 in

    Results not typical. Your results may vary.

  4. Why can't sound waves be polarized?
    Sound waves can't be polarized because they are longitudinal.