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Index of refraction, angular seperation

  1. Jul 23, 2006 #1
    I can't seem to figure these out guys......maybe I'm missing a few equations too.


    1. If light of wavelength 600nm in air enters a medium with index of refraction 1.50, its wavelength in this new medium is...
    A. unchanged
    B. 900nm
    C. 400nm
    D. 602nm
    ---I know that n=c/v....but I don't know how to relate that to find wavelength.

    2. A 200m radio telescope is used to investigate sources emitting a 21cm wavelength. The minimum angular seperation resolvable for this system is...
    A. 0.154deg
    B. 0.073deg
    C. 0.0026deg
    D. 0.0013deg

    I have no clue on this one....we haven't even covered this stuff, and he tested us on it.....trying to figure out how to do it, before the final(this week). Again, any help is great!!!

    Thanks again,
    Brad
     
  2. jcsd
  3. Jul 23, 2006 #2
    1) n = v1/v2 = lambda1 / lambda 2
     
    Last edited: Jul 23, 2006
  4. Jul 23, 2006 #3
    sorry, your image didn't load.......
     
  5. Jul 23, 2006 #4

    Andrew Mason

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    What is the speed of the light wave in the medium? What is the relationship between wavelength and speed for a wave of a particular frequency?

    This is essentially a single slit diffraction problem: What is the angular separation of the first minimum for a 200 m wide slit? This works out to: [itex]\sin\theta = \lambda/D[/itex] where D is the diamater of the telescope. It should be the same for radio or optical telescopes. For small angles, [itex]\sin\theta \approx \theta[/itex]

    AM
     
  6. Jul 23, 2006 #5
    Ok, so for #1.....I still can't get it figured out...might need a little bit more help. V for the light in the medium is 2x10^8m/s. Using v=c/n. Then the relationship you asked for, is it v=wavelength x frequency. If so, I don't understand how to use that to get my answer.

    for #2.....so basically, I take .21m/200m.....which = .00105 I'm not sure why that answer isn't up there....did I do it right?
     
  7. Jul 23, 2006 #6

    Andrew Mason

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    Does the frequency change? So how does wavelength vary with speed?
    They may be using the Raleigh diffraction formula which is [itex]sin\theta = 1.22\lambda/D[/itex]

    Also remember: you must put your answer in degrees. Your answer is in radians (or [itex]sin\theta[/itex]).

    AM
     
  8. Jul 23, 2006 #7
    ok andrew....first let me say thanks......

    so for #1, I ended up taking frequency=speed of light/my v(2x10^8).....which = 3/2. I took 3/2 x 600nm = 900nm...answer B.

    And for #2.....I did the math over(and used your new formula)....and came up with answer B.

    Are these correct??

    Thanks again,
    Brad
     
  9. Jul 23, 2006 #8

    Andrew Mason

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    Why is v = 3/2? The speed of light in air is 3 x10^8 m/sec. So the speed in the medium is 2/3 of the speed through air.

    #2 looks right.

    AM
     
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