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What is the parallel shift of the light beam?

  1. Aug 7, 2006 #1
    a parallel glass plate with a thickness of 30 mm and a refractive
    index n = 1.65. A light beam hits the surface under an angle of β = 45°.
    (a) What is the parallel shift of the light beam?

    what is the meaning of parallel shift??

    pls help
     
  2. jcsd
  3. Aug 7, 2006 #2
    I may be wrong but parallel shift is the final light beam path subtracted by the original light beam path the light would of travelled without the glass plate, so that you end up with a difference from the original beam if you trace your light beam straight through the glass plate. You should see that the 2 beams are parallel. Then use trigonometric properties and solve for the shift.

    Levi
     
  4. Aug 7, 2006 #3
    do u have any diagram for it??
    i cannot understand them.pls explain a bit further

    thanx
     
  5. Aug 7, 2006 #4
    so,how should i calculate it??
     
  6. Aug 7, 2006 #5

    berkeman

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    Staff: Mentor

    Draw the diagram that IrAlien is suggesting. Draw the plate and the incident beam. Draw the angular shift upon entering the plate. Extend that beam through the plate until you hit the glass-->air boundary, then draw the angular shift there and draw the final air beam out a ways. Then notice that the beam on the other side of the glass plate has the same direction as the incident beam, but is just shifted to the side a bit. That's when you use trig to figure out how far it is shifted. The thicker the glass, the more offset the shift is.
     
  7. Aug 7, 2006 #6
    [ The thicker the glass, the more offset the shift is.]
    may i know what does this mean??

    i have drawn out the diagram,but i still do not know how to find the angular shift with the vlaues given as above
    may i know what does this mean??
     
  8. Aug 7, 2006 #7
    is it i should use the eqn sin(tita1) / sind(tita2) = n(2) / n(1) ??

    then find tita2 ??
     
  9. Aug 7, 2006 #8

    berkeman

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    Staff: Mentor

    Yes you need to use Snell's Law (twice in fact). Here is the wikipedia summary of it:

    http://en.wikipedia.org/wiki/Snells_law

    And click on the "Refraction" link to see a picture of what you are trying to calculate (offset) -- see what the beam does in going through the clear plastic block?
     
  10. Aug 7, 2006 #9
    yes,i saw it,so is it the formula i stated above correct for this calculation??

    {use the eqn sin(tita1) / sind(tita2) = n(2) / n(1) .then find tita2 }
     
  11. Aug 7, 2006 #10

    berkeman

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    Staff: Mentor

    I usually remember it as

    [tex]n_1 sin(\theta_1) = n_2 sin(\theta_2)[/tex]

    But either way is fine. It's an important formula to memorize, so memorize it whatever way is easiest for you to remember for a long time...
     
  12. Aug 7, 2006 #11
    the full question is -Given is a parallel glass plate with a thickness of 30 mm and a refractive
    index n = 1.65. A light beam hits the surface under an angle of β = 45°.
    (a) What is the parallel shift of the light beam? (b) What changes when
    the experiment is performed in water??

    if i use snell's law to calculate for part a) then what should i do in part b) ??

    because i assume part a) is between water and refractive index n = 1.65
    and also may i know what should i do with the thickness given??
     
  13. Aug 7, 2006 #12

    berkeman

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    Staff: Mentor

    What is the n for air in -a- ?

    What is the n for water in -b- ?


    And as for the thickness, please just look at the picture and look at your drawing -- it really should be obvious.
     
  14. Aug 7, 2006 #13
    so in part a) i should use n for air and in part b) i should use n for water
     
  15. Aug 8, 2006 #14
    is the glass plate with a thickness of 30 mm use in this calculation??

    pls help
     
  16. Aug 8, 2006 #15

    mukundpa

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    Homework Helper

    when the ray of light is passes through a parallel plate the emergent ray (after two refractions) is parallel to the incident ray.(apply snell's law on both surfaces). The perpendicular distance between the incident ray and the emergent ray is called parallel shift. First of all draw a ray digram for air and glass, produce the incident ray and use geometry and trigonometry to find perpendicular distance between the rays in terms of the thickness given.
     
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