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Coin in water pool, how does coin appear? Optics

  1. Dec 20, 2005 #1
    5.A coin is resting on the bottom of a water pool (nwater = 1.33) 1.00m deep. On top of the water floats a layer of benzene (nbenzene = 1.50). which is 20.0cm thick. Looking down perpendicularly, how far beneath the topmost surface does the coin appear.

    Since the angle of incidence is zero, can I assume that the distance between the topmost surface with the coin = 120cm?
     
  2. jcsd
  3. Dec 20, 2005 #2

    Doc Al

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    No need to assume that, it's given.
     
  4. Dec 20, 2005 #3
    So, the answer should be 120cm ?
     
  5. Dec 20, 2005 #4

    Doc Al

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    No. 120 cm is the actual distance between the top surface and the coin. The apparent distance between the surface and the coin's image is something else. To find that, consider refraction at the surfaces for slightly off-axis light.
     
  6. Dec 20, 2005 #5
    What does it mean? Is it consider the incident angle does not equal to zero? If so, why? Since the question mention the angle of incidence is zero.
     
  7. Dec 20, 2005 #6

    Doc Al

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    It takes more than one "ray" to locate an image. Consider the coin as a point source of light. Figure out how the rays of light bend as they refract from the surfaces. To find the apparent depth, trace back the rays to find the location of the coin's image. (If you only consider the ray with angle of incidence exactly = 0, you won't get far since that ray doesn't get refracted.)
     
  8. Dec 20, 2005 #7
    Sorry, I cant get what you mean.
     
  9. Dec 21, 2005 #8

    Doc Al

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  10. Dec 22, 2005 #9

    daniel_i_l

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    If you think about it, one ray doesn't tell you anything about the coins position. For example, if I have a ray of light going to the right, it could be coming from one meter away, one kilometer away, one lightyear away, and you wouldn't be able to tell from the direction. But when two rays cross, then there is only one point were those rays could have both come from.
     
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