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Thin len problem

  1. Apr 19, 2005 #1
    This is a true of false question from my homework:
    Three light rays start out from a candle flame at the same time. To the right of the candle there is a lens, and to the right of the lens there is a screen. Ray A goes through the edge of the lens. Ray B goes through the center of the lens. Ray C goes through an intermediate part of the lens. All three rays are focussed on the screen. Ray B arrives at the screen first because it travels the shortest path.

    I am wondering, does it matter if the len is converged or diverged?
    And I don't get how Ray A would travel after passing the edge of the len. From what I have read from the book, i only know how the ray exactly travel if it is traveling, parallel to the axis, through the center of len, and the focus.
    So where would Ray A go after passing through the len?
  2. jcsd
  3. Apr 19, 2005 #2


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    You are TOLD that the 3 rays are "focused" on the screen. That means they all go to the same point on the screen.
  4. Apr 19, 2005 #3
    Yeah you need to understand the difference between focused and focus. (sorry if I gave away too much here :biggrin:)
  5. Apr 19, 2005 #4
    does the picture that i have attached interpert the question?
    I think this is a nice question to check my understanding of the topic

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  6. Apr 20, 2005 #5


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    Your diagram is a fair representation of the problem. Ray A could have been through either edge of the lens, and ray C could have been through the lower half of the lens, but your choices are as good as the alternatives. Now, can you answer the question?
    Last edited: Apr 20, 2005
  7. Apr 20, 2005 #6
    Yes, Ray B is obvioulsy travelling in the shortest distance.

    But as we know that light is traveling at the speed of 3 * 10^8 m/s, what is significant about this problem? ;-)
  8. Apr 20, 2005 #7


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    Shortest distance does not necessarily mean shortest time. What causes the bending of light in a lens?
  9. Apr 20, 2005 #8
    the index of refraction
  10. Apr 20, 2005 #9
    But in this question, we have air- len (glass)(very short distance) - air, so the index of refraction doesnt affect much to the distance right
  11. Apr 20, 2005 #10


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    The index of refraction is related to the speed light travels in the glass. Light changes direction because it slows down in the glass. Which ray spends the longest time in the glass?
  12. Apr 20, 2005 #11
    Oh, ray B travels in the shortest path but it takes the longest time because it stays more time in the len !
  13. Apr 20, 2005 #12


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    Well, you may have gone "too far" in changing your opinion :smile: There is a notion called "optical path length" that is used to account for the change in direction of waves due to their slowing in some medium. If you think in terms of "wave fronts" instead of rays, the direction of propegation of a wave is perpendicular to the wave front. A wave front means a surface over which the light's electric and magnetic fields remain in phase. These are surfaces to which light must travel from a source in the same amount of time, regardless of variations in speed along the path that got them there. The surfaces distort because of slowing and speeding effects. We draw rays perpendicular to these surfaces to represent the direction of propegation of the light.

    It takes some pondering, no doubt, but what this amounts to is that the light coming out of the lens that all arrives at one point on the screen does so because there is a surface equidistant from that point on the screen over which all the light is in phase. Since it all came from the same source, it must have been in phase all along. We say that the optical path length from the source to this surface is the same for all rays that get the light from the source to that surface, and what that means is that it takes the same amount of time for light travelling along any ray to get from the source to that surface.

    If you can absorb all of that, you will realize that the surface is going to collapse down to a point on the screen, with all of the rays arriving in phase. That will lead you to the answer to the question.
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