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Refraction of a beam of parallel light in a thick-walled wine goblet

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A thick-walled wine goblet sitting on a table can be considered to be a hollow glass sphere with an outer radius of 4.00 cm and an inner radius of 3.40 cm. The index of refraction of the goblet glass is 1.50.

    A) A beam of parallel light rays enters the side of the empty goblet along a horizontal radius from the left. Where, if anywhere, will an image be formed?

    B) The goblet is filled with white wine (n=1.37). Where is the image formed?

    2. Relevant equations
    n1sinθ1=n2sinθ2
    1/s+1/s'=1/f


    3. The attempt at a solution

    I don't know where to start with this problem. I'm confused with the beam of parallel light rays, do they mean just one ray? Or do they mean many rays that cover the entire left side of the sphere? I also considered this to be a thin lens, but there was no object. I'm really just looking for a direction to take this problem. Thanks
     
  2. jcsd
  3. Apr 8, 2012 #2

    Doc Al

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    Hint: When parallel rays hit a thin lens, where do they focus? (Yes, treat the glass as a thin lens and the beam as a pencil of light.)
     
  4. Apr 8, 2012 #3
    When the object is at an infinite distance from the lens, the image distance equals the focal length.

    I have two equations now, f=R/2, and 1/f=(n-1)(1/R1-1/R2) (lens makers' equation).

    Using the lens makers equation:

    1/f=(1.50-1)(1/4.0cm-1/3.4cm)
    f=-45.3 cm

    Using the first equation f=4cm/2=2cm

    I don't believe the first equation is the right one to use here, but I'm still confused with the lens makers equation. Do you see my error?
     
  5. Apr 8, 2012 #4

    Doc Al

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    Good.

    That first equation is for mirrors. The second is the one you want.

    Good. So that's the focal length of the 'lens' and the position of the image from the first side of the glass. Keep going. (There's the other side of the glass to worry about.)

    No good.

    I don't see an error so far. But you're not done.
     
  6. Apr 8, 2012 #5
    As the light continues along its path it reaches the right side of the sphere. Using the same equation:

    1/f=(1.5-1)(1/3.4-1/4.0)
    f=+45.3 cm

    So there will be an image on either side of the glass sphere, 45.3 cm away. The question asks where will the image be formed to the left from the left rim of the goblet.

    Do these two images interfere with each other? Or am i supposed to use this image as my new object and calculate a new image distance?
     
  7. Apr 8, 2012 #6

    Doc Al

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    Careful with signs. The focal length of each 'lens' should be the same. (The two sides are mirror images of each other.)
    Use the image from the first lens as the object for the second.
     
  8. Apr 8, 2012 #7
    Oh ok, so the focal length of the second 'lens' will be -45.3 cm as well. Using the image from the first lens as the object:

    1/f=1/s+1/s'

    s=45.3cm + Outer radius + inner radius
    s=45.3+4+3.4
    s=52.7 cm

    1/-45.3=1/52.7+1/s'

    s'=-24.4

    Final image is -24.4+7.4 cm = 17 cm to left of left rim of goblet?
     
  9. Apr 8, 2012 #8

    Doc Al

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    That's what I would say. We're making a thin lens approximation; the distance between the lenses would be the distance from their midpoints (which is what you called 'outer radius + inner radius').
     
  10. Apr 8, 2012 #9
    Hm, my answer of 17 cm is incorrect. Could the separation of the lenses be too big for the approximation to hold true?
     
  11. Apr 8, 2012 #10

    Doc Al

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    Now that I think about it, the distance from the left side of the glass and the center of second 'lens' would be 4 + 3.7 = 7.7 cm. Which would change your answer slightly.
     
  12. Apr 8, 2012 #11
    Still no luck using 16.7 cm, can you see any other error? I appreciate your help a lot this is a difficult question for me
     
  13. Apr 9, 2012 #12

    Doc Al

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    Well, maybe they want you to treat the sides as thick lenses instead of thin lenses. (I always try the easier thing first. I'm lazy.) A bit more of a pain, but doable.

    Treated as a thick lens, the lens-maker's equation is a bit different.

    What book is this problem from?
     
  14. Apr 9, 2012 #13
    The question is from the Mastering Physics program, the textbook is University Physics 12th Edition (Young and Freedman).

    Would i use 1/f=(n-1)(1/R1-1/R2+(n-1)d/nR1R2) ?
     
  15. Apr 9, 2012 #14

    Doc Al

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    Ah... I'm familiar with that one.
    Yes, I think that's right. (I'd have to look it up to double check.)

    But instead of that, go back to basics. (Where we should have started.) Just treat each interface as a spherical refracting surface. (That should be in the book, just before lenses.) That's the easy way, instead of messing with thick lens formulas.
     
  16. Apr 9, 2012 #15
    I read the section on spherical refracting surfaces, and the equation for object-image relationship is:

    n1/s+n2/s'=(n2-n1)/R

    For this question, the object is at infinity and n1 is air so the equation reduces to

    n2/s'=(n2-1)/R

    1.5/s'=0.5/4

    s'=12 cm to the right of left edge of sphere? Then is it the same as before I use this image as my object for the right of the sphere?
     
  17. Apr 9, 2012 #16

    Doc Al

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    Exactly!
     
  18. Apr 9, 2012 #17
    Using this as my object for the right side of the sphere, i get s=12-8=4cm to right of right side of sphere.

    n1/s+n2/s'=(n2-n1)/R

    1/4+1.5/s'=(1.5-1)/4

    s'=-12 cm, or 12 cm to the left of the right side of the sphere. So the final image is 4 cm to the left of the left side of the sphere?

    I am confused with what happens to the beam of light after it reaches the left side of the sphere, does it not continue on to the inner radius sphere and create another image?
     
  19. Apr 9, 2012 #18

    Doc Al

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    What you found was the image location from the first interface (air to glass) at the left side of the goblet. Use that image as the object for the second interface (glass to air) still on the left side of the goblet. What's the distance to that second surface?

    And then do the same for the third and fourth interfaces at the right side of the goblet.
    See above.
     
  20. Apr 9, 2012 #19
    Oh I see. The distance of that image to the second surface (glass to air on left side of sphere) is -8 cm.

    n1/s+n2/s'=(n2-n1)/R

    1.5/-8+1/s'=(1-1.5)/3.4

    s'=24.7cm or 24.7 cm to the right of the second surface.

    If this is correct, continuing onto the third surface (air to glass on right side of sphere), the image distance would be 24.7-(3.4*2)=17.9cm.

    1/-17.9+1.5/s'=(1.5-1)/3.4
    s'=7.4 cm to the right of third surface.

    Onto the fourth surface (glass to air on right side of sphere), the image distance would be 7.4-0.6=6.8cm to right of fourth surface.

    1.5/-6.8+1/s'=(1-1.5)/4
    s'=10.5 cm to right of fourth surface.

    I'm still a little confused with the sign convention, trying to understand it. I know i have messed up somewhere because i know the final image is to the left of the first surface.
     
  21. Apr 9, 2012 #20

    Doc Al

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    How did you get this? The image from the first surface is 12 cm to the right of that first surface. The thickness of the glass is 4 - 3.4 = 0.6 cm. Thus the distance to the second surface is 0.6 - 12 = - 11.4 cm.
     
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