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Another Index of Refraction Problem ! HELP! PLEASE

  1. Jan 9, 2008 #1
    [SOLVED] Another Index of Refraction Problem!!!!!!!!!!! HELP! PLEASE

    1. The problem statement, all variables and given/known data

    When the rectangular metal tank in Figure 33-49 is filled to the top with an unknown liquid, observer O, with eyes level with the top of the tank, can just see the corner E. A ray that refracts toward O at the top surface of the liquid is shown. If D = 1.45 m and L = 1.13 m, what is the index of refraction of the liquid?
    The Figure 33-49 is attached to this thread!

    2. Relevant equations

    Sin(Theda)N1= Sin(Theda2)N2
    I don't know if this equation will help or not?!

    3. The attempt at a solution

    180 degrees, but that didnt work
     

    Attached Files:

  2. jcsd
  3. Jan 9, 2008 #2

    Shooting Star

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    The eqn to be used is Snell's law and you should use it because it describes the relationship between angles of incidence and refraction and the RIs. You are using the correct eqn.

    Find i and r, the two angles mentioned above. Use geometry to find sin i and sin r.
     
  4. Jan 9, 2008 #3

    Shooting Star

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    i and r are the angles of incidence and refraction respectively, what you insist on calling theda and theda2. If it is going from liquid to water, the angle i is the angle the incident ray makes with the normal at the point of incidence on the interface. I hope you know by now what angle r is.

    I didn't want to say it, but doesn't r look like 90 deg to you?
     
  5. Jan 9, 2008 #4
    OOOOO now i get what you mean!
    SRY it took me a while. okay so to find sin i i have to do sin(i) = 1.13m , right?
     
  6. Jan 9, 2008 #5

    Shooting Star

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    Not right. Sin of an angle lies between -1 and +1.

    In your diagram, sin(i) = L/diagonal.

    (sin i)/(sin r) is also equal to Nair/Nliquid.

    Now can you do it?
     
  7. Jan 9, 2008 #6
    SO;

    1.13/1.83831= sin (i)
     
  8. Jan 9, 2008 #7

    Shooting Star

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    Yes. Please finish off the rest of it.
     
  9. Jan 9, 2008 #8
    (sin i)/(sin r) is also equal to Nair/Nliquid.

    so

    sin(37.93) / sin(37.93) = Nair/Nliquid
    would
    does Nliquid= 1.13?
     
  10. Jan 9, 2008 #9

    Shooting Star

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    Quick, what is the value of r?
     
  11. Jan 9, 2008 #10

    Shooting Star

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    r not eq to i. r=90 deg. Look at the other thread.
     
  12. Jan 9, 2008 #11
    oooo okay so sin(37.93) / sin(90) = Nair/Nliquid

    is Nliquid= 1.13?
     
  13. Jan 9, 2008 #12

    olgranpappy

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    if you could start thinking about the problem a bit on your own... THAT would be great.
     
  14. Jan 9, 2008 #13

    hage567

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    What are you using for Nair?
     
  15. Jan 9, 2008 #14

    Astronuc

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    Staff: Mentor

    Use the dimensions D and L to find the angle of the light in the water with respect to the vertical (normal to interface). The index of refraction in air is approximately 1, or 1.0003.

    Use Snell's Law as you wrote it.

    Refer to http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/refr.html

    I think it was discussed earlier that the angle (between the ray and the normal) in air was a right angle.
     
  16. Jan 9, 2008 #15
    sin(37.93) / sin(90) = 1/Nliquid
     
  17. Jan 9, 2008 #16

    hage567

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    Yes, but be careful with your significant figures.
     
  18. Jan 9, 2008 #17
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