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Fish has gone missing in a tank

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    At the aquarium where you work, a fish has gone missing in a 8.00mdeep, 8.80m diameter cylindrical tank. You shine a flashlight in from the top edge of the tank, hoping to see if the missing fish is on the bottom. What’s the smallest angle your flashlight beam can make with the horizontal if it’s to illuminate the bottom?

    2. Relevant equations
    n1sinΘ1 = n2sinΘ2
    index of refraction of water: nw = 1.333
    index of refraction of air: na = 1.000293

    3. The attempt at a solution
    I thought that the minimum angle would be when the beam hit the edge of the tank and formed a right triangle so the angle in the water would be:
    Θ2 = arctan(8/8.8)
    n2 = nw = 1.333
    n1 = na = 1.000293
    Θ1 = arcsin[(1.333/1.000293)sin(arctan(8/8.8)] = 63.69°
    So then
    1.333sin(42.27°) = 1.000293sin(63.69°)
    But this is wrong. I also tried to subtract it from 90° but that is wrong as well, to measure from the vertical instead. Thanks in advance.
     
  2. jcsd
  3. Nov 16, 2014 #2

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    Sketch it out, and think over the trig definitions.
     
  4. Nov 17, 2014 #3
    Okay, I am confused what is meant by illuminate the bottom. We model the wave fronts of the light source as a ray, so to illuminate the bottom, will this ray have to reach the center of the bottom of the tank, or just an edge of the bottom of the tank where it meets the wall?
     
  5. Nov 17, 2014 #4

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    Edge-wall boundary is good enough. Think, what is the definition of θ2 in Snell's law?
     
  6. Nov 17, 2014 #5
    Hmm, Θ2 in Snell's Law is the angle of refraction of the light ray measured with respect to the normal to the surface of the refracting medium, or "to the vertical." Ah, so I swapped the sides of the right triangle (which is what you had hinted at in outlining), and I was getting the cotangent of the angle Θ2. Θ2 = tangent (opp./adj.) = tan(diameter/height) = tan(8.8/8). So then I can apply Snell's Law to get:
    1.333sin(arctan(8.8/8)) = 1.000293sinΘ1
    So if Snell's Law can be used to find Θ1, the angle of incidence of the incident light ray with respect to the normal to the surface, I can take the complement of the angle to get the angle measured with respect to the horizontal, right?
    I can also use this identity that I came across:
    sin(arctan(x)) = x/√(1 +x^2)
     
  7. Nov 17, 2014 #6

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    Good hunting --- we'll hope "Nemo" just ducked out for a burger.
     
  8. Nov 17, 2014 #7
    :p I will be finding Dory now... Thanks a bunch for the posts! I need to be more careful with what the formulas mean in words, so then I will be less likely to apply them blindly and incorrectly.
     
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