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Homework Help: Resolving Power/Angle etc.

  1. Dec 6, 2011 #1
    Hello all, I'm little stumped on this question. Have tried multiple times and can not find the right answer to it. Here it is:

    1. Question

    The pupil of an eagle's eye has a diameter of 5.93 mm. Two field mice are separated by 0.0124 m. From a distance of 186 m, the eagle sees them as one unresolved object and dives toward them at a speed of 18.1 m/s. Assume that the eagle's eye detects light that has a wavelength of 565 nm in a vacuum. How much time passes until the eagle sees the mice as separate objects?

    3. The attempt at a solution

    565e-9/5.93e-3 = 9.527824620573355e-5 or approx 1.00x 10^-4 which is the resolving angle
    .0124/186 = 6.666666666666667e-5 which is approx 2/3 of resolving angle (this is angle when eagle starts)

    if taking 2/3 of 186m I get 124.

    124m / 18.1 m/s = 6.85s

    However, the answer is wrong. Any advice/help/etc.?
  2. jcsd
  3. Dec 6, 2011 #2


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    Why don't you just solve for the distance at which the angular separation of the mice is equal to the resolution limit?

    (0.0124 m)/(d) = (565 nm)/(5.93 mm)

    Solve for d.

    Your approach was fine, and equivalent, except that you introduced a huge rounding error by rounding that first angle up.
  4. Dec 7, 2011 #3
    I actually had 7.19s as my first answer which is what you get when you solve for d. This answer is apparently wrong though.. :confused:
  5. Dec 7, 2011 #4


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    Hopefully you realized that your method and "solving for d" are not actually different methods. It's just that your ratio should have been closer to 0.7 rather than 0.667 --- an error that was due to rounding.

    In any case, as for why you are getting the wrong answer still: the only think that I can think of is that the problem explicitly mentions that 565 nm is the vacuum wavelength of the light that the eagle is sensitive to. But the eagle and the mice are not in a vacuum. I don't know why else this information would have been specified in the first place, if not so that you could convert this to the wavelength in the appropriate medium (in this case air).
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