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Optics Problems.

  1. Aug 7, 2006 #1
    A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 32 degrees relative to the normal.
    If the indices of refraction of air, water, and glass are 1.0, 1.33, and 1.4 respectively, at what angle does the light leave the glass (relative to its normal)? Answer in units of Degree.

    This is how I did it...(1.0)(Sin32)=(1.4)(SinX)
    22.24 Degrees = Sin X

    Where did I go wrong?

    And my other question is:
    A concave mirror has a focal length of 19 cm. What is the position of the object if the image is erect and is 2 times larger than the object? Answer in units of cm.

    I have no idea what to do for this question.
  2. jcsd
  3. Aug 7, 2006 #2
    remember that the ray of light first is refracted from water to glass and next from glass to air.

    so we get:

    [tex]n_{w}\cdot sin(\theta_{w}) = n_{g}\cdot sin(\theta_{g})[/tex]

    then since the refracted ray has an angle [tex] \theta[/tex] equal to the angle after refraction by the glass, the next equation will be:

    [tex]n_{g}\cdot sin(\theta_{g}) = n_{air}\cdot sin(\theta_{air})[/tex]

    And thus:

    [tex]n_{w}\cdot sin(\theta_{w}) = n_{air}\cdot sin(\theta_{air})[/tex]
    Last edited: Aug 7, 2006
  4. Aug 7, 2006 #3
    Alright that helped. Thanks.

    Any ideas for the other question :) ?

    and there's one more...lol

  5. Aug 7, 2006 #4
    use the fact that the image is enlarged 2 times. Use the realtion:

    [tex] M = \frac {i} {o} = \frac {L_{i}} {L_{o}} = 2.0[/tex] with M = magnification, i = imagedistance, o = objectdistance and L = length.

    Now we know from this relation that the image distance i is twice as large as the object distance o --> i = 2o

    Now use the lens equation and substitute i = 20 and solve the equation for o (f is already known):

    [tex] \frac {1} {19} = \frac {1} {2o} + \frac {1} {o}[/tex]
  6. Aug 7, 2006 #5
    o = 28.5

    Plug that in.....i/0 = 2.0

    Li = 57?
  7. Aug 7, 2006 #6
    They ask for the position of the object, thus in my opinion the object distance.
  8. Aug 7, 2006 #7
    Hmm...maybe I did the calculations right but 28.5 for object distance is wrong.
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