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Refraction angle

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data

    I need to calculate refraction angle of light, if light ray is falling to the surface of aquarium water from the bottom in the angle of 60 degrees, n for water = 1,33
    2. Relevant equations
    sinφ2=(n1×sinφ1) /n2

    3. The attempt at a solution
    So i get sinφ2=1,33*0,866( meant as sin60) / 1 = 1,15178

    this is where i ended and teacher didn`t liked it... Is there a mistake in calculations? Whats the final refraction angle?
     
  2. jcsd
  3. May 26, 2015 #2

    Merlin3189

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    So, if you have found sin φ2=1.15178 why have you not "calculate refraction angle of light," ?

    What does this tell you?
     
  4. May 26, 2015 #3
    sin for 1,1517 is something like 99 degrees - is this correct answer? I thought that there are some problems with calculations, because in this test it wasn`t compulsory to calculate exact values in degrees
     
  5. May 26, 2015 #4

    Merlin3189

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    No it isn't 99o
    Since they asked for the angle, I think you should give an angle, not just a sine.
    BUT there is a problem here. You can't calculate the angle from your result!
    Your calculation is ok, but it gives you a funny answer. WHY?

    Edit: BTW. Try drawing a diagram and put in your ray at 99 degrees (which is wrong, but may help you see why.)
     
  6. May 27, 2015 #5
    99 degrees - this is too wide... So the correct answer is - there will be no refraction?

    Or it will reflect backward and the answer is 9 degrees? This is confusing
     
  7. May 27, 2015 #6

    ehild

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    What do you mean? you have to get the inverse sine of 1.1517 - is there any angle with sine greater than 1?
    If refraction angle does not exist then can the light ray enter into air?
    Light partly refracts and partly reflects at an interface between two media. What do you know about the angle of reflection?
     
  8. May 27, 2015 #7

    Merlin3189

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    You are getting there!
    There is no sine greater than 1, so the equations beloved of PF do not just give you the answer. You have to understand what happens.

    Have you heard of "Total Internal Reflection" or "Critical Angle"?
     
  9. May 27, 2015 #8
    As I`m from non-english speaking country I may be incorrect in terms but total internal reflection is when (n1 > n2) - than it`s possible for the ray to reflect& refract, unless... unless sinφ1>n2/n1

    in this case sin60°>1/1,33 leads us too 0,866>0,751

    So the answer ir - there will be no refraction?
     
  10. May 27, 2015 #9

    Merlin3189

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    Yes, there will be no refraction. The light will be reflected and angle of incidence = angle of reflection, no sines!

    The Critical Angle is the angle of incidence which would give 90o angle of refraction,
    so sin( critical angle) = sin(90o) * nless dense/nmore dense = 1 * 1/1.33 = 0.7519
    so critical angle = arcsin(0.7519) =48.8o

    If the angle of incidence from the more dense medium to the less dense is greater than 48.8o then the light is totally internally reflected.

    In your case, angle of incidence was 60o, so ray is reflected.

    You should also remember that no sine or cosine can be greater than 1 (nor less than -1)

    Incidentally, sin(99o) = 0.9877 = sin(81o) After sine gets to 1, it starts coming down again.
     
  11. May 28, 2015 #10
    Thanks!
     
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