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Snell's law, critical angle, and angle of incidence

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A pair of students measure the refraction of light in passing from acetone into air for several angles. When the angle of incidence is 30, the angle of refraction is 42.

    b) What is the angle of incidence (in degrees) when the refracted angle is the critical angle?


    2. Relevant equations
    Snells law: Ni*sin(i)=Nr*sin(r)


    3. The attempt at a solution

    I found the index of refraction for air to be 1 and for acetone to be 1.36. By snell's law setting r=90:
    sin(i crit) = nr/ni * sin(90) = nr/ni = 1/1.36

    icrit = arcsin(1/1.36) = 55.73 degrees

    Is my answer correct? I'm a bit confused about the part that says "...when the refracted angle is the critical angle". Isn't the angle of incidence (when it creates a refraction of 90 degrees) the critical angle?
     
    Last edited: Oct 6, 2009
  2. jcsd
  3. Oct 6, 2009 #2
    I see what you mean, indeed it's worded poorly...

    although realistically it would be a stupid question asking you to work out the angle of incidence where the angle of refraction is the critical angle...i would go with your workings and ask your teacher/whatever for clarification
     
  4. Oct 6, 2009 #3

    kuruman

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    Yes, it is. You went only half-way answering part (b) which is to find the critical angle. Now answer the question. You have a refracted angle of 55.73 degrees, what is the angle of incidence?
     
  5. Oct 7, 2009 #4
    Why wouldn't it just be 55 degrees?
     
  6. Oct 7, 2009 #5

    rl.bhat

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    55.73 is the angle of refraction in the air. For that what should be the angle of incidence in acetone?
     
  7. Oct 8, 2009 #6
    I thought that in calculating the critical angle to be 55.73 it would be equal to the angle of incidence, not the refracted angle. Isn't the refracted angle just 90?
     
  8. Oct 8, 2009 #7

    jambaugh

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    It appears in the problem that the indexes of refraction used are not the standard values you looked up. (I get a lower ratio)

    From the original angles you can use Snell's law to calculate the ratio of the indexes of refraction. Then as explained find the incident angle which refracts to 90deg and that is the critical angle.

    The problem is worded OK in my opinion but instead of "acetone" the problem should have stated "some unknown liquid".

    EDIT: Hint: I get an angle of _ _ . _ _ _ _ 1711717478....
    And it is significantly less than 55.73
     
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