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Iinterference in thin films due to reflected light

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data


    dear revered members,
    please find the attached image for the topic interference in thin films due to relection.

    2. Relevant equations
    my questions are
    1) angle between the incident ray or reflected ray or refracted ray is called angle of incidence, angle of reflection and angle of refraction, as the case needs. so <AON is angle of incidence. how <ACN is angle of incidence? because CN is normal line and AC IS POINT OF CONTACT, then how it is angle of incidence?
    is there any geometry involcd in it?
    ANY HELP IN THIS REGARD WILL BE GREATLY APPRECIATED, REVERED MEMBERS


    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2009 #2

    kuruman

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    Yes, there is a theorem in Euclidean geometry that says that if two angles are less than 90o and their sides are mutually perpendicular, then the angles are equal. Label the top of the dotted line that is perpendicular to the interface as Q. Then angle SAQ is equal to QAT by the law of reflection. Note that line QA is perpendicular to AC and line AT is perpendicular to line NC.
     
  4. Aug 30, 2009 #3
    sir, i understood the concept. but, in my attachment SA is incident ray and NA is the normal, so <SAN is angle of incidence. but, NC is normal to AT and CQ. <i is between AC and CN. how can it be termed as angle of incidence. because both AT and CQ, according to my attachment are reflected rays or transmitted rays. so they should be <r. Why <i there?
     
  5. Aug 30, 2009 #4

    kuruman

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    I am not disputing that SA is the incident ray. The angle of incidence is SAI. This angle in degrees is equal to the reflection angle INT in degrees and also equal to angle ACN in degrees. I never said that ACN is the angle of incidence; I said it is equal to the angle of incidence in degrees. Do you see the difference?
     
  6. Sep 2, 2009 #5
    thanks you sir. but on what basis they are equal? adjacent angles? alternate angles? or something else.
     
  7. Sep 2, 2009 #6

    kuruman

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    They are equal on the basis shown above.
     
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