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An interference problem

  1. Dec 13, 2006 #1
    1. The problem statement, all variables and given/known data
    a soap film of refractive index 1.333 is illuminated with white light at an angle of 45.the light refracted by it is examined and two bright bands are focussed corresponding to wavelengths 6.1 X 10 to the power -5 and 6.0 X 10 to the power -5.find thickness "t" of the film


    2. Relevant equations

    since it is given as destructive interference,formula will be

    2 Xrefractive index(myu) X t Xcosr=n Xlamda

    3. The attempt at a solution

    2 X 1.333 X t X cos45=?

    what is the wavelength i should take.there are two.its bit confusing for me......please help me.........
     
    Last edited: Dec 13, 2006
  2. jcsd
  3. Dec 13, 2006 #2

    OlderDan

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    Why do bright bands lead you to think in terms of destructive interference? If two different wavelengths are bright under the same geometric conditions, then the thickness must be creating the same interference condition for both wavelengths.
     
  4. Dec 14, 2006 #3
    so should i take lamda1 and lamda2,then is this is the formula
    2 X mu X t Xcosr=n X lamda
    or
    2 X mu X t X cosr=(2n-1) X lamda/2
     
  5. Dec 14, 2006 #4

    OlderDan

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    I'm not sure what you mean by r, and I have never seen μ used for index of refraction, but if that is what you mean by mu, then it does come into play.

    The optical path difference (number of wavelengths) between two rays incident at 45º needs to be determined. The rays change direction when entering the liquid, and there is a phase revesal at one surface (which one?) assuming air is on both sides of the film. It is not clear to me that you have the path difference correct in either equation. It appears you have not accounted for the extra path length in air while the one ray is in the liquid. I could be wrong and will look at it moe carefully if you tell me you think you have.
     
    Last edited: Dec 14, 2006
  6. Dec 15, 2006 #5
    actually "r" here means the angle and "mu" is the refractive index as i attempted in the solution.as per given question refractive index is 1.333 and r=45.
     
  7. Dec 15, 2006 #6

    OlderDan

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    I don't think r is 45º. If the light is incident on the first surface at 45º then it is refracted to a new angle within the higher inex film. The light reflected at the first surface has a longer path length in air than the light that is reflected at the second surface. It is not apparent to me that you have accounted for that extra path length in air.

    Draw your two parallel incident rays with a line perpendicular to these rays (which represents a wave front) that intersects the first point of contact with film and do the same at the exit point of the penetrating ray. The optical path difference from the entry front to the exit front is what determines the final phase difference between the rays.
     
  8. Jan 4, 2007 #7

    vij

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    You are considering the transmitted system (and not the reflected system), I presume. The condition for brightness in the transmitted system is the one you have written. You will get two values for ‘n’ on substituting the two given values of the wave length. What are you going to calculate?
    If you are given that the angle of refraction is 45 degrees you can substitute it in the equation. If the given angle (45 degrees) is the angle of incidence, you will have to find cos r using the equation, refractive index = sini/sinr.
     
  9. Jan 6, 2007 #8

    vij

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    Since you want to calculate the thickness ‘t’ of the film using the transmitted system, use the equation, 2 n t cosr = mλ where ‘n’ is the refractive index ‘m’ is the order of the interference band and ‘λ’ is the wave length.
    The lower order (m) will be for the longer wave length (λ1 = 6.1×10^ –5 cm). Understand that the order ‘m’ for the longer wave length implies that the thickness of the film is such that there has to be a path difference of ‘m’ wave lengths of this light. The condition for constructive interference for the light of shorter wave length (λ2 = 6.0×10^ –5 cm) is established when the order increases by 1. Therefore, the order of the band produced by the shorter wave length light is (m+1).
    So, you have, mλ1 = (m+1)λ2 from which m = λ2/( λ1– λ2) .
    Now you have, 2 n t cosr = m λ1 and 2 n t cosr = (m+1)λ2. Use one of these equations to calculate the thickness ‘t’, substituting for n (= 1.333) and cos r.
    Since the angle of incidence i = 45 degrees, the value of ‘r’ will wok out to very nearly 32 degrees using n = sini/sinr. Now, calculate 't'.
     
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