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An exercise about thin-film interference

  1. Aug 2, 2011 #1
    1. The problem statement, all variables and given/known data

    We have a thin film of glass which has thickness [itex]t=\lambda[/itex] and [itex]n=1.5[/itex] and light ([itex]\lambda[/itex]) passing through it with an angle [itex]\theta_0[/itex].
    We have to find the minimum angle that allows us to see both constructive and destructive interference.


    2. Relevant equations

    Max: [itex] 2n\cos(\theta_1) = (2m+1) [/itex]
    Min: [itex] 2n\cos(\theta_1) = m [/itex]

    Where [itex]\theta_1[/itex] is not the incidence angle but the angle of refraction.
    In the previous formulas I have already taken into account that [itex]t=\lambda[/itex]

    3. The attempt at a solution



    Max: [itex] \cos(\theta_1) = \frac{2m+1}{3} [/itex]
    Min: [itex] \cos(\theta_1) = \frac{m}{3} [/itex]

    So, i have to get both "kinds" of interference and I need the minimum angle:

    [itex] \frac{2m+1}{3} = \frac{m}{3} [/itex] which leads me to m=-1

    Is that right? I thought m could only be 0, 1, 2...

    If that is right, i get [itex] \theta_1 = 109.47°[/itex]

    How to find the incidence angle? Using Snell's formula:

    [itex]\sin (\theta_0) =1.5 \sin(109.47°) [/itex]

    which can't be solved because the second member is something like 1.4

    Where am i wrong?

    Thank you in advance and sorry for my bad english ...
     
  2. jcsd
  3. Aug 2, 2011 #2

    PeterO

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    My reading would be that when you were told

    "We have to find the minimum angle that allows us to see both constructive and destructive interference".

    It means what angle gives constructive interference, and what angle gives destructive interference - two different angles.
     
  4. Aug 2, 2011 #3
    Hey, thank you so much, I think u are right, but still, ahaha, maybe it's me but new problems here!

    Max: [itex] 2n\cos(\theta_1) = (2m+1) [/itex]

    it becomes [itex] cos(\theta_1)=\frac{2m+1}{3}[/itex]

    The first solution i found is when [itex]m=1[/itex], indeed

    [itex] cos(\theta_1)=1[/itex] so that

    [itex]\theta_1=0°[/itex],

    now, to find the angle I need,

    [itex] \sin(\theta_0)=1.5\sin(0°)[/itex], so [itex]\theta_0=0°[/itex]

    Is that right?
     
  5. Aug 2, 2011 #4

    PeterO

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    That looks correct - if the angle of incidence = 0° then the angle of refraction = 0°.

    So you have the angle for the maximum: What is the angle for a minimum?

    Note: re thickness of the glass: was it "one wavelength in air" thick, or " one wavelength in glass" thick?
     
  6. Aug 2, 2011 #5

    gneill

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    Staff: Mentor

    A convenient interpretation of the question would be that the wavelength inside the medium is λ and the angle with the normal in the medium is θ. Beware of the phase shift that occurs upon reflection from the upper surface (medium has higher index of refraction than surrounding air).

    Since the thickness of the film is fixed you really don't need to use the 'm = 0,1,2...' expressions. You should be able to state, by inspection, what the path length needs to be for both cases. Just find an expression for the path length w.r.t. angle and plug in the required path lengths, solving for angle.
     
  7. Aug 2, 2011 #6
    @PeterO:

    Finishing what I have started:

    Min: [itex] \cos(\theta_1) = \frac{m}{3}[/itex]

    which has the "first" solution when m=3,

    it gives [itex] \theta_1 = 0 [/itex] ... which I think is wrong.... :\



    @ both:

    The problem says the light has been emitted with [itex] \lambda [/itex] and then it incides on the medium...

    @ gneill:

    The formulas I have been using to solve this problem have already taken into account that phase shift so i think they are not wrong, do you agree?

    I don't get what you mean when you say "by insepction"...
     
  8. Aug 2, 2011 #7

    gneill

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    Staff: Mentor

    This will complicate matters somewhat, because the wavelength inside the medium won't be [itex] \lambda [/itex]; a ray with angle of incidence 0° will not see a total path length of twice its wavelength inside the medium... there will be a phase shift for this minimum length path.
    I agree, but the formulas were written assuming that you would be looking for an unknown thickness (and possibly several thicknesses: m = 0,1,2,...) of the medium. Here the medium is of a given, fixed thickness. So the 'm' mechanism is not required (it doesn't make it wrong! It's just an unnecessary added complication).

    I mean, you should be able to look at a diagram of the scenario and write down an appropriate equation without requiring derivations.

    If the wavelength really is not [itex] \lambda [/itex] in the medium, then the perpendicular incident ray will not be a solution for either case. [EDIT: I take that back. With n = 1.5 the straight path can be exactly 2.5 wavelengths in the medium!]
     
    Last edited: Aug 2, 2011
  9. Aug 2, 2011 #8

    PeterO

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    My reading of this says we are looking at the light passing through the glass not reflecting so the phase change of reflected light at the first surface is unimportant.

    Light passing straight through, would interact with light having two internal reflections before passing out. Although there there is no phase change with these reflections, it wouldn't matter if there was, since two reflections are involved so they would cancel.

    Edit: I would not work out the problem by substituting into a formula, rather create a necessary geometric situation to solve - but that's me.
     
  10. Aug 2, 2011 #9

    PeterO

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    I was wondering what m was myself. You should be concentrating more on the path difference I feel - or required path difference.

    What does the path difference need to be for constructive interference?

    What does the path difference need to be for destructive interference?

    What is the minimum path difference with this piece of glass? [the path difference can be made longer by having light incident at an angle, but can't be made shorter]

    EDIT: Arguably you can't see light reflected with an angle of incidence of zero, since either your head will block the light source, or the light source will block the reflection to your eye, depending which is closer to the glass.
     
    Last edited: Aug 2, 2011
  11. Aug 2, 2011 #10

    gneill

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    Staff: Mentor

    Allowing for possible language translation issues, and assuming that the problem is likely to strongly resemble the "usual" thin film sort of problem, I imagined that the scenario would be as follows:

    attachment.php?attachmentid=37713&stc=1&d=1312330976.gif

    The idea being to find the minimum angles of incidence that produce constructive and destructive interference when viewed from above.
     

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