1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Index of Refraction, reflection diminishment

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A glass lens is coated on one side with a thin film of magnesium floride (MgF2) to reduce reflection from the lens surface. The index of refraction of MgF2 is 1.38; that of the glass is 1.50. Assume that the light is perpendicular to the lens surface and that the thickness of the coating is the least possible needed to eliminate the reflections of light of wavelength 550 nm at normal incidence.

    hrw7_35-20.gif

    (a) Calculate the percentage by which reflection is diminished by the coating at 445 nm.
    (b) Calculate this percentage for a wavelength of 695 nm.

    2. Relevant equations
    maxima: 2L=(my)/n
    minima: 2L=((m+0.5)y)/2

    [y is lambda)


    3. The attempt at a solution
    What is meant (mathematically) by a diminished reflection?
    This is an extension of a sample problem in my textbook; in the book, the problem is to find the thickness of MgF2 that'll minimize reflection (99.6nm). But the only equations I can think of relating to this system use the quantities: thickness of MgF2, n-air, n-MgF2, n-glass, wavelength, and integer m. It doesn't seem, to me, that any of these are what I should be comparing in the 99.6nm thickness and 445nm thickness situations in order to find the percentage by which reflection is diminished.
     
  2. jcsd
  3. Nov 29, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    "Diminished" here must mean reduction of reflection from the usual 7% or so of the light. This is quite a difficult problem! As wavelength varies away from the value for which the coating was optimized, the destructive interference of the two reflected waves will be imperfect because the coating thickness will not be exactly 1/4 wavelength. AND the amount of reflection at the two surfaces will not be equal (if it was equal at the optimum wavelength) due to variation in the index of refraction with wavelength. Interesting reading at
    http://www.edmundoptics.com/technical-support/optics/anti-reflection-coatings/ [Broken]
    and http://www.mellesgriot.com/products/optics/oc_2_2.htm
    including graphs showing the "answer" for real optical coatings.
    I don't see how you can calculate an accurate answer unless you assume equal reflection from the two surfaces and just do the destructive interference part. I don't know how to do even that - perhaps an integral over the sum of two sinusoidals, one phase shifted relative to the other by the double trip through the coating.
     
    Last edited by a moderator: May 4, 2017
  4. Nov 29, 2009 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is not difficult to derive a formula for the reflectance of a thin film coating if you are familiar with the theory of thin layer optics, but it is not undergraduate level! You can find such expression as below under the topics "thin layer optics":

    n0 is the refractive index of air, n0=1,
    n1 is the refractive index of the layer, n1=1.38,
    n2 is the refractive index of the glass substrate, n2=1.50,
    [itex]\lambda[/itex] is the wavelength, d is the thickness of the layer.


    [tex]R=\frac{(n_0-n_1)^2(n_1+n_2)^2+(n_0+n_1)^2(n_1-n_2)^2+2(n_0-n_1)(n_1-n_2)\cos(4 \pi/\lambda\cdot n_1d)}{(n_0+n_1)^2(n_1+n_2)^2+(n_0-n_1)^2(n_1-n_2)^2+2(n_0-n_1)(n_1-n_2)\cos(4 \pi/\lambda\cdot n_1d)}[/tex]

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook