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Penetration depth

  1. Mar 29, 2016 #1
    This is a review problem for an upcoming exam. I'm pretty sure a problem very similar to this will be on the exam. And surprise surprise I'm freaken lost.

    1. The problem statement, all variables and given/known data

    A high index prism is used to launch an evanescence wave at the water/prism interface as shown. A He-Ne laser (λ = 623 mm) is used as the light source.
    a) find the required angle to excite the evanescence wave.
    b) calculate the penetration depth, the distance into the water at which the amplitude of the evanescence wave has dropped to a value of 1/e of its maximum value at the interface, into the water.
    This is the picture connected to the problem.
    2. Relevant equations

    3. The attempt at a solution
    For part (a) I thought sinθ*n1 = sin90*n2, where n1 = 1 and n2 = 2, solve for θ would give me a solution. It did for for one of his homework assignments. But it doesn't work here.

    For (b) I think I need to get the answer to part (a) first but even then I'm still stuck. My book had two sentences related to penetration depth so it's no help at all.
  2. jcsd
  3. Mar 29, 2016 #2
    ok for (b) i found that to get the drop by e-1 the wave has to propagate a distance of y = 1/α. where α = 2ωn1/c.

    if i solve for ω and plug it in the i get y = 5x10-8.

    does that sound right?
  4. Mar 29, 2016 #3
    for part (a) is n1 = 2 and n2 = 1.33? i get a solution with that. 41.6 degrees
  5. Mar 29, 2016 #4


    User Avatar

    Staff: Mentor

    I think they're looking for the angle of incidence θ at the air-prism boundary. So you need to deal with both interfaces. One to find the critical angle of incidence for the prism-water boundary, the second to find the angle of incidence required at the air-prism boundary.

    I'm not too familiar with the details of penetration depth. But those expressions only reference one refractive index, n1. Is there some context surrounding the expressions that would clarify the scenario?

    From your information, y = 1/α = (c/ω)/(2n1) = (λ/n1)(1/2)

    That would make the penetration depth a half wavelength of the light in whatever medium n1 is associated with.
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