Unformatted text preview: have a lot of incentive for writing a great problem! 1 6.007 Spring 2011 Problem Set 8: Electromagnetic Waves at Boundaries Problem 8.1 – Snell’s Law
In ray optics, it is useful to use Snell’s Law at an interface between two materials: n1 sin(θ1 ) = n2 sin(θ2 ).
(a) Imagine a (collimated) beam of light being shone down to an airwater interface from 5 cm above the
water at 45◦ from the normal. The index of air is taken to be 1 and the index of water is taken to be
1.33. A ﬁsh is swimming at 15 cm horizontal distance away from the light as shown. At what depth
will the ﬁsh see the beam of light? 45±
5 cm
n1=1
n2=1.33 d 15 cm (b) Now the ﬁsh is 15 cm deep, looking up at the water at 30◦ as shown and there is a ﬂy skimming the
water surface 1 cm above the water and changing his position. Can the ﬁsh see the ﬂy? Please explain
your answer. 1 cm
cm n1=1
n2=1.33
15 cm 30± 2 6.007 Spring 2011 Problem Set 8: Electromagnetic Waves at Boundaries Problem 8.2 – Frustrated Total Internal Reﬂection
This problem explores the phenomenon of frustrated total internal reﬂection and the more general math that
goes with it.
In lecture, we discussed what happens when total internal reﬂection is frustrated by bringing a second
medium (e.g., glass) into the evanescent ﬁeld of the reﬂected wave. Figure 1 shows a schematic of the
physical setup of frustrated internal reﬂection, where light which would be reﬂected internally inside a glass
waveguide is able to transmit across an air gap into another piece of glass. Figure 1: Schematic of frustrated internal reﬂection. To simplify our modeling of the above system, we’ll look only in the direction across the air gap, assuming
that the incoming angle is such that total internal reﬂection occurs inside the ﬁrst piece of glass, and therefore,
the ﬁeld in the air gap is evanescent. Figure 2 shows a schematic in 1D of the glassairglass transition. I II III ~ ~ ~
n1 n1 n2 Incident
Wave z 0 d x Figure 2: 1D schematic of coupling via evanescent ﬁeld. We set up the following equations for the electric ﬁeld in the various regions in Figure 2, setting the incident
wave’s magnitude to 1 so that the reﬂection and transmission coeﬃcients (rn and tn ) can be solved for
directly: Ey,I (x, t) = ej (ωt−k1x x) +...
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 Spring '11
 VladimirBulovic
 Light, Polarization, Electromagnet, electromagnetic waves, air gap

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