4500HWR03 - Reflection and Refraction at a Boundary - 4...

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Unformatted text preview: Reflection and Refraction at a Boundary - 4 Light of freespace wavelength 587.6 nm is propagating in BK-7 glass. This cor- responds to the D3 line of a sodium discharge lamp. This light is then incident upon a planar interface with air. Both TE and TM polarizations are present in the incident light wave. For the above BK-7/ air interface, calculate, showing all work, the Brewster an- gle (if it exists) and the critical angle (if it exists). For these calculations, use 1.000 for the refractive index of air. Express angles (if they exist) in degrees accurately to within 0.0001°. If an angle does not exist, write “does not exist.” Put your final answers in the spaces provided. The incident light has an angle of incidence is 60° as measured counter—clockwise from the normal. For this case for the reflected waves, calculate, showing all work, the quantities on the attached sheet. For these calculations, use 1.000 for the refractive in- dex of air. Express angles in degrees accurately to within 0.0001°. Express fractions ac— curately to Within 0.00001. Write your final answers in the spaces provided. Brewster angle (if it exists) 2 Critical angle (if it exists) : TE Polarized Reflected Wave Angle of reflected wavevector (with respect to normal to boundary) 2 Fraction of amplitude reflected :— Phase shift upon reflection 2 Fraction of power reflected TM Polarized Reflected Wave Angle of reflected wavevector (with respect to normal to boundary) 2 Fraction of amplitude reflected = Phase shift upon reflection 2 Fraction of power reflected Reflection and Refraction at a Boundary - 4 A = 587.6 nm n1 = 1.51633 112 = 1.0000 01 = 60° 63 =tan"1(n2/n1) = 33.40435 90 = sin”1(n2 /n1) = 41.2607 6° Therefore, total internal reflection will occur. TE Polarized Reflected Wave Angle of reflected wavevector (with respect to normal to boundary) 2 60.0000° from law of reflection Fraction of amplitude reflected = 1.00000 from total internal reflection Phase shift upon reflection from Fresnel’s equations 2 ' 2 0 __ 2 4m = 2tan—1< "1 8m 1 "2) = 96.61312° n1 cos 91 Fraction of power reflected : 1.00000 from total internal reflection TM Polarized Reflected Wave Angle of reflected wavevector (with respect to normal to boundary) 2 60.0000° from law of reflection Fraction of amplitude reflected = 1.00000 from total internal reflection Phase shift upon reflection from Fresnel’s equations 2 2 "1 sin2 61 — n2 ¢TM = 2tan’1(n1 ) = 137.64596° 2 . n2 cos 01 Fraction of power reflected = 1.00000 from total internal reflection ...
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4500HWR03 - Reflection and Refraction at a Boundary - 4...

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