{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4500HWR03

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}