{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4500PR04

# 4500PR04 - Reﬂection and Refraction at a Boundary 1 Light...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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 - 1 Light of freespace wavelength 546.1 nm (e line) is incident in air upon a planar surface of BK-7 glass. The angle of incidence is 60° measured counter-clockwise from the normal. Both TE and TM polarizations are present in the incident light wave. For the transmitted and reﬂected waves7 calculate, showing all work, the quanti— ties on the attached sheet. Express angles in degrees accurately to within 0.0001°. Ex— press fractions accurately to within 0.00001. Write your ﬁnal answers in the spaces pro— vided. TE Polarized Transmitted Wave Angle of transmitted wavevector (with respect to normal to boundary) = Fraction of amplitude transmitted 2 Phase shift upon transmission 2 H Fraction of power transmitted 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 Transmitted Wave Angle of transmitted wavevector (with respect to normal to boundary) = Fraction of amplitude transmitted = Phase shift upon transmission 2 Fraction of power transmitted TM Polarized Reﬂected Wave Angle of reﬂected wavevector (with respect to normal to boundary) = Fraction of amplitude reﬂected = Phase shift upon reﬂection = Fraction of power reﬂected Reﬂection and Refraction at a Boundary - 1 A = 546.1 nm 71 = 1.51825 01 = 60° 02 = sin_1[(n1/n2)sin01] = 34.77884° 03 = tan—1(n2/n1) = 56.62898° From Fresnel’s equations TE Polarized Transmitted Wave Angle of transmitted wavevector (with respect to normal to boundary) 2 34.77884° Fraction of amplitude transmitted = 0.57240 Phase shift upon transmission 2 0° Fraction of power transmitted = 0.81716 TE Polarized Reﬂected Wave Angle of reﬂected wavevector (with respect to normal to boundary) 2 60.0000° Fraction of amplitude reﬂected = —O.42760 Phase shift upon reﬂection : 180.0000° Fraction of power reﬂected = 0.18284 TM Polarized Transmitted Wave Angle of transmitted wavevector (with respect to normal to boundary) = 34.77884° Fraction of amplitude transmitted = 0.63272 Phase shift upon transmission = 0° Fraction of power transmitted = 0.99845 TM Polarized Reﬂected Wave Angle of reﬂected wavevector (with respect to normal to boundary) = 60.0000° Fraction of amplitude reﬂected = ——0.03938 Phase shift upon reﬂection = 180.0000° Fraction of power reﬂected = 0.00155 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

4500PR04 - Reﬂection and Refraction at a Boundary 1 Light...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online