The angles made with the plane of the interface are found from E,.a, = |E¡|cos(90° – 8,) - 5= V38 sin 6,• 8; = 54.2° E,.a, = |E,lcos(90° – 0,) - 2= v17 sin 6, - Oz = 29.0° A useful relation can be obtained from E tan 0, =- E2 tan 02 =- by division of these two equations gives tan 0 Er2 tan 62 Er1
The angles made with the plane of the interface are found from E,.a, = |E¡|cos(90° – 8,) - 5= V38 sin 6,• 8; = 54.2° E,.a, = |E,lcos(90° – 0,) - 2= v17 sin 6, - Oz = 29.0° A useful relation can be obtained from E tan 0, =- E2 tan 02 =- by division of these two equations gives tan 0 Er2 tan 62 Er1
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Transcribed Image Text:Prove the last law?
The angles made with the plane of the interface are found from
E,. a, = [E;|cos(90° – 6,) o
5 = v38 sin 6, - 0, = 54.2°
%3D
Ez. a, = |E,lcos(90° – 0,) • 2= V17 sin 62
Oz = 29.0°
%3D
A useful relation can be obtained from
Pale
tan 6, =
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poreent
Palesad
E2
tan 62
VE)* + (E,„)* /Ea)* + (E,)'
by division of these two equations gives
tan 8 Er2
tan 02 Er1
%3D
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