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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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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, = %3D poreent Palesad E2 tan 62 VE)* + (E,„)* /Ea)* + (E,)' by division of these two equations gives tan 8 Er2 tan 02 Er1 %3D
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, = %3D 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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