snell's law. how to solve for 1.0009sin30=n2sin20?

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snell's law. how to solve for 1.0009sin30=n2sin20?

2006-11-12 06:53:37 · 3 answers · asked by jrd 2 in Science & Mathematics ➔ Mathematics

3 answers

Given:
1.0009 sin30° = n2 sin20°
1.0009(0∙5) = n2(0∙342 020 143)
0∙500 45 = n2(0∙342 020 143)
0∙500 45 = n2(0∙342 020 143)
n2 = (0∙500 45)/(0∙342 020 143)
n2 = 1∙463 179 12

2006-11-12 07:05:28 · answer #1 · answered by Brenmore 5 · 1⤊ 0⤋

1.0009sin30=n2sin20
n=1.0009sin30/2sin20
n=1.0009x0.5/2x0.3420
=0.5/0.6840
=0.72

2006-11-12 15:52:23 · answer #2 · answered by Cab302 2 · 0⤊ 0⤋

n = 1.0009(sin30)/(2sin20) ~= 0.73

2006-11-12 14:56:46 · answer #3 · answered by jacinablackbox 4 · 0⤊ 0⤋