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 .  ’’ a    a    ’ + a = 90 Therefore:   =   ’  

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Presentation on theme: " .  ’’ a    a    ’ + a = 90 Therefore:   =   ’  "— Presentation transcript:

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5  ’’ a    a    ’ + a = 90 Therefore:   =   ’  

6  ’’ a    b    ’ + b = 90 Therefore:   =   ’  ’’ b  

7  a  b Sin     hyp   Sin     hyp Therefore: sin    =   sin   

8 Since v is proportional to sin      v 1 = constant sin     v 2 sin     sin     This constant is the relative refractive index, n

9 sin      v 1 = n 2 = n sin     v 2 n 1 Snell's Law


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