ray of light
Asked by chandant | 30th Jan, 2011, 03:14: PM
Using snell's law,
n ref.ind X 0.7 = n air X sin(angle)
1.6 X 0.7 = 1 X sin (angle)
Sin -1 ( 1.12 ) = 1 X angle
This is greater than 90 deg.
i here is 45 deg.
Hence, The sine of the angle of refraction of the emergent ray will be greater than 0.7.
Hope this helps.
Answered by | 31st Jan, 2011, 02:05: PM
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