The speed of light in a material is 0.48 c. What is the critical angle of a light ray at the interface between

the material and a vacuum?





a) 23 degrees


b) 29 degrees


c) 26 degrees


d) 20 degrees

The speed of light in a material is 0.48 c. What is the critical angle of a light ray at the interface between
The nearest answer is b.) : 29 degrees





That is because the angles of incidence ' i ' and refraction ' r ' are connected by Snell's Law:





n_i sin i = n_r sin r ; (N)





here, ' n_i ' and ' n_r ' are the corresponding "indices of refraction." Indices of refraction themselves satisfy:





c_n = c / n,





wher c_n is the speed of light in some given medium.





Because of this inverse speed relationship, (S) can be rewritten:





(sin i) / c_i = (sin r) / c_r , or sin i = (sin r) [c_i / c_r] . (S)





The critical angle ' i ' of a medium with a vacuum occurs when sin r = 90 deg., that is sin r = 1. So the angle is given by sin i = [0.48c] / c.





Thus i = sin^(-1) 0.48 = 28.685... degrees --- or 29 degrees, when rounded to the nearest degree.





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