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  • A ray of light is incident normally on one face of an equilateral prism of refractive index 1.5. What is the angle of deviation of the ray?

A ray of light is incident normally on one face of an equilateral prism of refractive index 1.5. What is the angle of deviation of the ray?

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best_answer

angle of deviation = δ = i + e - A

Here 

  i = 0, hence r1 = 0.

r_2=A-r_1=60^0

the critical angle for glass-air=\sin ^{-1}( \frac{1}{1.5})=41.81^0

So  r2 > critical angle for glass-air

 It means that the light ray is incident at an angle more than the critical angle.  The light ray is totally internally reflected back into the prism from the second face. 

 The angle of reflection will be 60 deg.  This ray will be incident normally on the base face of the prism.  Hence it will emerge normally perpendicular to the base of the prism. 

 i.e  It will emerge with the angle of emergence e = 0 deg.

δ = 0 + 0 - 60 deg  =  60 deg.

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avinash.dongre

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