# 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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by bhushanSengupta969

2014-12-21T00:47:57+05:30

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60 degrees

angle of deviation = δ = i + e - A

i = 0, hence r1 = 0.
Hence r2 = 60  deg  (angle of incidence of second face)

r2 > critical angle for glass-air = Sin ⁻¹ 1/1.5 = 41.81 deg.

It will be totally internally reflected and fall on the base of prism at 90 deg to it.  It will emerge with angle of emergence e = 0 deg.

δ = 0 + 0 - 60 deg  =  60 deg.

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angle of incidence on first face is 0 deg.  Hence angle of refraction is 0 on the first face.

Angle of incidence is 60 degrees on the second face.

angle of refraction =  r  =>
Sin i / Sin r  =  1 / 1.5
sin r =  1.5 * sin 60  = 1.299    > 1
It means that the light ray is incident at an angle more than the critical angle.  The light ray is totally internally reflected back in to 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.

The deviation of the light ray due to the prism is  then  60 degrees.
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