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Unformatted text preview: 59. (a) A ray diagram is shown below. Let θ1 be the angle of incidence and θ2 be the angle of refraction
at the ﬁrst surface. Let θ3 be the angle of incidence at the second surface. The angle of refraction
there is θ4 = 90◦ . The law of refraction, applied to the second surface, yields n sin θ3 = sin θ4 = 1.
As shown in the diagram, the normals to the surfaces at P and Q are perpendicular to each other.
The interior angles of the triangle formed by the ray and the two normals must sum to 180◦ , so
θ3 = 90◦ − θ2 and sin θ3 = sin(90◦ − θ2 ) = cos θ2 = 1 − sin2 θ2 . According to the law of refraction,
applied at Q, n 1 − sin2 θ2 = 1. The law of refraction, applied to point P , yields sin θ1 = n sin θ2 ,
so sin θ2 = (sin θ1 )/n and
n 1− sin2 θ1
=1.
n2 Squaring both sides and solving for n, we get
1 + sin2 θ1 . n= .....
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. θ 90 P Q θ θ θ 90 (b) The greatest possible value of sin2 θ1 is 1, so the greatest possible value of n is nmax = √ 2 = 1.41. (c) For a given value of n, if the angle of incidence at the ﬁrst surface is greater than θ1 , the angle
of refraction there is greater than θ2 and the angle of incidence at the second face is less than θ3
(= 90◦ − θ2 ). That is, it is less than the critical angle for total internal reﬂection, so light leaves
the second surface and emerges into the air.
(d) If the angle of incidence at the ﬁrst surface is less than θ1 , the angle of refraction there is less than
θ2 and the angle of incidence at the second surface is greater than θ3 . This is greater than the
critical angle for total internal reﬂection, so all the light is reﬂected at Q. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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