ch33-p067 - 67. (a) A ray diagram is shown below. Let 1 be...

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(c) For a given value of n , if the angle of incidence at the first surface is greater than θ 1 , the angle of refraction there is greater than 2 and the angle of incidence at the second face is less than 3 (= 90° – 2 ). 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 1 , the angle of refraction there is less than 2 and the angle of incidence at the second surface is greater than 3 . This is greater than the critical angle for total internal reflection, so all the light is reflected at Q . 67. (a) A ray diagram is shown below. Let 1 be the angle of incidence and 2 be the angle of refraction at the first surface. Let 3 be the angle of incidence at the second surface. The angle of refraction there is
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