p34_051 - θ 2 Now the angles in the triangle ABC must add...

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51. We label the light ray’s point of entry A , the vertex of the prism B , and the light ray’s exit point C . Also, the point in Fig. 34-49 where ψ is de±ned (at the point of intersection of the extrapolations of the incident and emergent rays) is denoted D . The angle indicated by ADC is the supplement of ψ ,sowe denote it ψ s = 180 ψ . The angle of refraction in the glass is θ 2 = 1 n sin θ . The angles between the interior ray and the nearby surfaces is the complement of θ 2 ,sowedenoteit θ 2c =90
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Unformatted text preview: θ 2 . Now, the angles in the triangle ABC must add to 180 ◦ : 180 ◦ = 2 θ 2c + φ = ⇒ θ 2 = φ 2 . Also, the angles in the triangle ADC must add to 180 ◦ : 180 ◦ = 2 ( θ − θ 2 ) + ψ s = ⇒ θ = 90 ◦ + θ 2 − 1 2 ψ s which simpli±es to θ = θ 2 + 1 2 ψ . Combining this with our previous result, we ±nd θ = 1 2 ( φ + ψ ). Thus, the law of refraction yields n = sin( θ ) sin( θ 2 ) = sin ( 1 2 ( φ + ψ ) sin ( 1 2 φ ) ....
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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.

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