21_Chapter 22 HomeworkCH22 Reflection and Refraction of Light

21_Chapter 22 HomeworkCH22 Reflection and Refraction of Light

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Unformatted text preview: Reflection and Refraction of Light 22.43 If θ c = 42 .0° at the boundary between the prism glass n and the surrounding medium, then sin θ c = 2 gives n1 q1 nglass nm = Surrounding Medium, nm b nglass 60.0° qr nm = sin 42.0° nglass or 253 a qc = 42.0° Surface 2 1 = 1.494 sin 42.0° From the geometry shown in the above figure, α = 90.0° − 42.0° = 48.0°, β = 180° − 60.0° − α = 72.0° and θ r = 90.0° − β = 18.0° . Thus, applying Snell’s law at the first surface gives nglass sin θ r −1 = sin ( 1.494 sin 18.0° ) = 27.5° nm θ1 = sin −1 22.44 (a) θ1′ = θ1 = 30.0° n1 sin θ1 n2 θ 2 = sin −1 ( 1.00 ) sin 30.0° = sin −1 1.55 air q1 q1¢ air q2 glass glass q2 Parts (a) and (c) = 18.8° n sin θ1 −1 ( 1.55 ) sin 30.0° (b) θ1′ = θ1 = 30.0° , θ 2 = sin −1 1 = sin = 50.8° 1.00 n2 q1 q1¢ Parts (b) and (d) ...
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