3/31/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 1/2Master ItThe light beam in the figure below strikes surface 2 at the critical angle. Determine the angle of incidence θ1.(Let θ2 = 42.1°.)Part 1 of 5 ConceptualizeWe estimate the value of the angle of incidence to be close to the value of the critical angle.Part 2 of 5 CategorizeWe will find the angle of incidence by applying Snell's law at the first interface where the light is refracted. Atthe second surface, knowing that the given angle is the critical angle, we can work backward to find the angleof incidence.Part 3 of 5 AnalyzeLet be the index of refraction of the surrounding medium and be the index of refraction for the prism.We can use the critical angle to find the ratio from Snell's law. We haveso thatPart 4 of 5 AnalyzeAs shown in the diagram below, we will call the angle of refraction at surface 1 The ray inside the prismforms a triangle with surfaces 1 and 2, so the sum of the interior angles of this triangle must be 180°. Thus,n1n2θ2n2/n1n2 sin θ2 = n1 sin 90° = n1, = = = = 1.49 .n2n11sin θ21sin 42.1 °1 0.67θ1r.3/31/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 2/2and solving for we obtainPart 5 of 5 AnalyzeApplying Snell's law at surface 1, we haveand solving for gives the following.FinalizeOur result is not as close to the given value of the critical angle as we expected. Solving this problemrequired that we apply Snell's law to two surfaces, determining angles of reflection and refraction at thesesurfaces. We used geometry to determine the relationship between the given critical angle and the angle ofrefraction at the first surface.(90.0° − θ1r) + 60.0° + (90.0° − θ2) = 180°,θ1r,θ1r = 60.0° − θ2 = 60.0° − 42.1 ° = 17.9 °.n1 sin θ1 = n2 sin θ1r,θ1θ1 = sin−1 = sin−1 1.49 sin 17.9 ° = 27.3
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