14_Chapter 22 HomeworkCH22 Reflection and Refraction of Light

# 14_Chapter 22 HomeworkCH22 Reflection and Refraction of Light

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Unformatted text preview: 246 CHAPTER 22 22.24 n From Snell’s law, sin θ = medium nliver But, so, sin 50.0° 12.0 cm 50.0° nmedium c vmedium v = = liver = 0.900 nliver c vliver vmedium h d nmedium q nliver q Tumor θ = sin −1 ( 0.900 ) sin 50.0° = 43.6° From the law of reflection, d= 22.25 12 .0 cm = 6.00 cm , and 2 h= d 6.00 cm = = 6.30 cm tanθ tan ( 43.6° ) As shown at the right, θ1 + β + θ 2 = 180° incident ray When β = 90° , this gives θ 2 = 90° − θ1 Air, n = 1.00 glass, ng Then, from Snell’s law sin θ1 = q1 nair = ng sin ( 90° − θ1 ) = ng cos θ1 sin θ1 = tan θ1 = ng or θ1 = tan −1 ng cosθ1 () 22.26 Given Conditions and Observed Results 26.5° Sheet 1 n1 b q2 refracted ray ng sin θ 2 Thus, when β = 90° , q1 reflected ray 26.5° Sheet 3 n3 Sheet 2 n2 31.7° Sheet 2 n2 36.7° Case 1 26.5° Sheet 1 n1 Case 2 For the first placement, Snell’s law gives, Sheet 3 n3 qR Case 3 n2 = n1 sin 26.5° sin 31.7° In the second placement, application of Snell’s law yields n1 sin 36.7° n sin 26.5° n3 sin 26.5° = n2 sin 36.7° = 1 sin 36.7° , or n3 = sin 31.7° sin 31.7° ...
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