14_Ch 22 College Physics ProblemCH22 Reflection and Refraction of Light

14_Ch 22 College Physics ProblemCH22 Reflection and Refraction of Light

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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° ...
View Full Document

This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online