Phys 202 1nd Edition Lecture 23Outline of Last Lecture I. Solving using Snell’s lawOutline of Current Lecture II. Example ProblemsCurrent LectureLight Wave Example Problems:Example 1:In the figure below, a light ray enters a slab made of an unknown transparent material at point A and then undergoes total internal reflection at point B. What minimum value for the index of refraction of the material can be inferred from this information, if θ = 35.8°? A θ1=35.8 n=1 Θ2 n B θ3=90-θ2 Θ4=90A. nsin(θ1)=nsin(θ2)B. Nsin(θ3)=nsin(90-θ2)=ncos(θ2)=11=ncos(θ2)=n√(1-sin2(θ2)) = n√(1-(sinθ1/n)2)=√(n2-sin2(θ1))N=√(1+sin2(35.8))=1.16These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Example 2:Radiation from the sun to the earth has an intensity of IE=1400W/m2. What is the intensity at mercury?The distance from the Earth to the sun is about 150x106kmThe distance from Mercury to the sun is about 57.9x106kmI=P/AA of a sphere=4πR2Intensity of Earth: IE=Ps/(4πR2E)=1400Intensity of Mercury: IM=Ps/(4πR2M)Therefore; IM= IE/R2IM=IE(RE/RM)2 = 1400(150x106km/57.9x106km)=9396W/m2Example 3:A laser beam of power 5.00 W and diameter 2.50 mm is directed upward at one circular face (of diameter less than 2.50 mm) of a perfectly reflecting cylinder, which is made to "hover" by the beam's radiation pressure. The cylinder's density is 2.00 g/cm3. What is the cylinder's height H?Momentum(p)=U/cIf it’s reflecting, then Δp=2pF=Δp/Δt = 2p/Δt = [2Δu/c]/Δt2*Δu/Δtc=mg 2(power/c)=mgFor this problem, m=volume=(πr2hd)(for a cylinder)2(power/c)=
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