Solutions to Additional Conceptual Practice Problems for PHYS 1112 Final Exam

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Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerand changing the absorption-reflection factorα S = 1 → α D = 2 = 2 × α S ;and changing the exposure cross-section area from a circle of radius R to a square of 4Rsidelength, i.e.,A S = πR 2 → A D = (4R) 2 = (16/π) × A S ,will change the required intensity for levitationI S → I D = 3 × (2) −1 × (16/π) −1 × I S = (3π/32)I Ssince the levitation intensity I is proportional to m, but inversely proportional to α and A.6
Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerCP 4.05+4.06:OllfTu.O 0..,.1. gI, _---..Side. 'VJ'ofl'e'1/-eol---r----r-:-~-r-:--......;--=---~-~"O""'lj ~t {V1.f e~s( te'~I - 02.1------_.........:.-I~ -:::::: 0--Clv-e ~'PIrnrr.-~~ _~1iD~Q~ FytJ £- Viri f1 +-5 :CoS( OiJ)=(f3')/J..CoS (bOO) = 2CP 4.05 (=Q1), Answer: (B)For unpolarized light of time-averaged (TA) intensity I 0 passing through polarization filter(PF) 1, the TA intensity after the filter is I 1 = I 0 〈cos 2 θ 1 〉 θ . Here, 〈...〉 θ means averaging overall possible values of the random angle between an incident waves packet’s electric field E ⃗ 0and transmission axis T 1 , denoted by θ 1 ≡ ̸ ( E ⃗ 0 , T 1 ). For uniformly random angle θ 1 (i.e.,any value of θ 1 between 0 o and 180 o is equally likely to be found among incident EM wavepackets), 〈cos 2 θ 1 〉 θ = 1/2 and hence I 1 = (1/2)I 0 .The wave packets passing through PF1 (i.e. between PF1 and PF2) are then polarized withelectric field vector E ⃗ 1 ‖ T 1 . But then the angle between E ⃗ 1 (incident upon PF2) and T 2 isθ 2 ≡ ̸ ( E ⃗ 1 , T 2 ) = 90 o , since T 2 ⊥ T 1 . Hence, by Malus’ Law, the TA intensity transmitted7
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Solutions to Additional Conceptual Practice Problems for PHYS 1112 Final Exam

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