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Q.

A beam of plane polarised light of large cross-sectional area and uniform intensity of 3.3 W m−2 falls normally on a polariser (cross sectional area 3×10−4 m2 ) which rotates about its axis with an angular speed of 31.4 rad/s. The energy of light passing through the polariser per revolution, is close to :

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a

1.5×10−4 J

b

1.0×10−4 J

c

1.0×10−5 J

d

5.0×10−4 J

answer is B.

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Detailed Solution

Malus law for polarised light ⇒ I=I0cos2θHere I0=3.3 W/m2θ=ωt where ω=10 π rad/sEnergy E=∫I A d t=I0 A∫cos2ωtdt=I0 Aω∫02πcos2ωt(ωt)=I0 Aωπ =(3.3)3×10-410=0.99×10-4 J

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