A beam of plane polarized light falls normally on a polarizer of cross sectional area  3×10−4m2. Flux of energy of incident ray in 10-3 W. The polarizer rotates with an angular frequency of 31.4 rad/sec. The energy of light passing through the polarizer per revolution will be

Using Malus law,  I=I0cos2θAs here polariser is rotating i.e. all the values of q  are possible. Iav=12π∫0 2πI dθ=12π∫ 0 2πI0cos2θ dθOn integration we get  Iav=I02where  I0=EnergyArea × Time=PA=10−33×10−4=103Wattm2\   Iav=12×103=53 Wattand Time period  T=2πω=2×3.1431.4=15sec\ Energy of light passing through the polariser per revolution  =Iav×Area×T=53×3×10−4×15=10−4 J.