Q: A beam of plane polarized light falls normally on a polarizer of cross sectional area 3 × 10-4 m2. Flux of energy of incident ray in 10-3 W. The polarizer rotates with an angular frequency of 31.4 rad/s. The energy of light passing through the polarizer per revolution will be
(a) 10-4 Joule
(b) 10-3 Joule
(c) 10-2 Joule
(d) 10-1 Joule
Ans: (a)
Sol: $ T = \frac{2 \pi}{\omega} $
$ T = \frac{2 \times 3.14}{31.4} = 0.2 sec $
Hence, Energy = 0.5 × power × time
Energy = 0.5 × 10-3 × 0.2
= 10-4 Joule