# A beam of plane polarized light falls normally on a polarizer of cross sectional area…

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