JEE Main20204 Sep 2020Morning ShiftPhysicsWave OpticsActual
A beam of plane polarized light of large cross-sectional area and uniform intensity of 3 .3 W m – 2 falls normally on a polarizer (cross-sectional area 3 × 10 – 4 m 2 ), which rotates about its axis with an angular speed of 31 .4 rad s - 1 . The energy of light passing through the polarizer per revolution, is close to:
Options
- A1 .0 × 10 − 5   J
- B1 . 0 × 10 − 4   J
- C1 . 5 × 10 − 4   J
- D5 . 0 × 10 − 4   J
Correct answer
B. 1 . 0 × 10 − 4   J
Step-by-step solution
Given, Intensity of plane polarized light of large cross-sectional area is I 0 = 3 . 3   W   m - 2 Area of cross-section of polarizer is A = 3 × 10 - 4   m 2 Angular speed of rotation is ω = 31 . 4   rad   s - 1 The intensity of light is the given by the definition, I = E A T     . . . ( 1 ) Intensity of light passing through the polarizer as polarizer rotates is given by I a v g = I 0 1 T ∫ 0 T cos 2 ω t d t ⇒ I a v g = I 0 1 T ∫ 0 T 1 + cos 2