JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
A beam of plane polarised light of large
cross-sectional area and uniform intensity
of 3.3 Wm⁻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 :
- (a) 1.0 × 10⁻⁵ J
- (b) 1.0 × 10⁻⁴ J
- (c) 1.5 × 10⁻⁴ J
- (d) 5.0 × 10⁻⁴ J
Answer and solution
Answer (b) 1.0 × 10⁻⁴ J
Solution
Physics sir · 4 steps
As the polariser spins, cos² θ averages to one half, so half the incident power gets through.
- Incident power. The polariser catches a power P₀ = IA = 3.3 × 3 × 10⁻⁴ = 9.9 × 10⁻⁴ W.
- Average over rotation. By Malus's law the transmitted power is P₀cos² θ, and θ = ω t keeps changing. Over one full revolution the average of cos² θ is 1/2, so the average transmitted power is 4.95 × 10⁻⁴ W.
E = P₀/2 × T = IA/2 × 2π/ω
- Time per revolution. T = 2π/ω = 6.28/31.4 = 0.2 s. So E = 4.95 × 10⁻⁴ × 0.2 = 9.9 × 10⁻⁵ J.
- So. The energy per revolution is about 1.0 × 10⁻⁴ J, option (b).
The idea to keep. When the angle in Malus's law keeps rotating, replace cos² θ by its average 1/2. Then energy is average power times the time you are asked about.
Answer from the NTA final answer key. Solution by PrepWiser.