A beam of plane polarized light of large cross-sectional area and uniform intensity of $3.3 \mathrm{Wm}^{-2}$ falls normally on a polarizer (cross sectional area $3 \times 10^{-4} \mathrm{~m}^{2}$ ) which rotates about its axis with an angular speed of $31.4 \mathrm{rad} / \mathrm{s}$. The energy of light passing through the polariser per revolution, is close to:

(D) $1.0 \times 10^{-4} \mathrm{~J}$

(A) $1.0 \times 10^{-5} \mathrm{~J}$

(B) $1.5 \times 10^{-4} \mathrm{~J}$

(C) $5.0 \times 10^{-4} \mathrm{~J}$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Wave Optics

A beam of plane polarized light of large cross-sectional area and uniform intensity of

3.3 Wm^{- 2}

falls normally on a polarizer (cross sectional area

3 \times 1 0^{- 4} m^{2}

) 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:

Answer: (D)

When plane polarised light passes through a polariser spinning about the beam axis, the transmitted intensity varies as

I cos^{2} θ

, and averaged over a full rotation

⟨ cos^{2} θ ⟩ = \frac{1}{2}

. The time for one revolution is

t = \frac{2 π}{ω} = \frac{2 π}{31.4} = 0.2

s. The energy passing through in that time is the incident intensity times the area times this time times the average factor,

E = I A t ⟨ cos^{2} θ ⟩ = 3.3 \times (3 \times 1 0^{- 4}) \times 0.2 \times \frac{1}{2}

, which comes to about

1.0 \times 1 0^{- 4}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option