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 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 :

Question

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 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 : (A) 1.0 × 10-5 J (B) 5.0 × 10-4 J (C) 1.0 × 10-4 J (D) 1.5 × 10-4 J

Tardigrade Admin · Accepted Answer

Intensity, I0=3.3 Wm -2 Area, A=3 × 10-4 m 2 Angular speed, ω=31.4 rad / s Average energy .=I0 A< cos 2 θ rangle ∵ langle cos 2 θ rangle=(1/2) per revolution ∴ Average energy =((3.3)(3 × 10-4)/2) ≃ 5 × 10-4 J

$1.0 \times 1 0^{- 5} J$

$5.0 \times 1 0^{- 4} J$

$1.0 \times 1 0^{- 4} J$

$1.5 \times 1 0^{- 4} J$

Intensity, $I_{0} = 3.3 W m^{- 2}$
Area, $A = 3 \times 1 0^{- 4} m^{2}$
Angular speed, $ω = 31.4 r a d / s$
Average energy $= I_{0} A < cos^{2} θ ⟩$
$∵ ⟨ cos^{2} θ ⟩ = \frac{1}{2}$ per revolution
$∴$ Average energy $= \frac{( 3.3 ) ( 3 \times 1 0 ^{- 4} )}{2} ≃ 5 \times 1 0^{- 4} J$

1.0×10−5J

5.0×10−4J

1.0×10−4J

1.5×10−4J

Intensity, I0​=3.3Wm−2 Area, A=3×10−4m2 Angular speed, ω=31.4rad/s Average energy =I0​A<cos2θ⟩ ∵⟨cos2θ⟩=21​ per revolution ∴ Average energy =2(3.3)(3×10−4)​≃5×10−4J

$1.0 \times 1 0^{- 5} J$

$5.0 \times 1 0^{- 4} J$

$1.0 \times 1 0^{- 4} J$

$1.5 \times 1 0^{- 4} J$

Intensity, $I_{0} = 3.3 W m^{- 2}$
Area, $A = 3 \times 1 0^{- 4} m^{2}$
Angular speed, $ω = 31.4 r a d / s$
Average energy $= I_{0} A < cos^{2} θ ⟩$
$∵ ⟨ cos^{2} θ ⟩ = \frac{1}{2}$ per revolution
$∴$ Average energy $= \frac{( 3.3 ) ( 3 \times 1 0 ^{- 4} )}{2} ≃ 5 \times 1 0^{- 4} J$