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 m²) which rotates about its axis with an angular speed of 31.4 rad s⁻¹. 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 1

To find the energy passing through a polarizer per revolution, multiply the power transmitted by the polarizer (calculated using intensity and area) by the time for one revolution (obtained from the angular velocity). The closest numerical answer to the calculated energy is 1.0×10⁻⁵ J.

Step-by-step explanation:

The problem involves calculating the energy of light passing through a rotating polarizer based on the intensity of the polarized light, the cross-sectional area of the polarizer, and the angular speed of rotation. First, we determine the amount of light energy that passes through the polarizer per second by using the intensity and the area:

Energy per second (Power) = Intensity × Area

Energy per second = 3.3 W/m² × 3×10⁻⁴ m² = 9.9×10⁻⁴ W

To find the energy per revolution, we need the time per revolution, which is given by:

Time per revolution = ω / (2π) where ω is the angular velocity

Time per revolution = 31.4 rad/s / (2π rad/rev) ≈ 0.1 s/rev

Now, we can find the energy per revolution by multiplying the energy per second by the time per revolution:

Energy per revolution = Energy per second × Time per revolution

Energy per revolution = 9.9×10⁻⁴ W × 0.1 s = 9.9×10⁻⁵ J

The answer closest to 9.9×10⁻⁵ J is 1.0×10⁻⁵ J, which corresponds to option A.