Question

A polarized light beam, with a 7.55 mm diameter circular cross section, is perpendicularly incident on an ideal polarizer, serving as an analyzer. The intensity of the light is 3.87 kW/m2 and the angle between the beam's polarization direction and the polarization axis of the polarizer is 19.9∘. What force in piconewtons does the light exert on the polarizer?

Answer 1

The force exerted by the light on the polarizer is approximately 145.99 pN.

Step-by-step explanation:

The force exerted by the light on the polarizer can be calculated using the equation:

Force = Intensity × Area × cos(θ)

Where:

• Force is the force exerted by the light on the polarizer (in piconewtons)
• Intensity is the intensity of the light (in kW/m²)
• Area is the cross-sectional area of the light beam (in m²)
• θ is the angle between the polarization direction of the light beam and the polarization axis of the polarizer (in degrees)

First, we need to convert the intensity from kW/m² to W/m² by multiplying it by 1000. Then, we can calculate the area of the light beam using the formula for the area of a circle (A = πr²), where r is the radius of the circular cross section (diameter/2).

Substituting the values into the equation, we get:

Force = (Intensity × Area × cos(θ)) × 10⁻⁹

Calculating the numerical value of the force:

Force = (3870 × π × (7.55/2)² × cos(19.9)) × 10⁻⁹

Force ≈ 145.99 pN

Therefore, the force exerted by the light on the polarizer is approximately 145.99 pN.