Question

If you have completely polarized light of intensity 150 W/m², what will its intensity be after passing through a polarizing filter with its axis at an 89.0º angle to the light’s polarization direction?

a) 0 W/m²
b) 15 W/m²
c) 75 W/m²
d) 150 W/m²

Answer 1

After applying Malus's Law, we can determine that the intensity of completely polarized light with an initial intensity of 150 W/m² will be approximately 0 W/m² after passing through a polarizing filter with its axis at an 89° angle to the light's polarization direction.

Step-by-step explanation:

The question deals with the concept of polarized light and its interaction with a polarizing filter. According to Malus's Law, the intensity of polarized light after passing through a polarizing filter is given by I = I_0 × cos^2(θ), where I_0 is the initial intensity of the light, and θ is the angle between the light's polarization direction and the axis of the filter.

Using the given values, I_0 = 150 W/m² and θ = 89.0°, we can calculate the intensity of the polarized light after passing through the filter as follows:

I = 150 W/m² × cos^2(89.0°)

Since cos(89.0°) is a very small number, cos^2(89.0°) will be even smaller, and when multiplied by 150 W/m², the resulting intensity is approximately 0 W/m², which represents an almost completely blocked light intensity due to the filter's orientation being nearly perpendicular to the light's polarization direction.