Unpolarized light passes through three polarizing filters. the first filter has its transmission axis parallel to the z direction, the second has its transmission axis at an angle of 30.0° from the z direction, and the third has its transmission axis at an angle of 60.0° from the z direction. if the light that emerges from the third filter has an intensity of 380.0 w/m2, what is the original intensity of the light? (assume both angles are measured in the same direction from the +z axis.)

 I₀ = 1351.1 W / m²

Explanation:

This problem can be solved using Malus's law.

          I = I₀ cos² θ

Where I is the transmitted intensity, Io is the incident intensity and θ is the angle between the polarization of the light and the polarizer

Let's use this equation for the third polarizer

          I₃ = I₂ cos² θ

The angle with respect to the light that reaches it is the angle of the polarized minus the angle with which the light comes

         θ = 60 - 30

         θ = 30º

We calculate the incident intensity on the third polarized

         I₂ = I₃ / cos² 30

         I₂ = 380.0 / cos² 30

         I₂ = 506.7 w / m²

We calculate the incident intensity on the second polarizer

        I₂ = I₁ cos² 30

        I₁ = I₂ / cos₂ 30

        I₁ = 506.7 / cos² 30

        I₁ = 675.6 W / m²

For the first polarizer the incident light is without polarization, so the polarizer lets half of the light pass, therefore, the light transmitted from the middle of the incident

        I₁ = I₀ / 2

        I₀ = 2 I₁

        I₀ = 2 675.6

        I₀ = 1351.1 W / m²

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we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

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