There are two polarizing filters that are out of alignment by 30 degrees. If the light of intensity 1.00 W/m^2 and initially polarized half-way between the two filters passes through the two filters, what is the intensity of the transmitted light?

Question

There are two polarizing filters that are out of alignment by 30 degrees. If the light of intensity 1.00 W/m^2 and initially polarized half-way between the two filters passes through the two filters, what is the intensity of the transmitted light?

I know the answer is 0.700 W/m^2, but I have no clue how to find that answer.
Thank you for any help!

Answer 1

The fraction that gets through the first fiter is (cos15)^2 = 0.9330

The fraction of that which gets through the second filter is (cos30)^2 = 0.7500

Now take the product.

Actually, sheet polarizing "Polaroid" films lose even more than that, due to absorption.

Answer 2

The intensity of the light isn't halved after it passes through each filter? That's what I don't understand.

Thank you for your help!

Answer 3

To find the intensity of the transmitted light, you need to consider the behavior of polarized light passing through polarizing filters.

When polarized light passes through a polarizing filter, its intensity is reduced according to Malus' law. Malus' law states that the intensity of light passing through a polarizing filter is given by the equation:

I = I₀ * cos²θ

Where I is the intensity of the transmitted light, I₀ is the initial intensity of the incident light, and θ is the angle between the direction of polarization of the incident light and the transmission axis of the polarizing filter.

In this case, the initial intensity of the incident light is given as 1.00 W/m^2. The incident light is initially polarized halfway between the two filters, so the angle θ is 0 degrees.

However, as the two polarizing filters are out of alignment by 30 degrees, we need to account for this misalignment in the calculation. We need to find the angle between the direction of polarization of the incident light and the transmission axis of the first filter.

Since the filters are out of alignment by 30 degrees, the angle between the direction of polarization of the incident light and the transmission axis of the first filter is 30 degrees.

Substituting the values into Malus' law equation:

I = 1.00 * cos²(30)

Now, you can simplify and calculate the result:

I = 1.00 * (cos(30))^2
I = 1.00 * (0.866)^2
I = 0.749 W/m^2

So, the intensity of the transmitted light is approximately 0.749 W/m^2, which is equivalent to 0.749 * 1000 = 749 mW/m^2.