Question

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at angles of 24.0 and 58.0 °, respectively, to that of the first. If unpolarized light is incident on the stack, the light has an intensity of 62.0 W/cm after it passes through the stack.

If the incident intensity is kept constant but the second polarizer is removed, what is the intensity of the light after it has passed through the stack? Express your answer with the appropriate units.
I=_____

Answer 1

Removing the second polarizer from a stack of three, the intensity I of light after it passes through the remaining two filters can be found by using Malus's Law, considering that the angle between the first and third filters is now 58.0° and the intensity after the first filter is 50% of the initial intensity.

Step-by-step explanation:

If we have three polarizing filters and we remove the second polarizer, which was originally at 24.0° with respect to the first, we need to calculate the intensity I of light after it has passed through the remaining two filters. According to Malus's Law, the intensity of light after passing through a polarizing filter is given by I = I0cos2(θ), where I0 is the initial intensity and θ is the angle between the light's polarization direction and the filter's axis. In this case, the angle between the first and third filters is now 58.0°.

Unpolarized light that passes through the first filter is reduced to 50% of its original intensity. Then, when it goes through another filter at an angle θ, the intensity becomes I = 0.5I0cos2(58.0°). Using the given intensity of 62.0 W/cm² when all three filters are in place, we can find the initial intensity I0 because the second filter (which we removed) would have reduced the intensity by another factor of cos2(24.0°).

So, the initial intensity without any filters, I0, can be found from 62.0 = I0 * 0.5 * cos2(24.0°) * cos2(34.0°), and after finding I0, we then calculate I without the second filter: I = 0.5I0cos2(58.0°).