What was the initial polarization direction of the incident light?

[xQuin]

Homework Statement

Two polarizers are oriented at 40deg to each other and plane-polarized light is incident on them. If only 15% of the light gets through both of them, what twas the initial polarization directon of the incident light?

Homework Equations

The Attempt at a Solution

I think it ha something to do with Malus's Law but I'm not sure exactly how

 

[tiny-tim]

welcome to pf!

hi xQuin! welcome to pf!

yes, assume that the angle of polarization to the first polarizer is θ …

how much goes through the first polarizer, and how much of that goes through the second?

show us what you get ​

 

[xQuin]

Thanks tim, I applied the law twice and managed to get an answer of 59.6deg =D

 

[tiny-tim]

looks good!

 

[asteriatic]

to use it in this situation.

The initial polarization direction of the incident light can be determined by using Malus's Law, which states that the intensity of light transmitted through two polarizers is proportional to the cosine squared of the angle between their transmission axes. In this case, the angle between the two polarizers is 40 degrees, so the intensity of light transmitted through both polarizers would be 0.15 times the initial intensity.

Using this information, we can solve for the initial polarization direction by setting up the equation:

I = I₀ cos²θ

Where I is the intensity of light transmitted through both polarizers, I₀ is the initial intensity, and θ is the angle between the two polarizers (40 degrees).

Rearranging the equation, we get:

I₀ = I / cos²θ

Substituting in the given values, we get:

I₀ = 0.15 / cos²40

Solving for I₀, we get:

I₀ = 0.15 / 0.766 = 0.196

Therefore, the initial polarization direction of the incident light must have been at an angle of 40 degrees to the first polarizer.