Intensity of polarized light that has passed through two polarizing sheets

[Kant Destroyer]

Homework Statement

A beam of polarized light of intensity 43.0 W/m2 is sent through a system of two polarizing
sheets. Relative to the polarization direction of that incident light, the polarizing directions of the sheets
are at angles θ for the first sheet and 90 degrees for the second sheet. If the intensity of the final, transmitted light is 8.6 W/m2, what is the value of θ?

Homework Equations

I = I₀cos²(θ)

The Attempt at a Solution

I recognized that to solve the problem I had to work backwards to solve for θ.

I used the equation I = I₀cos²(θ) and found that I_final = I₁cos²(90-θ), where I₁ is the intensity of the light after passing through the first polarizing sheet. Now I'm unsure where to head with this because I have two unknown variables, I₁ and (90-θ). I am also lacking another equation to use a system of equations as far as I can tell.

 

[haruspex]

What fraction of light makes it through the first polarising sheet?

 

[Kant Destroyer]

What fraction of light makes it through the first polarising sheet?

The intensity of the light that is between the first and second polarizing sheets can be seen in my original post as I₁.

 

[haruspex]

The intensity of the light that is between the first and second polarizing sheets can be seen in my original post as I₁.

Sure, but how does it relate mathematically to l₀?

 

[Kant Destroyer]

Sure, but how does it relate mathematically to l₀?

The relationship between I₀ and I₁ is: I₁ = I₀cos²(θ). Are you suggesting that I substitute this in for I₁ in my equation and then solve for θ? I have tried this and I am not sure how to solve for theta in this instance.

 

[haruspex]

The relationship between I₀ and I₁ is: I₁ = I₀cos²(θ).

How is theta involved? It hasn't reached the second filter yet.

 

[Kant Destroyer]

How is theta involved? It hasn't reached the second filter yet.

As described in the question, theta is the angle of orientation of the first sheet relative to the polarization direction of the light entering the sheet. Per the relevant equation, this means that theta is used to find the intensity of the light after it has traveled through the first polarizing sheet.

 

[haruspex]

As described in the question, theta is the angle of orientation of the first sheet relative to the polarization direction of the light entering the sheet.

You have I₀ initially, I₁ between the two sheets, I₂ finally.
The last two are related by $I_2 = I_1 \cos^2(\theta)$.
I'm asking how I₁ relates to I₀. Remember that the light between the two filters doesn't 'know' it's about to go through a second filter, so the angle theta can have nothing to do with this relationship. If it helps, mentally throw the second filter away and answer my question.

 

[Kant Destroyer]

You have I₀ initially, I₁ between the two sheets, I₂ finally.
The last two are related by $I_2 = I_1 \cos^2(\theta)$.
I'm asking how I₁ relates to I₀. Remember that the light between the two filters doesn't 'know' it's about to go through a second filter, so the angle theta can have nothing to do with this relationship. If it helps, mentally throw the second filter away and answer my question.

I think you misunderstand my original post because I failed to use another variable for the angle between the two polarizing sheets and instead used (90-θ). Theta is meant to represent the angle of orientation of the first polarizing sheet, and per the relevant equation it is used to find I₁.

 

[haruspex]

I think you misunderstand my original post because I failed to use another variable for the angle between the two polarizing sheets and instead used (90-θ). Theta is meant to represent the angle of orientation of the first polarizing sheet, and per the relevant equation it is used to find I₁.

Sorry, dealing with too many threads at once. let me start again.
You have I_final = I₁cos²(90-θ), I₁ = I₀cos²(θ) and the value of I_final / I₀ . So your difficulty is in simplifying the trig, yes?
What trig formulae do you know which involve expressions like sin(θ)cos(θ)?

 

[Kant Destroyer]

Sorry, dealing with too many threads at once. let me start again.
You have I_final = I₁cos²(90-θ), I₁ = I₀cos²(θ) and the value of I_final / I₀ . So your difficulty is in simplifying the trig, yes?
What trig formulae do you know which involve expressions like sin(θ)cos(θ)?

The only trig formulas I remember are sin²+cos² = 1 and cos²(θ) = 1/2 + 1/2cos(2θ).

 

[haruspex]

The only trig formulas I remember are sin²+cos² = 1 and cos²(θ) = 1/2 + 1/2cos(2θ).

You need the one for expanding $\sin(2\theta)$, and one for $\cos(90-\theta)$. You don't know those?

 

[Kant Destroyer]

You need the one for expanding $\sin(2\theta)$, and one for $\cos(90-\theta)$. You don't know those?

I'm not even sure how sin(2θ) plays into the mix in this problem seeing as they are both cos² functions, but no I do not know those. If I remember correctly the cos(90-θ) is actually just sin(θ) but I'm not sure about that.

 

[haruspex]

cos(90-θ) is actually just sin(θ)

Right. So making that substitution, write out the equation for I_final as a function of I₀.

 

[Kant Destroyer]

Right. So making that substitution, write out the equation for I_final as a function of I₀.

I_final/cos²(θ)sin²(θ) = I₀

EDIT: Misread your response. I_final = I₀cos²(θ)sin²(θ)

 

[haruspex]

I_final = I₀cos²(θ)sin²(θ)

Right. You ought to know that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.

 

[Kant Destroyer]

so cos²(θ)sin²(θ) = 1/4sin²(2θ) ?

 

[haruspex]

so cos²(θ)sin²(θ) = 1/4sin²(2θ) ?

If you mean (1/4)sin²(2θ) , yes. Now you can use the two trig equations you posted earlier to get rid of the remaining quadratic.

 

[Kant Destroyer]

If you mean (1/4)sin²(2θ) , yes. Now you can use the two trig equations you posted earlier to get rid of the remaining quadratic.

Thank you very much for taking the time to help me.

 

[haruspex]

Thank you very much for taking the time to help me.

OK. Sorry about my earlier misdirection.