Solving for Theta & Io: An Exploration

[bbbbb]

Homework Statement

A ray of light is plane polarized. A sheet of Polaroid is held at an angle of 60 to the ray. What fraction of the original intensity of the ray is transmitting through the Polaroid at an angle of 60.

Relevant Equations

I=IoCos^2theta

I am not sure if theta will be 60 degrees or 30 degrees in this case and Io is given.

 

[kuruman]

I agree that the problem is ambiguous. Rays are thought of as straight lines along the direction of propagation and polarizers are usually held perpendicular to that direction but rotated about the ray axis. I would assume the obvious and set θ = 60^o in the relevant equation that you quoted.

 

[bbbbb]

If the angle is 60, how will I calculate I when Io is unknown.

 

[kuruman]

If the angle is 60, how will I calculate I when Io is unknown.

You don't. The expected answer is a fraction of $I_0$, e.g. $I=\dfrac{1}{\pi}I_0~~$ (not the correct answer). The problem is asking for the fraction, i.e. what multiplies $I_0$ in the answer.

 

[bbbbb]

so, it will be I/Io=Cos60
which is I/Io=0.5

 

[kuruman]

so, it will be I/Io=Cos60
which is I/Io=0.5

No it won't be that. Look again at your relevant equation in post #1.

 

[bbbbb]

I am looking for the fraction of the original intensity.
I am back to square one of finding Io, which is needed to solve the question.

 

[kuruman]

I am looking for the fraction of the original intensity.
I am back to square one of finding Io, which is needed to solve the question.

Maybe you are unclear about what "fraction of the original intensity" means. The fraction $f$ of the original intensity is defined as $f=\dfrac{I}{I_0}$. Solve your "relevant equation" for that and substitute the value for the angle. What do you get?

 

[bbbbb]

We are in a way saying the same thing; I need to find Io first then substitute in the formula above after finding I, the formula you said can now work.

 

[kuruman]

We are not saying the same the thing. I am saying that you are looking for an expression for $\dfrac{I}{I_0}$. You are saying that you need a number for $I_0$ before you can find an expression for the ratio. These are not the same thing. You do not need any numbers to find what I am asking. Do it with symbols and you will see where I am going with this. You sort of did it in post #5, but not quite.

 

[bbbbb]

I have solved in respect to angle 60. I made I/Io the subject of formula
I=IoCos^2theta
I/Io=cos(2*60...I/Io= -0.5

 

[kuruman]

I have solved in respect to angle 60. I made I/Io the subject of formula
I=IoCos^2theta
I/Io=cos(2*60...I/Io= -0.5

IoCos^2theta is not[/color] cos(2*60, What does the uparrow "^" signify?

 

[bbbbb]

is it that the transmitted electric field is reduced by a factor cos (60) and the intensity is reduced by the square of that amount

 

[kuruman]

is it that the transmitted electric field is reduced by a factor cos (60) and the intensity is reduced by the square of that amount

Not quite. The trnasmitted electric field component is reduced to E = E₀ cos(60^o) not by a factor of cos(60^o).

Yes, the intensity is reduced by a factor of cos²(60^o). I think you got it now.

 

[bbbbb]

'^' means square which is I/Io= 0.25.

 

[kuruman]

Yes!

 

[bbbbb]

Is that the final answer.

 

[kuruman]

I said, "Yes" with confetti, didn't I? Yes, it is the final answer.

 

[bbbbb]

oh okay, thank you.