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physics fundamental: modern physics lesson 1 Chapter 24 practice quiz #8&9&10

 
 
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physics fundamental: modern physics lesson 1 Chapter 24 practice quiz #8&9&10
by Hongyu Shi - Thursday, 25 February 2021, 8:17 PM
 

I have no idea how to solve question #8, 9, 10

 
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Re: physics fundamental: modern physics lesson 1 Chapter 24 practice quiz #8&9&10
by John Lensmire - Friday, 26 February 2021, 12:49 PM
 

Hi Hongyu. I'll reply with a few hints and ideas later tonight. Thanks for your patience!

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Re: physics fundamental: modern physics lesson 1 Chapter 24 practice quiz #8&9&10
by John Lensmire - Saturday, 27 February 2021, 3:32 PM
 

Here's some hints.

For question 8, the angle is actually not needed. Because the light is initially unpolarized (so the waves could have all different angles), the intensity of the light is always decreased by a factor of $2$.

The angle in question 8, however, is relevant to question 9. Recall when the polarization changes by an angle of $\theta$, the intensity is given by the equation $I = I_0\cdot \cos^2(\theta)$.

Overall, and as a hint for $10$, let's see what happens with a simple example: What happens when initially unpolarized light with an intensity of $400$ passes through polarizers with angles $0^\circ$, $30^\circ$, and $90^\circ$.

  1. Since the light starts unpolarized, passing through the first polarizer gives a new intensity of $400\cdot \dfrac{1}{2} = 200$.
  2. The change in angle from the first to second polarizer is $30^\circ$, so after the second polarizer the intensity is $200\cdot \cos^2(30^\circ) = 200\cdot \dfrac{3}{4} = 150$.
  3. The change in angle from the second to third polarizer is $60^\circ$, so the final intensity is $150\cdot \cos^2(60^\circ) = 150\cdot \dfrac{1}{4} = 37.5$.