Brewsters angle

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Homework Statement



The problem about this is, which Fresnel Equation am I supposed to use?

Show that the brewster's angle is

[tex]\tan(\theta)=\frac{n_{2}}{n_{1}}[/tex]

but which Fresnel equation do you use

Problem: Using the correct Fresnel Equation using (plugging in) the transmitted angle [tex]\theta_{2}[/tex]

Homework Equations



I figured, If I know the correct Fresnel equation, I would be able to just set my other angle equal to 90 and then I would be able to just solve for [tex]\theta[/tex] and then it would yield [tex]\tan[/tex] of [tex]\theta_{2}[/tex] using Snell's Formula

show or rather derive the Brewster's angle


The Attempt at a Solution

 
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Answers and Replies

  • #2
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Can someone just please verify if this is the correct procedure?

so this is what I wrote on my paper

[tex]\frac{\tan(\Theta_{1}-\Theta_{2})}{\tan(\Theta_{1}+\Theta_{2})}[/tex]

[tex]\Theta_{1}+\Theta_{2}=90 Degrees[/tex]

Using Snell's Law then yields [tex]n_{1}\sin(\theta_{1})=n_{2}\sin(\theta_{2})[/tex]

and then that yields plugging in for [tex]\Theta_{2}=90-\Theta_{1}[/tex]

yields the Brewster's Angle formula by setting [tex]R_{p}[/tex] equal to [tex]0[/tex] which is the Reflection Coefficient?

and since [tex]\sin(\theta_{2})=\sin(90-\theta_{1})=\cos(\Theta_{1})[/tex]

Makes it [tex]n_{1}\sin(\Theta_{1})=n_{2}\cos(\Theta_{1})[/tex]

solving for the [tex]\Theta[/tex] value yields

[tex]\frac{\sin(\Theta_{1})}{cos(\Theta_{1})}=\frac{n_{2}}{n_{1}}=\tan(\Theta_{B})[/tex]

Which is just equal to [tex]\tan(\Theta_{B})[/tex]
 
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  • #3
Redbelly98
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Welcome to Physics Forums.

I didn't follow all your details, but your final answer and starting arguments (θ1 + θ2 = 90o, Snell's Law) are correct.
 

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