# Brewsters angle

## Homework Statement

Show that the brewster's angle is

$$\tan(\theta)=\frac{n_{2}}{n_{1}}$$

but which Fresnel equation do you use

Problem: Using the correct Fresnel Equation using (plugging in) the transmitted angle $$\theta_{2}$$

## 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 $$\theta$$ and then it would yield $$\tan$$ of $$\theta_{2}$$ using Snell's Formula

show or rather derive the Brewster's angle

## The Attempt at a Solution

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

so this is what I wrote on my paper

$$\frac{\tan(\Theta_{1}-\Theta_{2})}{\tan(\Theta_{1}+\Theta_{2})}$$

$$\Theta_{1}+\Theta_{2}=90 Degrees$$

Using Snell's Law then yields $$n_{1}\sin(\theta_{1})=n_{2}\sin(\theta_{2})$$

and then that yields plugging in for $$\Theta_{2}=90-\Theta_{1}$$

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

and since $$\sin(\theta_{2})=\sin(90-\theta_{1})=\cos(\Theta_{1})$$

Makes it $$n_{1}\sin(\Theta_{1})=n_{2}\cos(\Theta_{1})$$

solving for the $$\Theta$$ value yields

$$\frac{\sin(\Theta_{1})}{cos(\Theta_{1})}=\frac{n_{2}}{n_{1}}=\tan(\Theta_{B})$$

Which is just equal to $$\tan(\Theta_{B})$$

Last edited:
Redbelly98
Staff Emeritus