Getting Brewster's Formula from Snell's Law

[coldjeanz]

From Snell’s Law equation (2) get the Brewster’s angle formula given by equation (4) when θincident+θrefracted=90°.

equation 2: n1 sin θ1 = n2 sinθ2

equation 4: tanθB = n2/n1

This is kind of foreign to me as I haven't studied it yet, not sure how you go from Eq 2 to E4, how does tan come from an equation with 2 sin in it?

 

[RoshanBBQ]

You know that θ1 + θ2 = 90. Try subing in θ2 = 90 - θ1.

 

[coldjeanz]

so then I would eventually end up with:

sinθ1/sin(90-θ1) = n2/n1

still not sure how the left side turns into tanθb though...

 

[ehild]

Remember the definition of sine and cosine of the angles in a right triangle.

ehild

 

[RoshanBBQ]

Since you know the answer, you can take a peek at it and then look back at what you currently have. What must sin(90-θ1) to transform where you are into the answer? The next question is "does sin(90-θ1) equal what it needs to equal?"