Solving Brewster's Angle: Finding θ_1 from tanθ = n_2/n_1

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unscientific
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Homework Statement



I have solved the first two parts, I'm having trouble with the final part.

Given the following relations, show the following relation of brewster angle:

[tex]tan θ = \frac {n_2}{n_1}[/tex]

bh10rd.png


Homework Equations


The Attempt at a Solution



Using:

[tex]\frac {n_2}{n1} = \frac {sin θ_1}{sin θ_3}[/tex]

Starting from:

[tex]sin 2θ_3 = sin 2θ_1[/tex]
[tex]sin θ_3 cos θ_3 = sin θ_1 cos θ_1[/tex]
[tex]\frac {sin θ_1}{sin θ_3} = \frac {cos θ_3}{cos θ_1}[/tex]
[tex]\frac {n_2}{n_1} = \frac {\sqrt {1 - sin^{2}θ_3}}{cos θ_1}[/tex]
[tex]\frac {n_2}{n_1} = \sqrt { \frac {1}{sin^{2}θ_1} - ( \frac {n_1}{n_2} )^{2} } tan θ_1[/tex]

How do i show the square root term at the bottom = 1?
 
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unscientific said:
Starting from:

[tex]sin 2θ_3 = sin 2θ_1[/tex]

Since ##\theta_1## and ##\theta_3## lie between 0 and ##\frac{\pi}{2}##, ##2\theta_1## and ##2\theta_3## lie between 0 and ##\pi##.

One way to solve ##\sin 2θ_3 = \sin 2θ_1## is to have ##\theta_1 = \theta_3##.

But there is also another relation between ##\theta_1## and ##\theta_3## that will satisfy ##\sin 2θ_3 = \sin 2θ_1##.
 
tsny said:
since ##\theta_1## and ##\theta_3## lie between 0 and ##\frac{\pi}{2}##, ##2\theta_1## and ##2\theta_3## lie between 0 and ##\pi##.

One way to solve ##\sin 2θ_3 = \sin 2θ_1## is to have ##\theta_1 = \theta_3##.

But there is also another relation between ##\theta_1## and ##\theta_3## that will satisfy ##\sin 2θ_3 = \sin 2θ_1##.

[tex]θ_1 = θ_3 + 2\pi[/tex]

Not sure if this helps at all..
 
unscientific said:
[tex]θ_1 = θ_3 + 2\pi[/tex]

Not sure if this helps at all..

We need to keep ##\theta_1## and ##\theta_3## less than ##\frac{\pi}{2}##, so ##2\theta_1## and ##2\theta_3## must lie between 0 and ##\pi##.

Sketch a graph of the sine function between 0 and ##\pi##. Draw a horizontal line that intersects the graph at two different angles. How are the two angles related?
 
TSny said:
We need to keep ##\theta_1## and ##\theta_3## less than ##\frac{\pi}{2}##, so ##2\theta_1## and ##2\theta_3## must lie between 0 and ##\pi##.

Sketch a graph of the sine function between 0 and ##\pi##. Draw a horizontal line that intersects the graph at two different angles. How are the two angles related?

[tex]θ_1 + θ_3 = \frac {\pi}{2}[/tex]
 
TSny said:
OK. Use this in Snell's law.

Ha ha, the answer just pops right out!