# Critical angle = brewster angle?

1. Jun 3, 2009

### fredrick08

1. The problem statement, all variables and given/known data
when the critical angle=brewster angle, what is the refractive index, if incident n is air=1.

arcsin(1/n)=arctan(n)=>n=1.272....????

how is this? i tried differentiating both side and solving but i just get into a big mess, can anyone help me if i am missing something?

Last edited: Jun 3, 2009
2. Jun 4, 2009

### fredrick08

ive tried pythagoras, and trig idendenties, but all i come up with is x=x??

3. Jun 4, 2009

### Cyosis

If you want to use identities you can take the sin on both sides and then use $\cos^2x+\sin^2x=1,\;1+\tan^2x=\sec^2x$ to simplify the right hand side.

4. Jun 4, 2009

### fredrick08

ok thanks i think i got it now... just confused about the algebra

5. Jun 4, 2009

### fredrick08

all solved, thanks very much

6. Feb 27, 2011

### larrykuhl

I'm still at a loss as to how n = 1.272.

so getting to:
1/n = sin(arctan(n))

then squaring and adding cos^2 gets:

1/n^2 - cos^2(arctan(n)) = 1

but any way I rearrange the above formula to get 1/cos^2 for sec^2 ends up resulting in n=1?