For n1 = 2.15 and n2 = 1.26, what is the critical angle so that all of

[Sloan650]

For n1 = 2.15 and n2 = 1.26, what is the critical angle so that all of the incident light, from medium 1 to medium 2 is reflected?



Im using Sin ic = n1/n2



For Sin ic i get =1.706

But the reverse Sin of that comes up math error?

HELP

 

[Doc Al]

Im using Sin ic = n1/n2

You're mixing up n1 and n2. Start from Snell's law and derive the expression for total internal reflection.

 

[Sloan650]

Thank you!

I need help rearranging the equation too, I am finding it impossible!

Nmax = 4a/lamda x sqaureroot (N1^2 - N2^2)

I need to find n2

 

[gash789]

Firstly you gave this equation:

Nmax = 4a/lamda x sqaureroot (N1^2 - N2^2)

Am I correct to think this is equivalent?

N_{max}=\frac{4a}{λ} \sqrt{n_{1}^{2}-n_{2}^{2}}

(I would suggest in future as opposed to using "x" to indicate multiplies, I would use "*". Simply for clarity)

If this is the case begin by squaring both sides of the equation, and then attempt to isolate the n_{2} term.

 

[Sloan650]

I have no idea how to rearrange this.

i know by squaring both sides the sqaure root will disappear. But on the other side Nmax = 1.

How does it rearrange so i get a positive number to square root to find n2?

 

[Doc Al]

i know by squaring both sides the sqaure root will disappear. But on the other side Nmax = 1.

So?

How does it rearrange so i get a positive number to square root to find n2?

Start by squaring both sides and then go from there.

 

[gash789]

Find an equation for n2, and then try to understand what this means.

 

[Sloan650]

Im so confused :(

every time i try to rearrange to find n2 i get math error!

 

[Doc Al]

every time i try to rearrange to find n2 i get math error!

Show what you did symbolically, step by step.

 

[gash789]

This is due to you doing it on a calculator. Rearrange it on paper, so that you have

n2= ...

 

[Sloan650]

Do I have to multiply out the brackets?

 

[gash789]

The term \sqrt{n_{1}^{2}-n_{2}^{2}}=\sqrt{(n_{1}^{2}-n_{2}^{2})}

So by squaring you will get

N_{max}^{2}=(\frac{4a}{λ})^{2} (n_{1}^{2}-n_{2}^{2})

It would be more convenient if you multiplied both sides of (\frac{λ}{4a})^{2} rather than multiplying out the brackets.