- #1

- 12

- 0

Im using Sin ic = n1/n2

For Sin ic i get =1.706

But the reverse Sin of that comes up math error?

HELP

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- Thread starter Sloan650
- Start date

- #1

- 12

- 0

Im using Sin ic = n1/n2

For Sin ic i get =1.706

But the reverse Sin of that comes up math error?

HELP

- #2

Doc Al

Mentor

- 45,254

- 1,614

You're mixing up n1 and n2. Start from Snell's law and derive the expression for total internal reflection.Im using Sin ic = n1/n2

- #3

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Thank you!

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

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

I need to find n2

- #4

- 59

- 0

Firstly you gave this equation:

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

Am I correct to think this is equivalent?

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

(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 [itex] n_{2} [/itex] term.

- #5

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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?

- #6

Doc Al

Mentor

- 45,254

- 1,614

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

Start by squaring both sides and then go from there.How does it rearrange so i get a positive number to square root to find n2?

- #7

- 59

- 0

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

- #8

- 12

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Im so confused :(

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

- #9

Doc Al

Mentor

- 45,254

- 1,614

Show what you did symbolically, step by step.every time i try to rearrange to find n2 i get math error!

- #10

- 59

- 0

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

n2= ...

- #11

- 12

- 0

Do I have to multiply out the brackets?

- #12

- 59

- 0

The term [itex] \sqrt{n_{1}^{2}-n_{2}^{2}}=\sqrt{(n_{1}^{2}-n_{2}^{2})} [/itex]

So by squaring you will get

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

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

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