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

  • Thread starter Sloan650
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  • #1
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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
 

Answers and Replies

  • #2
Doc Al
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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.
 
  • #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
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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
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45,017
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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.
 
  • #7
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Find an equation for n2, and then try to understand what this means.
 
  • #8
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Im so confused :(

every time i try to rearrange to find n2 i get math error!
 
  • #9
Doc Al
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every time i try to rearrange to find n2 i get math error!
Show what you did symbolically, step by step.
 
  • #10
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This is due to you doing it on a calculator. Rearrange it on paper, so that you have

n2= ...
 
  • #11
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Do I have to multiply out the brackets?
 
  • #12
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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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