richyw
  • richyw
Total internal reflection question.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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richyw
  • richyw
If was assume in the case of total internal reflection that there is no transmitted wave, we can reformulate\[r_{\perp} = \frac{n_i\cos{\theta_i}-n_t\cos{\theta_t}}{n_i\cos{\theta_i}+n_t\cos{\theta_t}}\]and\[r_{\parallel}=\frac{n_t\cos{\theta_i}-n_i\cos{\theta_t}}{n_i\cos{\theta_t}+n_t\cos{\theta_i}}\] such that\[r_{\perp}=\frac{\cos{\theta_i}-(n_{ti}^2-\sin^2{\theta_i})^{1/2}}{\cos{\theta_i}+(n_{ti}^2-\sin^2{\theta_i})^{1/2}}\]\[r_{\parallel}=\frac{n_{ti}\cos{\theta_i}-(n_{ti}^2-\sin^2{\theta_i})^2}{n_{ti}\cos{\theta_i}+(n_{ti}^{1/2}-\sin^2{\theta_i})^{1/2}}\]
richyw
  • richyw
oh and i'm pretty sure that the notation they used is \[n_{ti}=\frac{n_t}{n_1}\]
richyw
  • richyw
my attempt: \[r_{\perp} = \frac{n_i\cos{\theta_i}-n_t\cos{\theta_t}}{n_i\cos{\theta_i}+n_t\cos{\theta_t}}\]\[r_{\perp} = \frac{\cos{\theta_i}-n_{ti}\cos{\theta_t}}{\cos{\theta_i}+n_{ti}\cos{\theta_t}}\]\[r_{\perp} = \frac{\cos{\theta_i}-n_{ti}\sqrt{1-\sin^2{\theta_t}}}{\cos{\theta_i}+n_{ti}\sqrt{1-\sin^2{\theta_t}}}\]then from snell's law I get\[\sin{\theta_t}=n_{ti}\sin{\theta_i}\]\[\sin^2{\theta_t}=n_{ti}^2\sin^2{\theta_i}\]

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richyw
  • richyw
so then I get \[r\perp=\frac{\cos{\theta_i}-n_{ti}\sqrt{1-n_{ti}^2\sin^2{\theta_i}}}{\cos{\theta_i}-n_{ti}\sqrt{1-n_{ti}^2\sin^2{\theta_i}}}\]
richyw
  • richyw
which isn't what I wanted?
richyw
  • richyw
ah, I see now that I messed up snell's law. Nevermind!

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