Details of total internal reflection

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
Messages
2,802
Reaction score
605
Consider snell's law [itex]n_1 \sin{\theta_1}=n_2 \sin{\theta_2}[/itex]([itex]n_1[/itex] and [itex]n_2[/itex] are real).
We know that if [itex]n_2<n_1[/itex], there exists an incident angle called critical angle that gives a refraction angle of ninety degrees i.e. [itex]\sin{\theta_c}=\frac{n_2}{n_1}[/itex].

But if the incident angle is greater than the critical angle(i.e. [itex]\sin{\theta_1}>\frac{n_2}{n_1}[/itex]),Then:[itex]\sin{\theta_2}=\frac{n_1}{n_2}\sin{\theta_1}>1[/itex]

But we know that [itex]\sin{\theta}>1[/itex] can happen for no real [itex]\theta[/itex],so we say that [itex]\theta_2[/itex] should be complex:
[itex]\theta_2=\alpha+i \beta[/itex] and [itex]\sin{\theta_2}=\sin{(\alpha+i \beta)}=\sin{(\alpha)}\cos{(i \beta)}+\sin{(i \beta)}\cos{(\alpha)}=\sin{(\alpha)}\cosh{(\beta)}+i\cos{(\alpha)}\sinh{(\beta)}[/itex]

But from snell'w law,we know that [itex]\sin{\theta_2}[/itex] should be real and so we should always have [itex]cos{\alpha}=0 \Rightarrow \alpha=\frac{\pi}{2}[/itex] and so [itex]\sin{\theta_2}=\cosh{\beta}[/itex].

This means that the only variable which is capable of giving information about the Total reflected ray,is [itex]\beta[/itex]. But how?

Thanks
 
Physics news on Phys.org
Shyan said:
But from snell'w law,we know that [itex]\sin{\theta_2}[/itex] should be real
You cannot use a law in a parameter range where it does not apply.

This means that the only variable which is capable of giving information about the Total reflected ray,is [itex]\beta[/itex]. But how?
What else do you need? The angle is just the same as the incident angle.
 
No reason to use complex angles here. You simply defined [itex]\beta[/itex] such that [tex]\cosh \beta = \frac{n_1}{n_2}\sin \theta_1[/tex]
 
Now, what you do get from the treatment of this problem with complex angles is an expression for the evanescent waves [itex]F = A e^{i\vec k_2\cdot \vec x},[/itex] where [itex]\vec k_2 = cos\theta_2\hat i + sin\theta_2\hat j[/itex], and [itex]\vec x = x\hat i + y\hat j[/itex]. Now if you plug in your complex parametrization [itex]\theta_2 = \frac{\pi}{2} + \beta[/itex], than you get [tex]F = A exp [iy\,cosh\beta - x sinh\beta][/tex]