1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Details of total internal reflection

  1. Nov 11, 2013 #1

    ShayanJ

    User Avatar
    Gold Member

    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
     
  2. jcsd
  3. Nov 11, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You cannot use a law in a parameter range where it does not apply.

    What else do you need? The angle is just the same as the incident angle.
     
  4. Nov 11, 2013 #3
    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]
     
  5. Nov 11, 2013 #4
    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Details of total internal reflection
Loading...