# Incidence on a conductor at an angle greater than the critical angle

• PhDeezNutz

#### PhDeezNutz

Homework Statement
This is more so a conceptual question than a homework question, from my instructors notes;

(I'm assuming that the first medium is air, even though my instructor does not explicitly say so, I figure I can address this case first and then generalize)

(The subscript

"If for example the medium on which the radiation is incident is a conductor then the refractive index ##\eta_t## will be complex.

In the case that the angle of incidence is greater than the critical angle (##\theta_c##), then

$$\frac{ \sin \theta_i}{\sin \theta_t} = \frac{\eta_t}{\eta_i} = \sin \theta_{c} \Rightarrow \sin \theta_t = \frac{\sin \theta_i}{\sin \theta_c} \gt 1$$

and then the sine and cosine of the transmission angle must be complex. This serves to indicate that the radiation is rapidly attenuated inside the medium.
Relevant Equations
Snell's Law

$$\eta_i \sin \theta_i = \eta_t \sin \theta_t$$

Incident Angle

$$\theta_i = \sin^{-1} \left( \frac{\eta_t}{\eta_i} \sin \theta_t \right)$$

Transmitted Angle

$$\theta_t = \sin^{-1} \left( \frac{\eta_i}{\eta_t} \sin \theta_i \right)$$

Critical Angle (corresponds to ##\theta_t = \frac{\pi}{2}##)

$$\theta_{ic} = \sin^{-1} \left( \frac{\eta_t}{\eta_i}\right)$$
I for one don't see how ##\sin \theta_t \gt 1## is possible, even when you extend into the complex numbers. Is there even a way to order the complex numbers? Does he mean to say that the magnitude is greater than 1?

Anyway here's my attempt at interpreting what my instructor is trying to say. Forgive the egregious handwaving and heuristic arguments, I'm not well versed in complex analysis.

If ##\theta_i \gt \theta_{ic}## then

$$\sin^{-1} \left( \frac{\eta_t}{\eta_i} \sin \theta_t \right) \gt \sin^{-1} \left( \frac{\eta_t}{\eta_i}\right)$$

##\sin^{-1}## is strictly increasing over one period so it follows that if ##\sin^{-1} x \gt \sin^{-1} y## then ## x \gt y##

So we have

$$\left( \frac{\eta_t}{\eta_i} \sin \theta_t \right) \gt \left(\frac{\eta_t}{\eta_i} \right)$$

Multiply through by ##\frac{\eta_i}{\eta_t}## and we get

$$\sin \theta_t \gt 1$$

But how can ## \sin \theta_t## be greater than ##1## (even if we extend into the complex plane), and how does that imply ##\theta_t## must be complex and how does that in turn imply "rapid attenuation inside the medium"?

I'm at an impasse because I don't know enough complex analysis to resolve this in my opinion, nor do I know enough about attenuation to see the connection.

As always any helpful hints and insight is greatly appreciated.

Alright I think I figured out why ##\left| \sin \theta_t \right| \gt 1 \Rightarrow \theta_t \in \mathbb{C}## But I haven't figured out how attenuation is implied.

Equivalently

$$\left| \sin \theta_t \right| \gt 1 \Rightarrow \left| \sin \theta_t \right|^2 \gt 1$$

Here we go

We know that if ##\theta_t \in \mathbb{R}## that ##\sin \theta_t## cannot be greater than 1 so we assume that ##\theta_t = c + id## and show that this allows for ##\sin \theta_t## to greater than 1.

$$\sin \theta_t = \sin \left(c + id \right) = \sin(c)\cos(id) + \cos(c)\sin(id)$$
$$= \sin(c)\cosh(d) + i \cos(c) \sinh(d)$$

find the square of the magnitude of ##\sin \theta_t##

$$\left| \sin \theta_t \right|^2 = \left[\sin \theta_t \right] \left[\sin \theta_t \right]^{*}$$
$$= \left[ \sin(c)\cosh(d) + i \cos(c) \sinh(d)\right] \left[ \sin(c)\cosh(d) - i \cos(c) \sinh(d)\right]$$
$$= \sin^2 (c) \cosh^2 (d) + \cos^2 (c) \sinh^2 (d)$$

And since ##\cosh## and ##\sinh## don't have restricted domains they are free to be as big as they want to be (I'm a bit iffy on this one) and therefore

$$\theta_t \in \mathbb{C} \Rightarrow \left| \sin\theta_t \right|^2 \gt 1 \Rightarrow \left|\sin \theta_t\right| \gt 1$$

##\theta_t \in \mathbb{C}## at least allows for ##\left| \sin \theta_t \right| \gt 1##

However, I don't see where to go from here in terms of how ##\theta_t## being complex implies attenuation.

Find a relationship between the refractive index and the wave vector ##\vec k## and see what that leads to.

• PhDeezNutz
Find a relationship between the refractive index and the wave vector ##\vec k## and see what that leads to.

Loosely speaking I think I understand

We have ##k_t = \frac{\eta_t \omega}{c}##, IF ##\eta_t## is complex then so is ##k_t## and then so is ##\vec{k_t}##

Plugging this complex ##\vec{k}## into the solution for the wave equation (directly beneath)

$$\vec{E_t} = \vec{E_{t,0}} e^{i \left(\vec{k_t} \cdot \vec{x} - \omega t \right)}$$

We see that if ##\vec{k_t} = \frac{\omega}{c} c_1 + i \frac{\omega}{c} d_1 ## is complex then the solution to the wave equation will then have a real valued "exponential die off factor" (i.e an attenuation constant)(in space, not time)

I need a little bit more time to work out the details but am I on the right track?

Yup!

• PhDeezNutz
So we could have substantiated rapid attenuation inside the medium without first substantiating theta_t had to be complex?

My only concern is what if d is negative? Would this correspond to an exponential growth factor?

edit: btw you're awesome!

edit 2: Is the mathjax processing software down? All my equations have disappeared.

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