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

In summary, the conversation discusses the possibility of ##\sin \theta_t \gt 1## in the complex numbers, and how it relates to the concept of attenuation inside a medium. The conversation also explores the implications of a complex refractive index on the wave vector and the solution to the wave equation. Ultimately, it is concluded that a complex ##\vec{k}## leads to a real-valued exponential die off factor, thus substantiating the possibility of rapid attenuation inside the medium.
  • #1
PhDeezNutz
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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.
 
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  • #2
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.
 
  • #3
Find a relationship between the refractive index and the wave vector ##\vec k## and see what that leads to.
 
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  • #4
vela said:
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?
 
  • #5
Yup!
 
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  • #6
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.
 
Last edited:

Related to Incidence on a conductor at an angle greater than the critical angle

What is the critical angle and why is it important?

The critical angle is the angle at which light traveling through a medium will be completely reflected rather than refracted. This phenomenon is important because it allows for total internal reflection, which has various applications in optics and telecommunications.

What happens to light incidence on a conductor at an angle greater than the critical angle?

When light is incident on a conductor at an angle greater than the critical angle, it will be totally reflected back into the medium rather than being transmitted through it. This is known as total internal reflection and is caused by the difference in refractive indices between the two mediums.

What factors affect the critical angle for a conductor?

The critical angle for a conductor is affected by the refractive indices of the two mediums involved, as well as the wavelength and polarization of the incident light. The angle of incidence also plays a role, as it must be greater than the critical angle for total internal reflection to occur.

Can the critical angle be calculated or measured?

Yes, the critical angle can be calculated using the refractive indices of the two mediums involved, as well as the Snell's Law of refraction. It can also be measured experimentally by varying the angle of incidence and observing the point at which total internal reflection occurs.

What are some practical applications of total internal reflection?

Total internal reflection has various applications in optics and telecommunications. It is used in optical fibers to transmit data over long distances, in prisms for splitting and reflecting light, and in reflective coatings for mirrors and lenses. It also plays a role in some natural phenomena, such as mirages and optical illusions.

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