Brewster Angle through Glass Slab

[samjohnny]

Homework Statement

Homework Equations

$θ_{brewster} = arctan(\frac{n_2}{n_1})$

The Attempt at a Solution

Hi all

I'm having difficulty on this part of a question Now, showing that all light passes through is to say that there be no reflection. This occurs when the angle of incidence on each of the two interfaces is at the required Brewster angle. As in the question, there will be no reflection at the first interface by definition since the beam is incident at the Brewster angle. For there to be no reflection at the second interface, it must be incident at $θ = arctan(\frac{n_{air}}{n_{glass}})$. However, how can I prove that this is the case? Snell's law shall of course come into play.

Many thanks

 

[samjohnny]

Anyone? :)

 

[blue_leaf77]

For part a start by writing the Snell's law for the first interface, from there you have
$$
n_a\sin\theta_1 = n_g\sin\theta_2
$$
with $\theta_1 = \arctan (n_g/n_a)$. Your next step would be to express $\tan\theta_2$ in terms of $n_a$ and $n_g$ only.

 

[samjohnny]

For part a start by writing the Snell's law for the first interface, from there you have
$$
n_a\sin\theta_1 = n_g\sin\theta_2
$$
with $\theta_1 = \arctan (n_g/n_a)$. Your next step would be to express $\tan\theta_2$ in terms of $n_a$ and $n_g$ only.

Ok here I'm using a hunch that $\theta_1+\theta_2=\frac{\pi}{2}$ . Doing so, and applying snells law at the bottom interface gives:

$$
n_g\sin\theta_2 = n_a\sin\theta_1
$$

But $\theta_1=-\theta_2+\frac{\pi}{2}$

Substituting this into the equation by Snells law gives:

$$
n_g\sin\theta_2 = n_a\cos\theta_2
$$

This yields:

$$
\tan\theta_2 = \frac{n_a}{n_g}
$$

Which gives the appropriate Brewster's angle at the second interface. However, I'm not certain at to why my hunch should be true.

 

[blue_leaf77]

Ok here I'm using a hunch that $\theta_1+\theta_2=\frac{\pi}{2}$

That's indeed the situation when the reflected P component vanishes, or in other words when the incident angle equals Brewster angle. To see why that's true, one can start from the reflection coefficient for P polarization. The numerator of this coefficient is $n_g\cos\theta_1 - n_a\cos\theta_2$. Since the reflection for this component vanishes, one has
$$
n_g\cos\theta_1 = n_a\cos\theta_2
$$
Together with Snell's law $n_a\sin\theta_1 = n_g\sin\theta_2$, one can form the following equation
$$
\cos\theta_2\sin\theta_2 - \cos\theta_1\sin\theta_1 = 0
$$
or
$$
\sin 2\theta_2 - \sin 2\theta_1 = 2\cos(\theta_1+\theta_2) \sin(\theta_2-\theta_1) = 0
$$
Given the allowed geometry of the system, the last equality can be satisfied if $\theta_1+\theta_2 = \frac{\pi}{2}$.

 

[samjohnny]

That's indeed the situation when the reflected P component vanishes, or in other words when the incident angle equals Brewster angle. To see why that's true, one can start from the reflection coefficient for P polarization. The numerator of this coefficient is $n_g\cos\theta_1 - n_a\cos\theta_2$. Since the reflection for this component vanishes, one has
$$
n_g\cos\theta_1 = n_a\cos\theta_2
$$
Together with Snell's law $n_a\sin\theta_1 = n_g\sin\theta_2$, one can form the following equation
$$
\cos\theta_2\sin\theta_2 - \cos\theta_1\sin\theta_1 = 0
$$
or
$$
\sin 2\theta_2 - \sin 2\theta_1 = 2\cos(\theta_1+\theta_2) \sin(\theta_2-\theta_1) = 0
$$
Given the allowed geometry of the system, the last equality can be satisfied if $\theta_1+\theta_2 = \frac{\pi}{2}$.

That makes sense; I set the condition for when the reflection coefficient is zero but failed to follow it through to the final condition for the thetas.

As for the second part of the question, by geometry we have that the angle of incidence at the second interface is $x = \theta_2 - \gamma$, where $\theta_2 = arctan(\frac{n_a}{n_g})$ from the previous part. Now essentially we're looking for a $\gamma$ such that $x$ is maximised. However, I'm unsure as to the condition for when there is only reflection and no transmission.

 

[blue_leaf77]

$x = \theta_2 - \gamma$

Check again your calculation, it should be $x = \theta_2 + \gamma$.

However, I'm unsure as to the condition for when there is only reflection and no transmission.

This is the case of the total internal reflection (TIR) which can only take place if the refractive index of the second medium is bigger than that of the first medium. TIR happens when
$$
\sin x = \frac{n_a}{n_g}
$$
From here you can find $\gamma$. By the way, note that by comparing the above form of $\sin x$ and $\tan\theta_2 = n_a/n_g$, you should see that $x > \theta_2$ asserting that the sign in front of $\gamma$ in the equation in the first line in this post should be plus.

 

[samjohnny]

Check again your calculation, it should be $x = \theta_2 + \gamma$.

This is the case of the total internal reflection (TIR) which can only take place if the refractive index of the second medium is bigger than that of the first medium. TIR happens when
$$
\sin x = \frac{n_a}{n_g}
$$
From here you can find $\gamma$. By the way, note that by comparing the above form of $\sin x$ and $\tan\theta_2 = n_a/n_g$, you should see that $x > \theta_2$ asserting that the sign in front of $\gamma$ in the equation in the first line in this post should be plus.

Ah yes, total internal reflection. Thank you. Also, this seems trivial but I'm now having trouble obtaining the angle $x$. I've attached my working.

 

[blue_leaf77]

You calculation is correct, but it depends on which incident angle you want to use in your analysis. I prefer to use the angle between the incoming ray and the normal line of the interface. If I define this angle to be $x$, then $x=\gamma+\theta_2$. Your $x$ in that calculation is the complementary angle of my $x$.

 

[samjohnny]

You calculation is correct, but it depends on which incident angle you want to use in your analysis. I prefer to use the angle between the incoming ray and the normal line of the interface. If I define this angle to be $x$, then $x=\gamma+\theta_2$. You $x$ in that calculation is the complementary angle of my $x$.

Conventions always throw me! Thanks a lot for the help.