Problem Set 7

(9.1)

Optics (as well as most of physics) can be derived from a global law as well as a local one, in this case Fermat’s Principle: a light ray chooses the path between two points that minimizes the time to travel between them. Apply this to two points on either side of a dielectric interface to derive Snell’s Law.

Let there be a dialectic interface along the y axis, between media with indices of refraction $n_1$ (left) and $n_2$ (right). The time it takes for light to go from a point $(x_1, y_1)$ on the left to $(x_2, y_2)$ on the right, passing through the interface at height $y$ is

$t = \frac{n_1}{c} \sqrt{x_1^2 + (y_1 - y)^2} + \frac{n_1}{c} \sqrt{x_2^2 + (y_2 - y)^2}$

The derivative of this with respect to $y$ is

$\begin{align*} \frac{dt}{dy} &= \frac{-n_1 (y_1 - y)}{c \sqrt{x_1^2 + (y_1 - y)^2}} + \frac{-n_2 (y_2 - y)}{c \sqrt{x_2^2 + (y_2 - y)^2}} \\ &= -n_1 \sin \theta_1 + n_2 \sin \theta_2 \end{align*}$

since

$\sin \theta_1 = \frac{y_1 - y}{\sqrt{x_1^2 + (y_1 - y)^2}}$

and

$\sin \theta_2 = \frac{y - y_2}{\sqrt{x_2^2 + (y_2 - y)^2}}$

(by SOHCAHTOA). Hence the time of flight is minimized when

$\frac{n_1}{n_2} = \frac{\sin \theta_2}{\sin \theta_1}$

(9.2)

(a)

Use Fresnel’s equations and the Poynting vectors to find the reflectivity and transmissivity of a dielectric interface, defined by the ratios of incoming and outgoing energy.

First note that

$\begin{align*} \sqrt{\frac{\epsilon}{\mu}} &= \sqrt{\frac{\epsilon_0 \epsilon_r}{\mu_0 \mu_r}} \\ &= \sqrt{\frac{\epsilon_0}{\mu_0}} \frac{n}{\mu_r} \\ &\approx \sqrt{\frac{\epsilon_0}{\mu_0}} n \end{align*}$

since most dielectric materials have $\mu_r \approx 1$ .

So for a plane wave in a homogeneous medium,

$\begin{align*} P &= E \times H \\ &= E \times \left( \sqrt{\frac{\epsilon}{\mu}} \hat{k} \times E \right) \end{align*}$

and

$\begin{align*} \langle | P | \rangle &= \frac{1}{2} \sqrt{\frac{\epsilon}{\mu}} E_\text{max}^2 \\ &\approx \frac{n}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2 \end{align*}$

There are two cases to consider. First, when the electric field is perpendicular to the plane of incidence, the reflected and refracted electric fields will be (respectively)

$\begin{align*} E_1 &= \frac{\sin(\theta_2 - \theta_0)}{\sin(\theta_2 + \theta_0)} E_0 \\ E_2 &= \frac{2 \cos \theta_0 \sin \theta_2}{\sin(\theta_0 + \theta_2)} E_0 \end{align*}$

So the reflectivity is

$\begin{align*} R &= \frac{E_1^2}{E_0^2} \\ &= \frac{\sin^2(\theta_2 - \theta_0)}{\sin^2(\theta_2 + \theta_0)} \end{align*}$

and the transmissivity

$\begin{align*} T &= \frac{E_2^2}{E_0^2} \\ &= \frac{n_2}{n_1} \frac{4 \cos^2 \theta_0 \sin^2 \theta_2}{\sin^2(\theta_0 + \theta_2)} \end{align*}$

When the electric field is in the plane of incidence,

$\begin{align*} E_1 &= \frac{\tan(\theta_0 - \theta_2)}{\tan(\theta_0 + \theta_2)} E_0 \\ E_2 &= \frac{2 \cos \theta_0 \sin \theta_2}{\sin(\theta_0 + \theta_2) \cos(\theta_0 - \theta_2)} E_0 \end{align*}$

$\begin{align*} R &= \frac{\tan^2(\theta_0 - \theta_2)}{\tan^2(\theta_0 + \theta_2)} \\ T &= \frac{n_2}{n_1} \frac{4 \cos^2 \theta_0 \sin^2 \theta_2}{\sin^2(\theta_0 + \theta_2) \cos^2(\theta_0 - \theta_2)} \end{align*}$

When the electric field has components out of and in the plane of incidence, the total reflectivity and transmissivity can be found via adding the values above in quadrature.

(b)

For a glass–air interface (n = 1.5) what is the reflectivity at normal incidence?

At normal incidence, there is no unique plane of incidence. So by symmetry it shouldn’t matter which reflection coefficients we use. Let’s choose the perpendicular case. We would like to evaluate

$R = \frac{\sin^2(\theta_2 - \theta_0)}{\sin^2(\theta_2 + \theta_0)}$

for $\theta = \theta_2 = 0$ . But this formula diverges, so instead we will compute its limit as $\theta_0$ and $\theta_2$ approach zero.

As they approach zero we must still respect Snell’s Law: $n_1 / n_2 = \sin \theta_2 / \sin \theta_0$ . Since we are interested in the limit, the small angle approximation is in this case not an approximation at all. So we can write this as

$\frac{n_1}{n_2} = \frac{\theta_2}{\theta_0}$

$\theta_2 = \frac{\theta_0 n_1}{n_2}$

Thus the reflectivity is

$\begin{align*} R &= \lim_{\theta_0 \to 0} \frac{\sin^2 \left( \theta_0 \left( \frac{n_1}{n_2} - 1 \right) \right)} {\sin^2 \left( \theta_0 \left( \frac{n_1}{n_2} + 1 \right) \right)} \\ &= \lim_{\theta_0 \to 0} \frac{\theta_0^2 \left( \frac{n_1}{n_2} - 1 \right)^2} {\theta_0^2 \left( \frac{n_1}{n_2} + 1 \right)^2} \\ &= \frac{\left( \frac{n_1}{n_2} - 1 \right)^2}{\left( \frac{n_1}{n_2} + 1 \right)^2} \\ &= \frac{\left( n_1 - n_2 \right)^2}{\left( n_1 + n_2 \right)^2} \end{align*}$

For an air to glass interface,

$\begin{align*} R &= \frac{\left( 1 - 1.5 \right)^2}{\left( 1 + 1.5 \right)^2} \\ &= 0.04 \end{align*}$

(c)

What is the Brewster angle?

$\theta_B = \tan^{-1} \left( \frac{n_2}{n_1} \right)$

So for air to glass,

$\theta_B = \tan^{-1} \left( \frac{1.5}{1} \right) = 0.98 \si{rad} = 56.3^\circ$

and for glass to air,

$\theta_B = \tan^{-1} \left( \frac{1}{1.5} \right) = 0.58 \si{rad} = 33.7^\circ$

(d)

What is the critical angle?

$\begin{align*} \theta_C &= \sin^{-1} \left( \frac{n_2}{n_1} \right) \\ &= \sin^{-1} \left( \frac{1}{1.5} \right) \\ &= 0.73 \si{rad} \\ &= 41.8^\circ \end{align*}$

for glass to air. There isn’t a critical angle for air to glass, since it requires $n_2 \leq n_1$ .

(9.3)

Consider a wave at normal incidence to a dielectric layer with index $n_2$ and thickness $d$ between layers with indices $n_1$ and $n_3$ .

(a)

What is the reflectivity? Think about matching the boundary conditions, or about the multiple reflections.

Using Snell’s Law and the small angle “approximation” as in the previous problem, we can find the the coefficients of reflection and transmission (this time for field strength instead of power, for reasons that will become clear). For this problem it’s important to keep track of the direction in which light crosses (or reflects from) the dielectric barriers, so I’ll use subscripts throughout.

$\begin{align*} R_{12} &= \lim_{\theta_0 \to 0} \frac{\sin \left( \theta_0 \left( \frac{n_1}{n_2} - 1 \right) \right)} {\sin \left( \theta_0 \left( \frac{n_1}{n_2} + 1 \right) \right)} \\ &= \lim_{\theta_0 \to 0} \frac{\theta_0 \left( \frac{n_1}{n_2} - 1 \right)} {\theta_0 \left( \frac{n_1}{n_2} + 1 \right)} \\ &= \frac{\left( \frac{n_1}{n_2} - 1 \right)}{\left( \frac{n_1}{n_2} + 1 \right)} \\ &= \frac{n_1 - n_2}{n_1 + n_2} \\ T_{12} &= \lim_{\theta_0 \to 0} \frac{2 \sin \left( \theta_0 \frac{n_1}{n_2} \right) \cos \theta_0} {\sin \left( \theta_0 \left( \frac{n_1}{n_2} + 1 \right) \right)} \\ &= \lim_{\theta_0 \to 0} \frac{2 \theta_0 \frac{n_1}{n_2}} {\theta_0 \left( \frac{n_1}{n_2} + 1 \right)} \\ &= \frac{2 \frac{n_1}{n_2}}{\left( \frac{n_1}{n_2} + 1 \right)} \\ &= \frac{2 n_1}{n_1 + n_2} \end{align*}$

Note that $R_{12} = -R_{21}$ . We’ll use this later.

Now the reflections can be modeled as an infinite sum, taking into account the phase shift that results from travel through the middle medium.

$\begin{align*} E^- &= E^+ R_{12} + E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d} + E^+ T_{12} R_{23}^2 T_{21} R_{21} e^{4 i k_2 d} \\ &= E^+ R_{12} + E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d} \sum_{n = 0}^\infty \left( R_{21} R_{23} e^{2 i k_2 d} \right)^n \\ &= E^+ R_{12} + \frac{E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d}} {1 - R_{21} R_{23} e^{2 i k_2 d}} \\ &= \frac{E^+ R_{12} \left( 1 - R_{21} R_{23} e^{2 i k_2 d} \right) + E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d}} {1 - R_{21} R_{23} e^{2 i k_2 d}} \\ &= E^+ \frac{R_{12} \left( 1 + R_{12} R_{23} e^{2 i k_2 d} \right) + T_{12} R_{23} T_{21} e^{2 i k_2 d}} {1 + R_{12} R_{23} e^{2 i k_2 d}} \\ &= E^+ \frac{R_{12} + \left( R_{12}^2 + T_{12} T_{21} \right) R_{23} e^{2 i k_2 d}} {1 + R_{12} R_{23} e^{2 i k_2 d}} \\ &= E^+ \frac{R_{12} + R_{23} e^{2 i k_2 d}} {1 + R_{12} R_{23} e^{2 i k_2 d}} \end{align*}$

The last line follows since

$\begin{align*} R_{12}^2 + T_{12} T_{21} &= \frac{(n_1 - n_2)^2}{(n_1 + n_2)^2} + \frac{2 n_1}{n_1 + n_2} \frac{2 n_2}{n_1 + n_2} \\ &= \frac{(n_1 - n_2)^2 + 4 n_1 n_2}{(n_1 + n_2)^2} \\ &= \frac{n_1^2 - 2 n_1 n_2 + n_2^2 + 4 n_1 n_2}{(n_1 + n_2)^2} \\ &= \frac{(n_1 + n_2)^2}{(n_1 + n_2)^2} \\ &= 1 \end{align*}$

So the total reflectivity in terms of power is

$\begin{align*} R &= \frac{(E^-)^2}{(E^+)^2} \\ &= \frac{(R_{12} + R_{23} e^{2 i k_2 d})^2}{(1 + R_{12} R_{23} e^{2 i k_2 d})^2} \end{align*}$

(b)

Can you find values for $n_2$ and $d$ such that the reflection vanishes?

To make $R = 0$ we need $R_{12} = -R_{23} e^{2 i k_2 d}$ . Since $R_{12}$ is a positive real number, the only way this can work is if $e^{2 i k_2 d} = -1$ and $R_{12} = R_{23}$ . (Consider that $e^{2 i k_2 d}$ lies on the complex unit circle. So to be real it has to be 1 or -1, and 1 would leave us with a negative number.) Thus $2 k_2 d = \pi$ , or

$d = \frac{\pi}{2 k_2} = \frac{\lambda}{4}$

since $k_2 = 2 \pi / \lambda$ . And $R_{12} = R_{23}$ implies

$\begin{align*} \frac{n_1 - n_2}{n_1 + n_2} &= \frac{n_2 - n_3}{n_2 + n_3} \\ (n_1 - n_2) (n_2 + n_3) &= (n_1 + n_2) (n_2 - n_3) \\ n_1 n_2 + n_1 n_3 - n_2^2 - n_2 n_3 &= n_1 n_2 - n_1 n_3 + n_2^2 - n_2 n_3 \\ n_2^2 &= n_1 n_3 \\ n_2 &= \sqrt{n_1 n_3} \\ \end{align*}$

(9.4)

Consider a ray starting with a height $r_0$ and some slope, a distance $d_1$ away from a thin lens with focal length $f$ . Use ray matrices to find the image plane where all rays starting at this point rejoin, and discuss the magnification of the height $r_0$ .

$\begin{align*} M &= \begin{bmatrix} 1 & d_2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1/f & 1 \end{bmatrix} \begin{bmatrix} 1 & d_1 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 & d_2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & d_1 \\ -1/f & -d_1/f + 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 - d_2/f & d_1 + d_2 - d_1 d_2/f \\ -1/f & 1 - d_1/f \end{bmatrix} \end{align*}$

$M \begin{bmatrix} r_0 \\ s_0 \end{bmatrix} = \begin{bmatrix} r_0 (1 - d_2/f) + s_0 (d_1 + d_2 - d_1 d_2/f) \\ -r_0/f + s_0 (1 - d_1/f) \end{bmatrix}$

The position (i.e. 0th element) doesn’t depend on $s_0$ if

$d_1 + d_2 - \frac{d_1 d_2}{f} = 0$

i.e.

$\frac{1}{d_1} + \frac{1}{d_2} = \frac{1}{f}$

Note that this doesn’t depend on $r_0$ , so there really is a focal plane.

At the $d_2$ satisfying the relation above, each light ray’s position is $r_0 (1 - d_2/f)$ . So the whole plane is stretched by a factor of $1 - d_2 / f$ .

(9.5)

Common CD players use an AlGaAs laser with a 790 nm wavelength.

(a)

The pits that are read on a CD have a diameter of roughly $1 \si{\mu m}$ and the optics are diffraction-limited; what is the beam divergence angle?

(b)

Assuming the same geometry, what wavelength laser would be needed to read $0.1 \si{\mu m}$ pits?

(c)

How large must a telescope mirror be if it is to be able to read a car’s license plate in visible light ( $\lambda \approx 600 \si{nm}$ ) from a Low Earth Orbit (LEO) of 200 km?