One-dimensional photonic crystal

NECESSARY ACCOMPANIMENT TO SOLUTION

Well well well

What do we have here?

Three-hundred-points for a little bitta shakespeare?

A bitta schoolin' on ya physical theories

And gettin down with ya geo-metric series

Right son, so we just entered heaven

Light passin' thru, bout naught-point-seven

Gettin' to the edge, but ya know whats coming back

Round about three-tenths on the same track

The story goes on, as ya might expect

And the same deal happens as the light reflects

You know i'm sayin', and its easy to see

Light coming westside as a G-P

So just sit back as i spit the automatic

Nothin' here but ya old-school mathematics.

$(0.7)^2[1 + (0.3)^2 + (0.3)^4 + (0.3)^6 +(0.3)^8 + ...]$

$(0.7)^2[1 + \frac{(0.3)^2}{1 - (0.3)^2}]$

$(0.49)(\frac{100}{91})$

$= 0.538$

$= 53.8\%$

If the light exits the crystal on the left, it must be transmitted through an interface twice, once when the light is initially shined into the crystal and once when the light exits the crystal on the left. It also must be reflected at an interface an even number of times, because if it were reflected an odd number of times it would be exiting on the right.

If the initial amount of light is $1$ , then after $n$ transmissions and $m$ reflections there is $t^n(1-t)^m$ amount of light left, because the amount of light transmitted at an interface is $t$ times the amount of light before the transmission, and the amount of light reflected at an interface is $1-t$ times the amount of light before the reflection.

We want to know the total amount of light that exits through the left, so we will take a sum over all the amounts of light after $n=2$ transmissions and an even number $m$ of reflections.

$$\sum_{m=0}^{\infty} {t^2(1-t)^{2m}}$$

Notice that this is an infinite geometric series with first term $a_0=t^2$ and common ratio $r=(1-t)^2$ , where $|r|<1$ , so its sum is given by the formula for the infinite geometric series:

$$\frac{a_0}{1-r}=\frac{t^2}{1-(1-t)^2}.$$

Substituting in $t=0.7$ gives

$$\frac{0.7^2}{1-(1-0.7)^2}=\frac{7}{13} \approx .5385=53.85 \%.$$

Since the light rays shine through the crystal perpendicularly, the angle of refraction is 0 degree.

Transmission coefficient is given by $t=\frac{I}{I_0}$ , where $I$ denotes the intensity of light rays before passing through a medium, $I_0$ denotes the intensity of light rays actually going into the medium.

So each time the light shines onto the crystal, or the light rays are coming from the crystal into the air, 70% of it passes through where as 30% of the light rays are perpendicularly reflected.

Hence, when the light shines onto the crystal for the first time, $(0.7)^2$ of light actually passes through the crystal and enters the air from the other side. Similarly, $(0.3)^2$ of light is reflected twice and ready to shine through the crystal in the original direction. To put it simply, the light rays are just bouncing inside the crystal, losing 70% of its intensity each time when it reaches the surface.

As such, the percentage of light leaving the crystal from the left hand side is given by,

$I_0=0.7^2\displaystyle \sum_{n=1}^\infty 0.3^{2n-2}=0.5385$

Thus the answer is $53.85$ in percentage.

The percentage of the light that exit from the left after:

Reflecting from right side 0 times: $t^2$ .
Reflecting from right side 1 times: $t^2 (1-t)^2$ .
Reflecting from right side 2 times: $t^2 (1-t)^4$ .
......
Reflecting from right side n times: $t^2 (1-t)^{2n}$ .

Therefore, the total percentage of the light that exit from the left is $t^2 (1+(1-t)^2+(1-t)^4+(1-t)^6+...)$ .
Consider $A=1+(1-t)^2+(1-t)^4+(1-t)^6+...$ , so $A (1-t)^2=(1-t)^2+(1-t)^4+(1-t)^6+...$ .

Then $A(1-(1-t)^2)=1$ or $A=\frac{1}{1-(1-t)^2}$ .
The total percentage of the light that exit from the left is $t^2.A=\frac{t^2}{1-(1-t)^2}=0.538=53.8\%$

The light must be reflected an even number of times (zero included) in order to exit the photonic crystal on the left side. The chance of the light exiting after $2k$ number of reflections is $(\frac{7}{10})^2 (\frac{3}{10})^{2k}$ .

The sum of the probabilities of the light exiting after a given even number of reflections gives the end probability that the light will exit on the left side. This sum is written as $P = \sum _{k=0}^{\infty } \left(\frac{7}{10}\right)^2 \left(\frac{3}{10}\right)^{2 k}$ . Note that $(\frac{3}{10})^{2k} = (\frac{9}{100})^k$ , so this becomes the sum of an infinite geometric series for $a = \frac{9}{100}$ .

$\sum _{k=0}^{\infty } \left(\frac{49}{100}\right) \left(\frac{3}{10}\right)^{2 k} = (\frac{49}{100})(\frac{1}{1-\frac{9}{100}})$

$\Downarrow$

$P = (\frac{49}{100})(\frac{100}{91}) = \frac{7}{13} \approx 58.85\%$

Let the intensity of entering light be $1$ , the intensity of direct (from right to left) light in glass be $A$ , the intensity of reverse light in glass be $B$ , the intensity of the light exiting from left be $C$ , the intensity of the light exiting from right be $D$ , the transmission coefficient be $m$ .
Then we have the linear equations:

1. $A = m*1+(1-m)*B$
   (since $A$ is the sum of transmitted $1$ and reflected $B$ )
   
2. $B = (1-m)*A$
3. $C = m*A$
4. $D = (1-m)*1+m*B$

From 1. and 2. $A = m + (1-m)^2*A$ , or $(2-m)*m*A = m \iff A = \frac{1}{2-m}$
Then $B = \frac{1-m}{2-m}$ , $C = \frac{m}{2-m}$ , $D = \frac{2*(1-m)}{2-m}$
We need $C = 0.7/1.3 \simeq 0.53846 = 53.846\%$

Any photon that emerges on the left side of the crystal needs to cross the air-glass interface two times, once to enter the crystal and another to exit.

It can achieve this result in an infinite number of ways. For instance it can cross the interface on the right then immediately cross the interface on the left hand side or it can enter the crystal, then reflect twice before exiting.

In general, it can reflect any even number of times inside the crystal before exiting.

If we label the transmission event $\mathbf{T}$ and the reflection event $\mathbf{R}$ , the photon can emerge on the left due to the following trajectories

$\mathbf{T}\mathbf{T}, \mathbf{T}\mathbf{R}^2\mathbf{T}, \mathbf{T}\mathbf{R}^4\mathbf{T}, \ldots$

Summing over all of these events, and writing the transmission probability as $t$ , we find that the probability for a photon to emerge on the left is given by

$\begin{aligned} \mathbf{P}_L &= t^2\left(1+(1-t)^2 + (1-t)^4 + \ldots\right) \\ &= t^2\frac{1}{1-(1-t)^2} \\ &= 0.7^2\frac{1}{1-0.3^2}\\ &= 0.538 \end{aligned}$

yielding a 53.8% chance of a given photon traversing the crystal.

One part of the entering light transmits through the crystal, while the other gets reflected. The light that entered comes to the second boundary, and transmits and reflects again. Then the reflected part can transmit and reflect again from the first boundary, and so on, the light reflects an infinite number of time before it exits the layer.

To see which light exits at which side, we can notice a pattern in reflections and transmissions. The light on the right can be either reflected once ( $r$ ), or it can enter the layer and then be transmitted twice (both times through the first boundary) and reflected once, three times, five times etc. Therefore, the light that exits on the right is proportional to $r + t^2(r + r^3 + r^5 + r^7 + ...)$ which contains the sum of the odd terms of infinite geometric series. We can calculate this to be $r + t^2 \frac{r}{1-r^2}$ .

The light on the left has to be transmitted twice (once at both first and the second boundary) and reflected even number of times. Therefore, it is proportional to $t^2(1 + r^2 + r^4 + r^6 + ...) = \frac{t^2}{1-r^2}$ .

Since the energy has to be conserved, $r + t = 1$ . To check if the right and left parts of the light sum up to the initial light we can write $r + (1-r)^2\frac{r+1}{1-r^2} = 1$ , so the energy is conserved. The part that exits on the left is equal to $\frac{(1-r)^2}{1-r^2} = \frac{1-r}{1+r} = 0.5385 = 53.85 \%$ .