Triple Pane Windows

Since everyone else is presenting a geometric progression summation approach, let me present one that approaches the problem by redefining it.

What we really care about is:
$100\%$ the amount of light entering pane 1,
$a$ the amount of light from pane 1 to pane 2,
$b$ the amount of light from pane 2 to pane 1,
$c$ the amount of light from pane 2 to pane 3,
$d$ the amount of light from pane 3 to pane 2,
$e$ the amount of light from pane 3 to the right.

From the problem, we have the following equations:

$\begin{array} { l l } a & = 0.7 \times 1 + 0.2 \times b \\ b & = 0.2 \times a + 0.7 \times d \\ c & = 0.7 \times a + 0.2 \times d \\ d & = 0.2 \times c \\ e & = 0.7 \times c \\ \end{array}$

Solving this system, we get that $a = 74.5\%, b = 22.5\%, c = 54.3 \%, d = 10.8 \%, e = 38.0 \%$ .

I decided to look for a solution that is easily generalized to other values and to more panes. Consider $n$ panes of glass with transmittancy $T = 0.7$ and reflectivity $R = 0.2$ .

Let $a_i$ describe the light traveling in the "forward" direction and $b_i$ the light traveling in the "backward" direction. We actually know more about the light near the last pane, because the implied assumption is that no light comes in from that side. Thus we call the last pane, "pane 1" and work backward. In other words, we are to find the ratio $\frac{b_0}{b_n}$ under the assumption $a_0 = 0$ . The optics of pane $i$ is defined by the equations $a_{i-1} = Ta_i + Rb_{i-1} \\ b_i = Tb_{i-1} + Ra_i$ Manipulating this, we get $\left(\begin{array}{c} a \\ b \end{array}\right)_i = M\ \left(\begin{array}{c} a \\ b \end{array}\right)_{i-1}\ \ \ \text{with}\ \ \ M = \left(\begin{array}{cc} \dfrac{T^2-R^2}T & \dfrac R T \\ -\dfrac R T & \dfrac 1 T\end{array}\right)$ and $\left(\begin{array}{c} a \\ b \end{array}\right)_n = M^n\ \left(\begin{array}{c} a \\ b \end{array}\right)_0.$ We find for the given values that $M = \frac1{14} \left(\begin{array}{cc} 9 & 4 \\ -4 & 20 \end{array}\right),$ and $\left(\begin{array}{c} a_3 \\ b_3 \end{array}\right) = \frac1{14^3} \left(\begin{array}{cc} 121 & 2580 \\ -2580 & 7216 \end{array}\right) \left(\begin{array}{c} 0 \\ b_0 \end{array}\right),$ so that $b_3 = \frac{7216}{14^3} b_0$ and $b_0/b_3 = 14^3/7216 \approx 0.380266 = \boxed{38.0266\%}$ .

For multiple window panes, it is more efficient to work with an eigenvalue decomposition of matrix $M$ . I will post more details in the comments.

Consider a $2$ -pane window, where the first pane transmits the proportion $p_1$ and reflects the proportion $q_1$ of the light incident on it, while the second pane's transmission and reflection coefficients are $p_2$ and $q_2$ . The proportion of light transmitted by the $2$ -pane window is $P_{12} \; = \; \sum_{n=0}^\infty p_1p_2(q_1q_2)^n \; = \; \frac{p_1p_2}{1-q_1q_2}$ (consider the number of times the light bounces back and forth between the two panes), while the proportion of light reflected is $Q_{12} \; = \; q_1 + \sum_{n=0}^\infty p_1^2q_2(q_1q_2)^n \; = \; q_1 + \frac{p_1^2q_2}{1-q_1q_2}$ A $3$ -pane window with relevant coefficients $p_1,q_2,p_2,q_2,p_3,q_3$ can be regarded as a $2$ -pane window with relevant coefficients $P_{12},Q_{12},p_3,q_3$ (thinking of the first two panes as a single unit), and so the overall proportion of transmitted light is $t \; = \; \frac{P_{12}p_3}{1 - Q_{12}q_3}$ For $p_1=p_2=p_3=p$ and $q_1=q_2=q_3=q$ we have $t \; = \; \frac{p^3}{(1-q^2)^2 - p^2q^2}$ With $p=0.7$ , $q=0.2$ this becomes $t = 0.380266$ , making the answer $\boxed{38.0266}$ %. Reassuringly, we obtain the same answer if we consider the second and third panes as a single unit first, obtaining the alternative formula $t \; = \; \frac{p_1P_{23}}{1 - q_1Q_{23}}$ for the transmission coefficient.

Also the same like double pane glass, but this time we have to consider the rays that return into the "first between" after successfully passing through the second pane.
First, we need to know the total lights that will manage to pass the second screen before going on to the third one, but we could not get them directly because the lights keep reflecting and transmitting back and forth.

Let X be the % of 2-paners lights (1-paners is 70%, and 3-paners is what the question wants).

X = (2-paners strictly from "first between") + (2-paners which have already reached "second between" but reflected at 3rd pane, transmitted at 2nd pane, reflected at 1st pane, and transmitted again at 2nd pane (for its a) and the rest of it) + (like the group before, but its a was reflected twice more before transmitted back into "first between") + .....

X = 49 / (1 - 0.04) + X * 0.2 * 0.7 * 0.2 * 0.7 / (1 - 0.04) + X * 0.2 * 0.2 * 0.2 * 0.7 * 0.2 * 0.7 / (1 - 0.04) + .......
0.96X = 49 + 0.0196X / (1 - 0.04).
X = 52.15.
Then just use X to get the 3-paners,
52.15*0.7/0.96 = 38.0266