1. Ray Diagram and Intensity Analysis

Let the intensity of the incident light be $I_0$. Let $r$ be the reflection coefficient (fraction of energy reflected) and $t = 1-r$ be the transmission coefficient at each interface.

The total transmitted intensity $I_{trans}$ is the sum of intensities of all emerging rays:

• First emerging ray: The light passes through two interfaces. It transmits at the first ($(1-r)$) and at the second ($(1-r)$).
  $$T_1 = I_0 (1-r)^2$$
• Second emerging ray: The light reflects internally twice (once at the bottom, once at the top) before exiting.
  $$T_2 = I_0 (1-r) \cdot r \cdot r \cdot (1-r) = I_0 (1-r)^2 r^2$$
• Third emerging ray: The light reflects internally four times.
  $$T_3 = I_0 (1-r)^2 r^4$$

2. Summing the Infinite Series

The total transmitted fraction $F$ is the sum of a geometric progression:

$$ F = \frac{I_{trans}}{I_0} = (1-r)^2 + (1-r)^2 r^2 + (1-r)^2 r^4 + \dots $$ $$ F = (1-r)^2 \left[ 1 + r^2 + r^4 + \dots \right] $$

The term in the brackets is an infinite geometric series with common ratio $r^2$. Since $r < 1$, the sum is $\frac{1}{1-r^2}$.

$$ F = (1-r)^2 \cdot \frac{1}{1-r^2} $$

Using the identity $1-r^2 = (1-r)(1+r)$:

$$ F = \frac{(1-r)^2}{(1-r)(1+r)} = \frac{1-r}{1+r} $$

3. Calculation

Given the reflection coefficient $r = 0.25$:

$$ F = \frac{1 – 0.25}{1 + 0.25} = \frac{0.75}{1.25} $$ $$ F = \frac{3/4}{5/4} = \frac{3}{5} = 0.6 $$