Even if you're not completely sure but think there might be an issue with my calculations, please don’t hesitate to contact me!

Datasheet

The following can be extracted from the LED datasheet:

Radiation Intensity: Min 3, Max 8 \(\frac{\text{mW}}{\text{sr}}\)
I choose 8 \(\frac{\text{mW}}{\text{sr}}\) as it represents the worst-case scenario, indicating the maximum output of the LED. This means that the LED emits 8 milliwatts of optical power per steradian² when operated at a forward current of 20 mA. Since the LED is limited to 2.4 mA due to the board design, and the datasheet indicates that the radiation intensity decreases linearly, we can calculate the adjusted radiation intensity as follows:

\[ \text{Adjusted Radiation Intensity} = \frac{8 \ \frac{\text{mW}}{\text{sr}} \times 2.4 \ \text{mA}}{20 \ \text{mA}} = 0.96 \ \frac{\text{mW}}{\text{sr}} \]

Half Light Angle: 120°
This indicates that the LED radiates light within a cone of 120°. The half light angle is the angle at which the emitted intensity falls to half of its maximum value. A wide angle provides broader illumination, making the LED suitable for applications requiring diffuse lighting.

² A steradian is a unit used to measure angles in three-dimensional space, similar to how a radian measures angles in a circle.

Solid Angle and Illuminated Area

To determine the LED's power per unit area at the eye, we need to calculate both the size of the illuminated area and the distance from the light source. The illuminated area is where the light cone intersects with the eye, which changes based on the distance from the light source.

We can calculate this area using the solid angle (\(Ω\)), which measures the amount of three-dimensional space the light occupies as it spreads from the LED (point source). This solid angle is defined in steradians (sr).

Calculating the Solid Angle (\(Ω\))

To calculate the solid angle ( \(Ω\) ) for a specific emission angle, we use the half angle ( \(θ\) ). The formula for calculating the solid angle for a light source with a specific emission angle is:

$$Ω = 2\pi \left(1 - \cos(θ)\right)$$

Given that the LED has an emission angle of \(120^\circ\), we need to divide it by 2 to get the half-angle:

$$θ = \frac{120^\circ}{2} = 60^\circ$$

The cosine of \(60^\circ\) is:

$$\cos(60^\circ) = 0.5$$

Now, inserting this value into the formula for the solid angle:

$$Ω = 2\pi \left(1 - 0.5\right) = \pi \ \text{sr}$$

Illuminated Area (\(A\))

The area (\(A\)) onto which the light from an LED is distributed at a certain distance (\(r\)) can be calculated using the following formula:

$$A = r^2 \times Ω$$

We have already calculated the solid angle \(Ω = \pi \ \text{sr}\) for the emission angle of \(120^\circ\). Assuming a distance of \(r = 1 \, \text{cm}\) from the eye to the LED, we can calculate the illuminated area as follows:

$$A = (1 \ \text{cm})^2 \times \pi \ \text{sr} = \pi \ \text{cm}^2$$

Thus, the area illuminated by the IR radiation at a distance of 1 cm from the LED is approximately:

$$A \approx 3.14 \ \text{cm}^2$$

These 3.14 cm² represent the area over which the IR radiation is distributed at a 1 cm distance from the LED. Since the LED is positioned outside of the eye, a significant portion of this radiation is lost to the sides. Furthermore, the eye's surface area is smaller than 3.14 cm², meaning only a fraction of the radiation actually reaches the eye.

If you've been paying attention, you might wonder: Why don't I use the smaller area that hits the eye for the further calculations, which would result in a much higher calculated power per cm² and could potentially exceed safe limits?

While this is a valid point, it only holds true when considering distance. At shorter distances, the light cone narrows, concentrating the radiation on a smaller area and thereby increasing the density, essentially scaling it.

However, in my calculations, I do not scale the radiation. Instead, I consider the total area of 3.14 cm², from which only a small fraction actually reaches the eye. For example, only about 0.14 cm² may illuminate the eye, while the remaining area dissipates into the environment. If I were to calculate using just 0.14 cm², it would inaccurately suggest that all the radiation concentrates on that small area, overlooking the significant portion that is lost.

Irradiance (Power Density)

Now we can convert the adjusted radiation intensity from \(\frac{\text{mW}}{\text{sr}}\) to \(\frac{\text{mW}}{\text{cm}^2}\):

The power density in \(\frac{\text{mW}}{\text{cm}^2}\) can be calculated using the formula:

$$\text{Power Density} \ [\frac{\text{mW}}{\text{cm}^2}] = \frac{\text{Adjusted Radiation Intensity} \ [\frac{\text{mW}}{\text{sr}}]}{A \ \text{[cm}^2]}$$

Now, substituting the given values into the formula:

$$\text{Power Density} \ = \frac{0.96 \ \frac{\text{mW}}{\text{sr}}}{3.14 \ \text{cm}^2}$$

Now, calculate the value:

$$\text{Power Density} \approx \frac{0.96}{3.14} \approx 0.305 \ \frac{\text{mW}}{\text{cm}^2}$$

Since four LEDs are used on each eye, this result must be multiplied by four.

$$\text{Total Power Density} = 0.305 \ \frac{\text{mW}}{\text{cm}^2} \times 4 \approx 1.22 \ \frac{\text{mW}}{\text{cm}^2}$$

The unit steradian (sr) is dimensionless because it represents the ratio of an area to the square of a radius. Since it is a pure ratio, it has no physical unit.

Result

If the LEDs are 1 cm away from your eye, the radiation they emit is 1.22 \(\frac{\text{mW}}{\text{cm}^2}\), which is well below the maximum limit of 10 \(\frac{\text{mW}}{\text{cm}^2}\). Even if the distance were halved (which is not physically possible) and the LEDs were 0.5 cm away, the radiation would only increase to 4.88 \(\frac{\text{mW}}{\text{cm}^2}\), still half the limit.