If an object is placed 1 m (3 ft) away from a light source and the lux level is 3200 lux, what is the lux level at a distance of 8 m (24 ft)?
Correct Answer: A
If an object is placed 1 meter (3 feet) away from a light source with a lux level of 3200 lux, the lux level decreases according to the inverse square law of light. At a distance of 8 meters (24 feet), the lux level can be calculated as:
New Lux Level=3200 lux(8/1)2=3200 lux64=50 lux\text{New Lux Level} = \frac{3200 \text{ lux}}{(8/1)^2}
= \frac{3200 \text{ lux}}{64} = 50 \text{ lux}New Lux Level=(8/1)23200 lux=643200 lux=50 lux Thus, at 8 meters, the light intensity drops significantly to 50 lux.
Reference: Axis Communications. "Light conditions and surveillance camera performance." Axis Light Conditions.