Question

На высоте 8 м над землей висит лампа с силой света 100кд.найти площадь участка,в пределах которого освещенность не меньше,чем 1 лк

Answer 1

8*100=800
800:1лк (400)
800:400=200

Answer 2

Calculation of the Illuminated Area

To find the area within which the illuminance is not less than 1 lux, we need to calculate the distance at which the illuminance drops to 1 lux from the lamp.

The illuminance (E) at a certain distance (d) from a point source of light can be calculated using the inverse square law:

E = I / (4πd^2)

Where: - E is the illuminance in lux - I is the luminous intensity of the lamp in candela (cd) - d is the distance from the lamp in meters

Given that the luminous intensity of the lamp is 100 cd and the height of the lamp from the ground is 8 meters, we can calculate the distance at which the illuminance drops to 1 lux.

Let's calculate the distance using the inverse square law:

1 = 100 / (4πd^2)

Simplifying the equation:

d^2 = 100 / (4π)

d^2 ≈ 7.9577

Taking the square root of both sides:

d ≈ 2.819 meters

Therefore, the distance at which the illuminance drops to 1 lux is approximately 2.819 meters.

To find the area within which the illuminance is not less than 1 lux, we need to calculate the area of a circle with a radius of 2.819 meters:

Area = π * (2.819)^2

Area ≈ 25.0 square meters

Hence, the area within which the illuminance is not less than 1 lux is approximately 25.0 square meters.

Please note that the above calculation assumes ideal conditions and does not take into account factors such as light absorption, reflection, or obstruction by objects in the environment.