Math Snap

STEP 1

1. The light source emits uniformly in all directions.
2. The illuminance at point $P$ is calculated using the inverse square law.
3. The angle $\theta$ and distance $d$ are related through trigonometry.

STEP 2

1. Calculate the distance $d$ from the light source to point $P$.
2. Use the inverse square law to find the illuminance $E$ at point $P$.

STEP 3

Calculate the distance $d$ from the light source to point $P$ using the Pythagorean theorem. The vertical distance from the light source to the bench is $2$ meters, and the horizontal distance to point $P$ is $1.8$ meters.
$$d = \sqrt{(2 \text{ m})^2 + (1.8 \text{ m})^2}$$ Calculate:
$$d = \sqrt{4 + 3.24}$$ $$d = \sqrt{7.24}$$ $$d \approx 2.69 \text{ m}$$

Use the inverse square law to find the illuminance $E$ at point $P$. The illuminance is given by:
$$E = \frac{I}{d^2}$$ where $I = 800 \text{ cd}$ and $d \approx 2.69 \text{ m}$.
Substitute the values:
$$E = \frac{800 \text{ cd}}{(2.69 \text{ m})^2}$$ Calculate:
$$E = \frac{800}{7.24}$$ $$E \approx 110.5 \text{ lux}$$ The illuminance at point $P$ is approximately:
$$\boxed{110.5 \text{ lux}}$$