Question

The lightbulb is 50 cm

from a mirror. It emits 1.0 W
of visible light. A small barrier blocks the direct rays of light from the bulb from reaching a sensor 70 cm
to the right, but not the reflected rays.
What is the light intensity at the sensor?

Answer 1

The light intensity at the sensor is approximately 0.69 W/m².

Step-by-step explanation:

The light intensity at the sensor can be calculated using the inverse square law for light. According to this law, the intensity of light decreases as the square of the distance from the source increases.

In this case, the lightbulb is 50 cm away from the mirror and the sensor is 70 cm to the right.

Since the reflected rays are not blocked by the small barrier, the distance between the lightbulb and the sensor can be considered as 50 cm + 70 cm = 120 cm or 1.2 m.

The initial power emitted by the lightbulb is 1.0 W.

Using the inverse square law, the light intensity at the sensor can be calculated as:

Initial Intensity / (Distance^2)

= 1.0 W / (1.2 m)^2

= 1.0 W / 1.44 m^2

≈ 0.69 W/m²