A 100.0 W light bulb hangs from the ceiling. What is the visible light intensity on the floor, 2.1 m directly below the bulb, if 10% of its output is visible light?
So I am attempting to use: I= Power/area
however, I don't have a radius. I know 10% of 100 is 10 so that would be the intensity for that and that the refraction for air should be used but I am missing something...
HELP :)
The radius is R = 2.1 m. The light is distributed uniformly over a sphere of that radius, at that distance. So divide 10 W by 4*pi*R^2 and you will have the visible light irradiance.
To calculate the visible light intensity on the floor, you need to determine the area over which the light is being spread. Since the light bulb is hanging directly above the floor, we can assume it is spreading the light uniformly in all directions, creating a spherical pattern.
To solve this, you can consider the surface area of a sphere at a certain distance from the source of light. The formula for the surface area of a sphere is A = 4πr^2, where A is the surface area and r is the radius of the sphere.
In this case, the light bulb is hanging 2.1 meters directly above the floor, so the radius of the sphere is 2.1 meters.
Now you can calculate the visible light intensity on the floor using the formula:
I = Power/ Area
First, calculate the surface area:
A = 4πr^2
= 4 * π * (2.1)^2
Now, calculate the visible light intensity:
I = Power / Area
= 100 W / [4 * π * (2.1)^2]
Solving this equation will give you the visible light intensity on the floor directly below the light bulb.