# What is the visible light intensity at the surface of the bulb?

Sep 5, 2016

#### Answer:

The equation needed is: I = P/(4 π d²)

Where P is the power of light emitted at the filament and d is the distance from filament to the surface of the bulb.

#### Explanation:

First some assumptions:
1. The bulb is a filament bulb.
2. The filament is a point source of light.

The first assumption means that the light travels a distance from filament to the surface of the bulb. The second means that the light spreads out across a sphere.

Define variables:
The distance from filament to the surface of the bulb is d. The electrical power converted to light at the filament is P. The intensity of light at the surface of the bulb is I.

As light travels from a source it spreads out to cover an area, A, at a given distance. In this case the source is a point source and the light spreads out in all directions to fill a sphere. So if the light has travelled distance d the radius of the sphere is d. All light that was emitted at the filament and travelled distance d will cover the surface area of a sphere of radius d.

The equation for light intensity is:
$I = \frac{P}{A}$

As we have seen above the relevant area for this case is the surface are of a sphere radius d:
A = 4 π d^2

So now combine the equations to get an equation for the light intensity at the surface of the bulb:
I = P/(4 π d^2)