# Video: Finding the Intensity of Light at a Distance from the Source

Ed Burdette

Two lightbulbs have electrical power inputs of 100 W and 75 W respectively. At what distance is the intensity of light produced by the 100 W lightbulb equal to the intensity of light 10.0 m away from the 75 W lightbulb? Assume that both lightbulbs have the same efficiency.

02:28

### Video Transcript

Two lightbulbs have electrical power inputs of 100 watts and 75 watts, respectively. At what distance is the intensity of light produced by the 100-watt lightbulb equal to the intensity of light 10.0 meters away from the 75-watt lightbulb? Assume that both lightbulbs have the same efficiency.

Weβll name the power of the first lightbulb, 100 watts, π sub one and the power of the second bulb, 75 watts, π sub two. Weβll also call the distance, 10.0 meters, that we measure the intensity from the 75-watt bulb π. We want to know what distance from the 100-watt bulb is the intensity from that bulb equal to the intensity from the 75-watt bulb 10.0 meters away. Weβll call that distance capital π·.

To start our solution, we can recall an equation for intensity, power, and distance. The intensity πΌ of light from a source is proportional to the power of that source divided by the distance π from the source squared. If we call our first bulb with a power rating of 100 watts B one and the second bulb B two, then we know that, at distance lowercase π away from B two, the light intensity from B two will equal the light intensity from B one at distance of capital π· from B one.

At that distance, we can write that πΌ one is equal to πΌ two. The light intensity from both bulbs is the same. Using our proportionality relationship for light intensity, we can therefore write that π one over capital π· squared is equal to π two over lowercase π squared.

Rearranging to solve for capital π·, we find that capital π· is equal to lowercase π times the square root of π one over π two. We know π one, π two, and lowercase π. So we can plug in these values and solve for capital π·.

When we do and introduce values on our calculator, we find that, to two significant figures, π· is 12 meters. Thatβs how far from bulb one we need to be in order for the intensity to match the intensity of bulb two at distance of 10.0 meters away.