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.
We’ll name the power of the first lightbulb, 100 watts, 𝑃 sub one and the power of the second bulb, 75 watts, 𝑃 sub two. We’ll also call the distance, 10.0 meters, that we measure the intensity from the 75-watt bulb 𝑑. 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. We’ll call that distance capital 𝐷.
To start our solution, we can recall an equation for intensity, power, and distance. The intensity 𝐼 of light from a source is proportional to the power of that source divided by the distance 𝑑 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 𝑑 away from B two, the light intensity from B two will equal the light intensity from B one at distance of capital 𝐷 from B one.
At that distance, we can write that 𝐼 one is equal to 𝐼 two. The light intensity from both bulbs is the same. Using our proportionality relationship for light intensity, we can therefore write that 𝑃 one over capital 𝐷 squared is equal to 𝑃 two over lowercase 𝑑 squared.
Rearranging to solve for capital 𝐷, we find that capital 𝐷 is equal to lowercase 𝑑 times the square root of 𝑃 one over 𝑃 two. We know 𝑃 one, 𝑃 two, and lowercase 𝑑. So we can plug in these values and solve for capital 𝐷.
When we do and introduce values on our calculator, we find that, to two significant figures, 𝐷 is 12 meters. That’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.