# Video: Pack 1 • Paper 3 • Question 18

Pack 1 • Paper 3 • Question 18

04:02

### Video Transcript

Tess is a distance 𝐷 away from a wall. Tess shines a torch at the wall. The intensity of the light on the wall is 𝐼. 𝐼 is inversely proportional to the square of 𝐷. The distance 𝐷 is decreased by 20 percent. Work out the percentage increase or decrease in the light intensity on the wall.

The first thing we need to consider here is the relationship between the intensity and the distance. We’re told that 𝐼 is inversely proportional to the square of 𝐷. If something is proportional to something else, for example, 𝑎 is proportional to 𝑏, we write this. However, if it’s inversely proportional, we say that 𝑎 is proportional to one over 𝑏.

Now, we have that the intensity is inversely proportional to the square of the distance. So we have that the intensity is proportional to one over the distance squared. We can also write this as the intensity is equal to 𝑘 over the distance squared, where 𝑘 is some constant. Now, let’s form an equation including the intensity and the distant for the initial distance of 𝐷. So the distance here is simply equal to 𝐷. And the intensity we can call 𝐼 one. So now, we can substitute these values into our equation for the intensity. And we obtain that 𝐼 is equal to 𝑘 over 𝐷 squared.

Now, let’s consider when the distance is decreased by 20 percent. If the distance is decreased by 20 percent, then the new distance will be 100 percent minus 20 percent which equals 80 percent of the original distance. And now, if we want to find 80 percent of something, we need to multiply it by 80 over 100 which is also equal to 0.8. So at this decreased distance, our distance will be equal to 0.8 timesed by the original distance 𝐷. And we can call our intensity 𝐼 two.

So again, we can substitute these values into our equation for the intensity. And we obtain that 𝐼 two is equal to 𝑘 over 0.8𝐷 all squared. We must be careful to put brackets around the whole of 0.8𝐷 since that’s the distance here and we need to square the whole distance. Now, we have formed equations for the intensities at both distances.

Let’s expand the brackets in the denominator of 𝐼 two. We get that 𝐼 two is equal to 𝑘 over 0.64𝐷 squared. And now we can split the fraction up to give one over 0.64 timesed by 𝑘 over 𝐷 squared. And now, we notice that 𝑘 over 𝐷 squared is simply 𝐼 one. And we can write the fraction of one over 0.64 as a decimal to obtain that 𝐼 two is equal to 1.5625𝐼 one. And this tells us that the intensity 𝐼 two is 1.6525 [1.5625] times larger than the intensity 𝐼 one.

So since 𝐼 two is greater than 𝐼 one, this means there has been a percentage increase here. Now, let’s find out what this percentage increase is. That’s right, the multiplier 1.5625 as a percentage. We simply multiply the multiplier by 100 and this gives us a percentage of 156.25. Therefore, we have that 𝐼 two is a 156.25 percent of 𝐼 one. And so therefore, this means there has been a 56.25 percent increase since we have got 100 percent of 𝐼 plus 56.25 percent of 𝐼. And so this leads to a 56.25 percent increase in the intensity.