### Video Transcript

Tess is a distance π· away from a
wall. Tess shines a torch at the
wall. The intensity of the light on the
wall is πΌ. πΌ is inversely proportional to the
square of π·. The distance π· 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. Weβre told that πΌ is inversely
proportional to the square of π·. If something is proportional to
something else, for example, π is proportional to π, we write this. However, if itβs inversely
proportional, we say that π is proportional to one over π.

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 π over the distance squared, where π is some constant. Now, letβs form an equation
including the intensity and the distant for the initial distance of π·. So the distance here is simply
equal to π·. And the intensity we can call πΌ
one. So now, we can substitute these
values into our equation for the intensity. And we obtain that πΌ is equal to
π over π· squared.

Now, letβs 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 π·. And we can call our intensity πΌ
two.

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

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

So since πΌ two is greater than πΌ
one, this means there has been a percentage increase here. Now, letβs find out what this
percentage increase is. Thatβs 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 πΌ two is a
156.25 percent of πΌ one. And so therefore, this means there
has been a 56.25 percent increase since we have got 100 percent of πΌ plus 56.25
percent of πΌ. And so this leads to a 56.25
percent increase in the intensity.