# Inverse Square relationship using paint problem confusion

I want to ask a question about the inverse square relationship using an aerosol paint spray mentioned in my book.

I am reading the book Advanced Physics by Steve Adams, and it mentions this in the book.

Imagine you are holding an aerosol paint spray at $50$cm from a wall.
By squirting it for one second, you make a circle of radius $10$cm.

Now, I am aware that I can find the area of the circle as follows:

$$A_{10} = \pi r^2 = \pi \times 10^2 = 100 \pi$$ The book next talks about increasing the distance from a wall:

Now imagine you move along the wall and stand twice as far from the wall - 100cm.
You squirt for the same length of time.
Because he is standing twice as far away from the wall, the radius of the patch is doubled to 20cm.

This therefore means that the area of the circle is now:

$$A_{20} = \pi r^2 = \pi \times20^2 = 400 \pi$$

Now, I have studied the inverse-law relationship ($I \propto \frac{1}{r^2}$ previously in regards to a light source, and I wanted to understand this concept more easily using this paint example.

However, I cannot understand why standing twice as far away from the wall, the radius of the patch is doubled to 20cm.

Can someone explain why this is the case please?

The spray comes out as a cone with opening angle $\alpha$ such that $$\tan\alpha=\frac{10cm}{50cm}=\frac{1}{5}$$

If you double your distance to the wall, then the opening angle $\alpha$ stays the same and therefore the radius of the paint patch $r$ is such that

$$\frac{1}{5} = \tan\alpha = \frac{r}{100cm}$$

If you solve this for $r$, you will see that it is now doubled, that is, $20cm$.

• Yes, you are perfectly correct, I will be surprised if the textbook did not include a similar illustration. Thanks for that. My post does not add anything to your answer. – user179430 Dec 30 '17 at 20:23
• Not sure, actually the picture in your answer was quite helpful! :) – Photon Dec 30 '17 at 20:24
• To the OP: this video makes the point fairly clearly. m.youtube.com/watch?v=KARCOGT95W0 – user179430 Dec 30 '17 at 20:26
• Brilliant answer. Straight to the point, no-nonsense explanation. Many thanks! – user55213 Dec 30 '17 at 21:00

It is square inverse intensity even if you can adjust the spread of spray like most real life paint sprayers to spray wider or narrower, even if the nuzzle has been damaged and sprays a wiggly circle like an amoeba with one or two spots even being sprayed outside of the main circle or holes left inside.

Let's assume that that this amoeba is measored horizontally 20 cm for a distance of 50 cm of sprayer from the wall. As long as we disregard the fact that spray shooting out of paint gun will go down gently in a parabolic path, we can assume the amoeba sprayed by this paint gun will be 40 cm wide at 100 cm distance.

Because the area of similar shapes is proportional to square of one of their sides, length, or width, the second amoeba will be four times bigger and hence will receive 1/4 paint per square cm.

As to the proof of why the area of similar shapes is proportional to square of their sides or length, let's assume we use tiles small enough to cover the suface of the first amoeba within acceptable precision and we call these unit tiles, the second amoeba will be covered by the same number and configuration of tiles except they are now each four times bigger than the original. Hence it's surface 4 times bigger.