# Thread: How to calculate inverse square law?

1. ## How to calculate inverse square law?

The original input is 70mW/cm2 for a light source.

I am trying to get to 10mW/cm2 for 10 minutes outputting about 6 J/cm2.

I understand each time I double the measurement, I quadruple the area covered so the power is 4x less...

But at what point do I start at, i.e. if I start at 0 and double the distance it won't compute. If I start at 1/2 inch the difference will be huge to if I start with 1 inch....

We're talking LED, 633nm and 830nm (but not together)

Thanks

2. Do you know the distance the original input is?

If so you can use the formula $\dfrac{I_1}{I_2} = 7 = \dfrac{(r_2-r_1)^2}{r_1^2}$. In this case $r_2 > r_1 \text{ and } r_2-r_1$ is the distance between sources

3. Originally Posted by e^(i*pi)
Do you know the distance the original input is?

If so you can use the formula $\dfrac{I_1}{I_2} = 7 = \dfrac{(r_2-r_1)^2}{r_1^2}$. In this case $r_2 > r_1 \text{ and } r_2-r_1$ is the distance between sources
Thanks.

But here in lies the problem. I'm trying to work out if your starting is from as close to the skin as possible and that is 70mW/cm2, then what is double that...? Double 0...

To be fair, it says no closer then 1/2 inch from skin in the book, but does say that reduces the power. Reading the booklet you'd think that 0 inches and 1/2 inch are both 70mW/cm2, which surely cannot be the case? I suppose I could work it out on the basis of 1/2 inch.

Let's say for example it was 1/2 inch is 70mW/cm2, then how to I get to 10mW/cm2 for 10 mins equalling 6J/cm2?

4. I would use the figure of half an inch which is $0.5 inch \cdot \dfrac{2.54 cm}{inch} = 1.27cm$

If you had 70mW/cm^2 at half an inch from the skin than you'd have an intensity of $70 \cdot 10^{-3} = \dfrac{k}{1.27^2}$ where k is some constant.

If you need to know the distance from the skin would give 10mW/cm^2 then you can say that $10 \cdot 10^{-3} = \dfrac{k}{d^2}$

Dividing the first equation by the second equation gives $7 = \dfrac{d^2}{1.27^2} \implies d = 1.27\sqrt{7} cm$ away from the skin. This is a reasonable answer since the power decrease by a factor of 7 we'd expect a distance increase of about $\sqrt{7}$

I have to admit the 10mW/cm^2 for 10 minutes given 6J/cm^2 confuses me. Energy is power multiplied by time $(E = Pt$ and your per area will cancel) and your 10mW/cm^2 acting continuously and constantly for 10mins gives an energy of

$\dfrac{E}{A} = 10 \cdot 10^{-3} \cdot 600 = 6000 \cdot 10^{-3} = 6 \text{J} \cdot \text{cm}^{-2}$

with units it becomes

$\dfrac{E}{A} (\text{J}\cdot{cm}^{-2}) = 10 \text{mW}\cdot \text{cm}^{-2} \cdot \dfrac{1 \text{W}}{1000 \text{mW}} \cdot 10 \text{mins} \cdot \dfrac{60 \text{secs}}{1 \text {min}} = 6 (\text{J} \cdot \text{cm}^{-2})$

5. Originally Posted by e^(i*pi)
I would use the figure of half an inch which is $0.5 inch \cdot \dfrac{2.54 cm}{inch} = 1.27cm$

If you had 70mW/cm^2 at half an inch from the skin than you'd have an intensity of $70 \cdot 10^{-3} = \dfrac{k}{1.27^2}$ where k is some constant.

If you need to know the distance from the skin would give 10mW/cm^2 then you can say that $10 \cdot 10^{-3} = \dfrac{k}{d^2}$

Dividing the first equation by the second equation gives $7 = \dfrac{d^2}{1.27^2} \implies d = 1.27\sqrt{7} cm$ away from the skin. This is a reasonable answer since the power decrease by a factor of 7 we'd expect a distance increase of about $\sqrt{7}$

I have to admit the 10mW/cm^2 for 10 minutes given 6J/cm^2 confuses me. Energy is power multiplied by time $(E = Pt$ and your per area will cancel) and your 10mW/cm^2 acting continuously and constantly for 10mins gives an energy of

$\dfrac{E}{A} = 10 \cdot 10^{-3} \cdot 600 = 6000 \cdot 10^{-3} = 6 \text{J} \cdot \text{cm}^{-2}$

with units it becomes

$\dfrac{E}{A} (\text{J}\cdot{cm}^{-2}) = 10 \text{mW}\cdot \text{cm}^{-2} \cdot \dfrac{1 \text{W}}{1000 \text{mW}} \cdot 10 \text{mins} \cdot \dfrac{60 \text{secs}}{1 \text {min}} = 6 (\text{J} \cdot \text{cm}^{-2})$

Great help. So 7cm away from my skin right? And what would it be for the IR at 830nm 55mW/cm2?

Re the calculation of J/cm2, this an example from a website and they way I was told to work it out.

Example:

A LED device has these specifications:

Red (wavelength 633 nm). Power output 8.4 mW/cm2. A mW is a milliWatt, or a thousandth of a Watt.

The light output is 8.4 mW per cm2 and your goal is 4 Joules per cm2 per treatment. One Joule is defined as one Watt per second, so in order to achieve that, you should know that J = W x S (Joule is Watt x Seconds)

In our case, we want to know how many seconds, so our equation becomes S = J / W. We just said 4 Joules is the target energy to be delivered to a square centimeter of skin, and the output of LED devices is stated in mW, so the formula to use is: Seconds of treatment per cm2 of skin = 4000 mJ / milliWatt per cm2 of skin.

4000 mJ divided by 8.4 = 476 seconds.

So each time you use the device, you should apply it for 476 seconds (8 minutes) on a skin surface area with the same size as the light emitting surface area as the device. This means that if you want to treat a skin area that is four times larger than the light emitting part of the LED device, that you need to move it over the skin for 4 x 8 = 32 minutes in total.

If the output per cm2 is not specified, look for the total output of the device and divide the total output by the total area of the panels with LEDs. This will give the Wattage per cm2. Example: Output area is 10 cm2, total Wattage is 50 mW. Wattage per cm2 is 50 divided by 10 = 5 mW.
.

In this example I was raking the 1 watt = 1 joule per second and calculating as. 6 (6 J/cm2) / 0.01 (10mW/cm2) = 600 seconds (10 minutes).

Or 600 (seconds) x 0.01 (mW/cm2) = 6 J/cm2

Is that not correct?

6. Originally Posted by moore778899
Great help. So 7cm away from my skin right? And what would it be for the IR at 830nm 55mW/cm2?
I got $1.27\sqrt{7} \approx 3.36cm$

For 55mW/cm^2 you can say that $I_2 = 55$ and so $55 \cdot 10^{-3} = \dfrac{k}{d^2}$

Hauling down the equation for $I_1 = 70$ from above: $70 \cdot 10^{-3} = \dfrac{k}{1.27^2}$

Dividing I1 by I2 gives: $\dfrac{70}{55} = \dfrac{d^2}{1.27^2}$. Solving for d gives $d= 1.27 \cdot \sqrt{\dfrac{14}{11}} \approx 1.43 \text{ cm}$ (We would expect this given that it has more power than the 10mW but less than 70)

Re the calculation of J/cm2, this an example from a website and they way I was told to work it out.

.

In this example I was raking the 1 watt = 1 joule per second and calculating as. 6 (6 J/cm2) / 0.01 (10mW/cm2) = 600 seconds (10 minutes).

Or 600 (seconds) x 0.01 (mW/cm2) = 6 J/cm2

Is that not correct?
I misunderstood the original question and thought you were given time. Your first calculation is fine since 0.01W = 10mW and would indeed given 10 minutes of time - $t = \dfrac{E}{P} = \dfrac{6}{0.01} = 600s$

For any laser with intensity I (mW/cm^2)

If you want to work out any power you can use the formula $d = 1.27 \cdot \sqrt{\dfrac{70}{I}} \text{ cm }$ which gives the distance d from the skin for any intensity I.

The amount of time necessary for this laser will be $t (\text{seconds}) = \dfrac{E (\text{J})}{P (\text{mW})} \cdot 1000 \dfrac{\text{mW}}{\text{W}}$

7. Hi,

Thanks.

But I'm a little confused as 3.36cm is only just over double the distance, so 4x less the power.

10 mins at 70mW/cm2 at 1/2 inch = 42 J/cm2

- 600 seconds / 0.07 = 42 J/cm2

But 42 / 4 = 10.2 J/cm2. Feels like it should be more like 4.5-cm.

Not doubting your calculation, just dont quite understand it.

8. It is measuring the distance from the skin. 3.36 is about 2.6 times bigger than 1.27 (half an inch) so you'd expect roughly a 2.6^2 drop which is about 6.75 which is closeish to 7 given that I have estimated heavily in this post.

nb: you seem to have put the divide sign when you actually (and correctly) multiplied by power to get energy.

edit: there is no problem in doubting my calculation, indeed it's important you can follow it

9. Originally Posted by e^(i*pi)
It is measuring the distance from the skin. 3.36 is about 2.6 times bigger than 1.27 (half an inch) so you'd expect roughly a 2.6^2 drop which is about 6.75 which is closeish to 7 given that I have estimated heavily in this post.
Ok, I must admit, I still don't understand it (and sorry for the error in my equation, it's very late here ).

Do you mean 6.75 J/cm2?