# Question about units in photonics calculation

My textbook has a photonics example problem, where the absorbed light energy density is calculated as

$$\Psi = 25 \times 9.55 = 238.75 \text{mJ/cm^3} = 2.38 \times 10^5 \text{J/m}^3$$

I'm new to physics, so I'm trying to understand how unit conversions work.

$$\Psi = 238.75 \text{mJ/cm}^3 = 2.38 \times 10^2 \text{mJ/cm^3} = 0.00238 \times 10^2 \text{J/cm^3} = 2.38 \times 10^5 \text{J/cm^3}$$

Now, given the difference between my (incorrect) solution and the textbook solution, I'm speculating that there is some sort of "dimensionality" issue with the units, where if I change the numerator to Joules, then to maintain the correct (let's call it) "dimensionality ratio", then I must also change the denominator to metres? What is the rational for this?

As I said, I'm new to physics, so I'm not sure what the correct terminology is here to articulate my thoughts, but I think what I've stated is close enough to be understandable.

I would appreciate it if people could please take the time to clarify this.

$$0.00238 \times 10^2 \frac{\text{J}}{\text{cm}^3} \neq 2.38 \times 10^5 \frac{\text{J}}{\text{cm}^3}$$
$$0.00238 \times 10^2 \frac{\text{J}}{\text{cm}^3} = 2.38 \times 10^{-3} \times 10^2 \frac{\text{J}}{\text{cm}^3} = 2.38 \times 10^{-1} \frac{\text{J}}{10^{-6}\text{m}^3} = 2.38 \times 10^{5} \frac{\text{J}}{\text{m}^3}$$
• Ahh, you're right, I did make an error. But shouldn't it be $0.00238 \times 10^2 \frac{\text{J}}{\text{cm}^3} = 2.38 \times 10^{-1}$? Apr 11, 2019 at 9:29
• Absolutely, that's what I wrote. You get the additional factor of $10^6$ from changing centimeters to meters in the denominator. Apr 11, 2019 at 9:31
• The units are changed like that $\text{cm}^3 = (0.01\, \text{m})^3 = (10^{-2} \text{m})^3 = 10^{-6}\text{m}^3$. Apr 11, 2019 at 9:38