So, let's look at another example here of how you might use these. We talked about the absorbers, we talked about the transmittance and there's lots of equations going on. And they can get a little bit a little bit confusing, but lets do an example to clarify this a little bit. So here we have an example of three important amino acids, and they absorb here in the, this is about 270 to 280 nanometers. And that's in the UV is the region of the spectrum, and you can see the tryptophan is the strongest absorber. So, [COUGH] the question is 280 nanometers passed through 1.0 millimeters of a solution that contained an aqueous solution of the amino acid, tryptophan. And the had tryptophan's a concentration of 0.5 millimoles per liter. So, you want to calculate the absorbance, and the molar absorption coefficient of tryptophan at this weight. And then you're asked on doing that to what would be the transmittance through a cell of thickness 2.0 millimetres is to be. So let's do that, and it'll perhaps clarify. [INAUDIBLE] So, I showed that question down here. So, let's sure I do this the right way. So, the first thing you're asked here is the light is reduced to 54% of its original value. What you want to calculate is the absorbance, and the molar absorption coefficient. So, 50% of its initial value. So, I over I0 is equal 54%, or equal to 0.54. So that's T, that's transmittance, that's what we're told. Now if you look back on the notes, we defined A, the absorbance is equal to minus log of T. So that's equal to the negative log of 0.54, and that's equal to, if you put that in your calculator, it's 0.27. So, you need to know for this question, you need to know the relationship between the absorbance and the transmittance. The absorbance equals negative log of the transmittance. The second best, the question was, that's the absorbance. Then you're asked what the molar absorption coefficient [COUGH] at this wavelength. So, here you need to use the equation that we developed for A. A was equal to epsilon, the molar absorption coefficient times the concentration times the [INAUDIBLE] So rearranging that equation, you can set that epsilon is equal to A over cl. So, you've worked out the absorbance above there, your A is equal to 0.27, you're given the concentration in the question, it was 0.5 millimoles per liter. And you were told the path length was, let's check above, path length 1 millilitre. So, you just plug these guys in so your absorbance is 0.27. You divide it by the concentration, which was 5.3 millimolar, so that's equal to, sorry, 0.5 millimolar. So, 0.5 millimolar, and that's 0.5 by 10 to the minus 3, you should know that milli means 10 to the minus 3. And that's given in mole per liter. And then you divide it by the l here, which is the path length, and that's 1 millimeter. But as I said before, we usually like to have it in centimeters. So to convert 1 millimeter to centimeters, you multiply it by 10 to the -1. So the path length is 1.0 by 10 to -1, and that's in centimeters. So, you plug them into your calculator, and you should come out with 5.3 or something like this, by 10 to the 3 and of course you bring these line now, so you've got have mol -1 liter, centimeter to the -1. So that's your [COUGH] extinction coefficient, or your molar absorption coefficient. And the last part of the questions was what would be transmittance through a cell of thickness 2 millimeters? So, you worked out for a cell of 1 millimeter. So, you have to do it for a cell of 2 millileters, and again, you need to look back at your equations transmittance. You could do this a few different ways, but transmittance as we define it, is the intensity of the light as it comes out after the sample, which is defined as I, divided by the intensity of the incident light. And if you look back to your definition of that, that's 10 to the minus epsilon cl. C is constration, l is a path length. And of course, the reason we're going to do this is now because we've worked out what the molar absorption coefficient is in the previous plan. So, again, if you plug that in you get 10 the minus, and first we have to write down the, this is just this the molar absorption coefficient up here, divide by 10 to the 3. Then keep in your units, moles to minus 1 liter, centimeter to the minus 1. And then you multiply, you have to bring it down here, multiply it by 5 0 by 10 to the minus 4, that's the concentration, moles liter minus 1. And then you multiply that, the new path length, which is 2 millimeters. So that's 2 by 10 the minus 1, and that's centimeters. So, all this is just 1 multiplied by each other. So, what you'd find here if you did it, you'd find that all these units cross out. This minus 1 liter, centimeters minus 1, so you actually get 10 to the minus 0.54, cuz the transmittance is unitless. So that's equal to 0.29, or 29%. So, when you increase the path length to 2, the transmittance drops to 29% from 24, from 54%.