[shafaifer]

Based on my attachment - I cannot understand how a sample absorbing 10% will have A = 1, When A is defined as:

Assumption: A = log(100 / %T) = 2 - log(%T) = 2 - log (10%) = 2 - log(0.1) = 4.3

Is there something wrong with my assumption?

[Corribus]

There's a bit of a language issue here.

The absorptivity A measured by a spectrometer is given by Beer's Law (or Beer-Lambert Law)

[tex]A = - log \frac {I}{I_0} = - log T = \epsilon c l[/tex]

(Logs are base 10). Let's work backwards. What does an absorptivity of 1 mean,for amount of light transmitted?

[tex] 1 = - log T[/tex]

Solving for T, T = 10^-1 = 0.1, or 10% transmittance when A = 1.

A sample absorbing 10% would have a transmittance of 90%, or T = 0.9. In this case, A ~ 0.046.

So, a sample absorbing 10% won't have a value of A = 1, but rather a sample transmitting 10% will have a value of A = 1. This is one reason why it's a good rule of thumb to keep absorbance values below A = 1, because at higher values of A, the amount of transmittence becomes very low, which can lead to sensitivity issues.

[shafaifer]

There's a bit of a language issue here.

The absorptivity A measured by a spectrometer is given by Beer's Law (or Beer-Lambert Law)

[tex]A = - log \frac {I}{I_0} = - log T = \epsilon c l[/tex]

(Logs are base 10). Let's work backwards. What does an absorptivity of 1 mean,for amount of light transmitted?

[tex] 1 = - log T[/tex]

Solving for T, T = 10^-1 = 0.1, or 10% transmittance when A = 1.

A sample absorbing 10% would have a transmittance of 90%, or T = 0.9. In this case, A ~ 0.046.

So, a sample absorbing 10% won't have a value of A = 1, but rather a sample transmitting 10% will have a value of A = 1. This is one reason why it's a good rule of thumb to keep absorbance values below A = 1, because at higher values of A, the amount of transmittence becomes very low, which can lead to sensitivity issues.

Thank you, Again, and sorry for the language problem. How did you get A to be 4.6 10^-2 if the transmittance is 90 %?

A = -log[T]=-log(0.9) = 0.11

Your method on this is something I greatly appreciate, with respect.

[Corribus]

Thank you, Again, and sorry for the language problem. How did you get A to be 4.6 10^-2 if the transmittance is 90 %?

A = -log[T]=-log(0.9) = 0.11

Your method on this is something I greatly appreciate, with respect.

I've done it on 3 calculators just to be sure. Are you positive you're not using natural logarithm instead of base 10?