The Student Room Group

What is this formula

Hi, can someone tell me what this formula is please?

A=log⁡10IoI=ϵ l cA = \log_{10}\frac{I_o}{I}=\epsilon \ l \ c

Tried typing it into wolfram alpha but it does not recognise it.

Thanks.
Reply 1
Avatar for alow
alow
19
https://www.wikiwand.com/en/Beer%E2%80%93Lambert_law

Specifically for uniform attenuation.
Original post by alow
https://www.wikiwand.com/en/Beer%E2%80%93Lambert_law

Specifically for uniform attenuation.


Thanks! Do you learn this in first year chem?
Reply 3
Avatar for alow
alow
19
Original post by AishaGirl
Thanks! Do you learn this in first year chem?


I learned it in my first year cell biology course.
Original post by alow
I learned it in my first year cell biology course.


Just a quick follow up question you might be able to help me with.

What is the best way to solve this first order ODE −dIdx=κI[A]-\frac{dI}{dx}=\kappa I[A] for intensity II if light transmitted through the length of solution ll where the original intensity is l0l_0

Should I integrate it using the boundary condition of the original light intensity? So I=I0I=I_0 at x=0x=0 ?

Thank you.
(edited 7 years ago)
Reply 5
Avatar for alow
alow
19
Original post by AishaGirl
Just a quick follow up question you might be able to help me with.

What is the best way to solve this first order ODE −dIdx=κI[A]-\frac{dI}{dx}=\kappa I[A] for intensity II if light transmitted through the length of solution ll where the original intensity is l0l_0

Should I integrate it using the boundary condition of the original light intensity? So I=I0I=I_0 at x=0x=0 ?

Thank you.


What is xx? And I assume you mean I0I_0 for the original intensity.

If xx is the coordinate along the length of the tube, then integration like you said would be fine:

∫I0I1I′dI′=−∫0lκ[A]dx\displaystyle \int_{I_0}^{I} \dfrac{1}{I'} \text{d}I' = - \int_{0}^{l} \kappa [A] \text{d}x
Original post by alow
What is xx? And I assume you mean I0I_0 for the original intensity.

If xx is the coordinate along the length of the tube, then integration like you said would be fine:

∫I0I1I′dI′=−∫0lκ[A]dx\displaystyle \int_{I_0}^{I} \dfrac{1}{I'} \text{d}I' = - \int_{0}^{l} \kappa [A] \text{d}x


Yeah it's through the total length.

Thanks.
Original post by alow
What is xx? And I assume you mean I0I_0 for the original intensity.

If xx is the coordinate along the length of the tube, then integration like you said would be fine:

∫I0I1I′dI′=−∫0lκ[A]dx\displaystyle \int_{I_0}^{I} \dfrac{1}{I'} \text{d}I' = - \int_{0}^{l} \kappa [A] \text{d}x


I think you made a typo there, isn't it meant to be ∫I0III′dI′\displaystyle \int_{I_0}^{I} \dfrac{I}{I'} \text{d}I' ?

You put a 1...
(edited 7 years ago)
Reply 8
Avatar for alow
alow
19
Original post by AishaGirl
I think you made a typo there, isn't it meant to be ∫I0III′dI′=−∫0lκ[A]dx\displaystyle \int_{I_0}^{I} \dfrac{I}{I'} \text{d}I' = - \int_{0}^{l} \kappa [A] \text{d}x ?

You put a 1...


Where are you getting that extra I from? II divided through by II then used I′I' as a dummy variable of integration.
Original post by alow
Where are you getting that extra I from? II divided through by II then used I′I' as a dummy variable of integration.


Yes sorry I was a bit lost. Starting to invent variables from thin air lol.

Thanks again.
Reply 10
Avatar for alow
alow
19
Original post by AishaGirl
Yes sorry I was a bit lost. Starting to invent variables from thin air lol.

Thanks again.


No problem :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you