Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    Hi, can someone tell me what this formula is please?

    A = \log_{10}\frac{I_o}{I}=\epsilon \ l \ c

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

    Thanks.
    Offline

    19
    ReputationRep:
    https://www.wikiwand.com/en/Beer%E2%80%93Lambert_law

    Specifically for uniform attenuation.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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?
    Offline

    19
    ReputationRep:
    (Original post by AishaGirl)
    Thanks! Do you learn this in first year chem?
    I learned it in my first year cell biology course.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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 -\frac{dI}{dx}=\kappa I[A] for intensity I if light transmitted through the length of solution l where the original intensity is l_0

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

    Thank you.
    Offline

    19
    ReputationRep:
    (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 -\frac{dI}{dx}=\kappa I[A] for intensity I if light transmitted through the length of solution l where the original intensity is l_0

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

    Thank you.
    What is x? And I assume you mean I_0 for the original intensity.

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

    \displaystyle \int_{I_0}^{I} \dfrac{1}{I'} \text{d}I' = - \int_{0}^{l} \kappa [A] \text{d}x
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by alow)
    What is x? And I assume you mean I_0 for the original intensity.

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

    \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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by alow)
    What is x? And I assume you mean I_0 for the original intensity.

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

    \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 \displaystyle \int_{I_0}^{I} \dfrac{I}{I'} \text{d}I' ?

    You put a 1...
    Offline

    19
    ReputationRep:
    (Original post by AishaGirl)
    I think you made a typo there, isn't it meant to be \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? I divided through by I then used I' as a dummy variable of integration.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by alow)
    Where are you getting that extra I from? I divided through by I then used I' as a dummy variable of integration.
    Yes sorry I was a bit lost. Starting to invent variables from thin air lol.

    Thanks again.
    Offline

    19
    ReputationRep:
    (Original post by AishaGirl)
    Yes sorry I was a bit lost. Starting to invent variables from thin air lol.

    Thanks again.
    No problem
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.