Mathemeticians - I'm being thick.

whyamihere
whyamihere Posts: 7,719
edited February 2011 in The Crudcatcher
I have 14160 reflections of a light ray in a fibre. At each reflection, 0.01% of the intensity is transmitted, rather than reflected. After the first reflection, there will be 99.99% remaining, after the second there will be 99.980001, after the third there will be 99.970003, and so on. Other than writing a program to make the computer loop through this 14160 times and work out the percentage left at the end (pretty easy now, but hard in an exam with no computer), does anyone know how to get the final percentage out?

Edit: I know it's an exponential decay, I just can't remember what form it takes.

Comments

  • welshkev
    welshkev Posts: 9,690
    why would you need to know this? my head hurts just reading it :lol:
  • Cferg
    Cferg Posts: 347
    Would a recurrance relation not work ?
  • whyamihere
    whyamihere Posts: 7,719
    Cferg wrote:
    Would a recurrance relation not work ?
    Certainly would. Still can't remember how the hell to set it up though...

    Edit: Actually, no, it wouldn't. The recurrence relation would be 
    I(n) = [99.99/100]I(n-1)

    I can't seed it with I(n-1), because I don't know it, unless I go through the entire thing, which I don't want to do.
  • whyamihere
    whyamihere Posts: 7,719
    Stopped being thick. For anyone interested, I(n) = [(99.99/100)^14160]I(0), leaving 24.3% at the end.
  • road_donut
    road_donut Posts: 284
    whyamihere wrote:
    Stopped being thick. For anyone interested, I(n) = [(99.99/100)^14160]I(0), leaving 24.3% at the end.

    Ahhhhh you just beat me to it :?
    Santa Cruz Blur TRc now a 5010
  • whyamihere
    whyamihere Posts: 7,719
    welshkev wrote:
    why would you need to know this? my head hurts just reading it :lol:
    Physics degree. This is the trivially easy maths, which is why I can't damn well remember it. Complicated stuff's easier, because I do it all the time, but ask me to add 2+3 and I'm lost.
  • heez29
    heez29 Posts: 612
    whyamihere wrote:
    ... but ask me to add 2+3 and I'm lost.

    6?
  • whyamihere
    whyamihere Posts: 7,719
    For sufficiently large values of 2 and 3, yes. ;)
  • Briggo
    Briggo Posts: 3,537
    whyamihere wrote:
    Stopped being thick. For anyone interested, I(n) = [(99.99/100)^14160]I(0), leaving 24.3% at the end.

    Cant believe you even had to ask the question in the first place.
  • Andy
    Andy Posts: 8,207
    whyamihere wrote:
    Stopped being thick. For anyone interested, I(n) = [(99.99/100)^14160]I(0), leaving 24.3% at the end.

    You seriously need to get laid :wink:
  • whyamihere
    whyamihere Posts: 7,719
    Briggo wrote:
    whyamihere wrote:
    Stopped being thick. For anyone interested, I(n) = [(99.99/100)^14160]I(0), leaving 24.3% at the end.

    Cant believe you even had to ask the question in the first place.
    I'm quite embarrassed to be honest.

    Andy - what's your sister's number? ;)
  • Andy
    Andy Posts: 8,207
    41
  • bennett_346
    bennett_346 Posts: 5,029
    Yeah you're being thick, what were you thinking!
  • addivllangirb
    addivllangirb Posts: 322
    Andy wrote:
    whyamihere wrote:
    Stopped being thick. For anyone interested, I(n) = [(99.99/100)^14160]I(0), leaving 24.3% at the end.

    You seriously need to get laid :wink:

    Well he is doing a physics degree......
    Eldridge Grade 08
  • Anonymous
    Anonymous Posts: 79,665
    heez29 wrote:
    whyamihere wrote:
    ... but ask me to add 2+3 and I'm lost.

    6?
    Don't be stupid, it's 23.
  • El Capitano
    El Capitano Posts: 6,400
    i-can-count-to-potato.jpg
    1 2 3 4 (Dead)5 6

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