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fblade
How did you guys find A and B that equation?


Take the log to base e (or ln)...ln A was the intercept and B was gradient..well that's what I did
libertine17
I'm really worried now, I was resitting it this june because I got a C in wave properties/ practical module last year,
and i forgot to put the frigging gradient as negative
****!


I say revise the wave properties well because it is weighted more than the practical.
Reply 42
londonkid89
Take the log to base e (or ln)...ln A was the intercept and B was gradient..well that's what I did

Nice one :biggrin:.
Reply 43
corbinator44
the gradient didnt have a unit as it was a ln against n graph so no units, thats what i think anyway


Why doesn't the ln V have a unit?

vb07
A good check of your accuracy was to do e^intercept ie. your value for A and comparing it to your value for V0. Ideally, they should be the same.

Yeah mMine was kinda near it +/- 3 V
Reply 44
fblade
Why doesn't the ln V have a unit

its because you are taking the log of it... its not voltage anymore. remember when you square something (e.g. distance) it becomes m2... the unit changes. i think its like that. ln V probably does have a unit, its just we're never thought what the unit is. thats my underesting...

EDIT: Actually I think it has something to do with the fact that powers dont have a unit. hmm...
Reply 45
Coolraj
its because you are taking the log of it... its not voltage anymore. remember when you square something (e.g. distance) it becomes m2... the unit changes. i think its like that. ln V probably does have a unit, its just we're never thought what the unit is. thats my underesting...

hmm... I'll ask my teacher. I was writing down the unit, volts, in th exam. But then my teacher told another guy it doesn't have units, I rushed to rub it out :P
Reply 46
OOps. My paper was a rushed, inaccurate, scruffy mess. Ah well. :redface:
fblade
hmm... I'll ask my teacher. I was writing down the unit, volts, in th exam. But then my teacher told another guy it doesn't have units, I rushed to rub it out :P


Your teacher can say bye bye to their job if that gets out.
Reply 48
i think i may have got my thicknesses for the second question wrong.

my thickness for just one came out to be 0.62mm, and i found i needed 14 slides.
Reply 49
nervosa
i didnt really understand the 50% light intensity..can someone help me...how to work this out, it was really bugging me, because my thickness ended being something like 5mm and i know thats not right. im gutted because i think i forgot the minus on the gradient, dammit easy marks down the bin!!!




This 1 was tricky... i couldnt do the maths on it. my original reading was 101.1mV so i found 50% which was 50.55mV. then i went back to the light detector and kept addding glass slides untill the voltage dropped to 50.55mV, this was 18 slides, 1 slide was 1.56mm....so 18x1.56 is 28.08mm of slides. thats how i got it!!

..:: Sim ::..
Simplest and most accurate way is to use a half life analogy

Half thickness = ln2 / -gradient
Reply 51
What was the answer to the micrometer percentage uncertainty one? Was it around 1% or 10%? I put 10% myself.

(Also it was slightly embarrassing to see that despite being physicists for 7 years in a fairly good grammar school, very few of us were actually able to use or read correctly a micrometer)
should have been 0.01/ your value x 100 so around 1 ish

o and the ln dont have units because you are taking the ln of a number measured in mV so it dissapears the unit thats why it was ln (V/mV)
Coolraj
its because you are taking the log of it... its not voltage anymore. remember when you square something (e.g. distance) it becomes m2... the unit changes. i think its like that. ln V probably does have a unit, its just we're never thought what the unit is. thats my underesting...

EDIT: Actually I think it has something to do with the fact that powers dont have a unit. hmm...


I thought that logs didn't have units? So whatever unit a quantity may have - when you log it, it becomes unitless.

fblade
hmm... I'll ask my teacher. I was writing down the unit, volts, in th exam. But then my teacher told another guy it doesn't have units, I rushed to rub it out :P


Isn't that called Gross Professional Misconduct?
Reply 54
tux01989
Isn't that called Gross Professional Misconduct?

Meh
Reply 55
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(Also it was slightly embarrassing to see that despite being physicists for 7 years in a fairly good grammar school, very few of us were actually able to use or read correctly a micrometer)

I measured the glass slide with my ruler, and then tried to interpret the micrometer so that my reading from it would be close to my estimate.
tux01989

Isn't that called Gross Professional Misconduct?


It certainly is.
Any teacher who helps students in an exam or breaks exam security would face disciplinary action at the least and could lose their teacher registration so they'd never work in teaching again , at the most. And yes, teachers do this and get caught every year. The pressure to get good results makes folk do crazy things.
Reply 57
tux01989
I thought that logs didn't have units? So whatever unit a quantity may have - when you log it, it becomes unitless.

yh.. i'm aware of that. but why? there must be a mathematical explanation.
Because you can only raise something to a power that is a pure number.

3m to the power of 5s doesnt make any sense.
Reply 59
if you look at the question it asked to take ln of (V/mV) or something along those lines....thats because the voltage (i.e the value of V) has the value and an associated unit namely mV so the mV's cancel and you're left with a pure number.

e.g. if you found your voltage as 200 mV (say) you can't do ln 200mV..so you do ln (200mV/mV) i.e. ln 200 :smile:

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