# Just took my final but I think there was an invalid question

Discussion in 'Odds & Ends' started by MarinaHighTennis, Dec 7, 2011.

1. ### MarinaHighTennisProfessional

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Last edited: Dec 7, 2011
2. ### TalkerHall of Fame

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Isn't I( initial ) just set to 100? No matter what it is (I) would be a percentage of that if I'm looking at this right.

Good luck!

3. ### MarinaHighTennisProfessional

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Sounds logical, I wish our teacher told us that. Brainfuse didn't even know how to do this problem
I already took my final, but I was stuck on this question.

4. ### thug the bunnyProfessional

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MHT, I think you are right, something is wrong with the question.

Transmittance should be unitless, either as a ratio <1, or as a percentage, not EU (is that Euros?). Absorbance is in units of AU (absorbance units). And, why ask what Io is, when it is always defined as 100%? Doesn't make sense...

5. ### MarinaHighTennisProfessional

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I hope the teacher didn't screw us over for the final, but EU is energy units so its not really a unit of measurement.

6. ### tolyHall of Fame

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By definition , see please http://en.wikipedia.org/wiki/Transmittance ,transmittance is

T=Iout/Iin (1)

where Iin is the intensity of the incident radiation and Iout is the intensity of radiation coming out of the sample.

The absorbance is

A=Log(Iin/Iout) (2)

We know that Iout = 10 EU and A = 2.
From (2) we have

Iin / Iout = 10** A = 10**2=100.

Thus T = 1/100 =1%

From (1)
Iin = Iout*100 = 1000 EU

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8. ### MarinaHighTennisProfessional

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Thanks for the help but one thing that puzzled me is that If A= log(1/T) and T=10 it comes out to be -1. Do we ignore the T=10?

9. ### tolyHall of Fame

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According to definition T=Iout/Iin.

Thus, T is always less or equal to 1 (0<T<1).

In your case: T=0.01=1%, Iout=10 EU, and Iin =1000 EU.

If T=1/100 than 1/T=100 and A=log(1/T)=log(100)=2.

EU is complicated energy unit.

Last edited: Dec 8, 2011
10. ### BaboFanRookie

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Correct me if I'm wrong but isn't 2 Max absorbtion? So t is? 0% and I 0

11. ### tolyHall of Fame

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If (0<T<1) than (∞>1/T>1)

A=log(1/T)

If 1/T=1 than A= log(1)=0

If 1/T=∞ than A=log(∞)=∞

So, (0<A<∞)

and Amax=∞

Last edited: Dec 8, 2011