Calculating Light Attenuation in Glass: A Homework Problem

[NotoriousNick]

Homework Statement

If a glass material has losses at 10^3 dB/km, what percentage of the original energy exists after moving 30m down the glass material.

Homework Equations

Intensity = Power / Area = (Energy / time ) / Area

dB = 10 log (Output Intensity / Input Intensity )

The Attempt at a Solution

Began by trying to cancel out the area and time in the Intensity equation to apply initial and final energy to the dB equation. Leaves:

dB = 10 log (Energy output / Initial Energy )

Took entire equation and divided it by km to make units jive.

Then (not sure if after previous step this next step is valid):

Solving for logarithm:

10^10^2 = Eo / Ei ( per kilometer)

Then, looking at the problem kind of in reverse, comparing the beginning of hte signal at the start of the kilometer to the end at the attenuated point, then does that mean that this is how much the signal is "amplified" from that attenuated point to the starting point.
(realizing because it's not linear this line of thinking is probably wrong)

Therefore, 10^100 x ( 1000 m / 1km ) (30m) = this is so wrong.

Help!

 

[mgb_phys]

It's much simpler than that - believe it or not dB was actually invented to make this simpler.
You lose 10^3 dB/km or 1dB/m so after 30m you have lost 30dB, from the definition of dB you have given what is the percentage of 30dB?