Answer:
<h3>a.</h3>
- After it has traveled through 1 cm :
![I(1 \ cm) = 0.5488 I_0](https://tex.z-dn.net/?f=I%281%20%5C%20cm%29%20%3D%200.5488%20I_0%20)
- After it has traveled through 2 cm :
![I(2 \ cm) = 0.3012 I_0](https://tex.z-dn.net/?f=I%282%20%5C%20cm%29%20%3D%200.3012%20I_0%20)
<h3>b.</h3>
- After it has traveled through 1 cm :
![od( 1\ cm) = 0.2606](https://tex.z-dn.net/?f=od%28%201%5C%20cm%29%20%3D%20%200.2606)
- After it has traveled through 2 cm :
![od( 2\ cm) = 0.5211](https://tex.z-dn.net/?f=od%28%202%5C%20cm%29%20%3D%20%200.5211)
Explanation:
<h2>
a.</h2>
For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient
the formula is:
![I(x) = I_0 e^{-\mu x}](https://tex.z-dn.net/?f=I%28x%29%20%3D%20I_0%20e%5E%7B-%5Cmu%20x%7D)
where I is the intensity of the beam,
is the incident intensity and x is the length of the material traveled.
For our problem, after travelling 1 cm:
![I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}](https://tex.z-dn.net/?f=I%281%20%5C%20cm%29%20%3D%20I_0%20e%5E%7B-%200.6%20%5Cfrac%7B1%7D%7Bcm%7D%20%5C%201%20cm%7D)
![I(1 \ cm) = I_0 e^{- 0.6}](https://tex.z-dn.net/?f=I%281%20%5C%20cm%29%20%3D%20I_0%20e%5E%7B-%200.6%7D)
![I(1 \ cm) = I_0 e^{- 0.6}](https://tex.z-dn.net/?f=I%281%20%5C%20cm%29%20%3D%20I_0%20e%5E%7B-%200.6%7D)
![I(1 \ cm) = 0.5488 \ I_0](https://tex.z-dn.net/?f=I%281%20%5C%20cm%29%20%3D%200.5488%20%5C%20I_0%20)
After travelling 2 cm:
![I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}](https://tex.z-dn.net/?f=I%282%20%5C%20cm%29%20%3D%20I_0%20e%5E%7B-%200.6%20%5Cfrac%7B1%7D%7Bcm%7D%20%5C%202%20cm%7D)
![I(2 \ cm) = I_0 e^{- 1.2}](https://tex.z-dn.net/?f=I%282%20%5C%20cm%29%20%3D%20I_0%20e%5E%7B-%201.2%7D)
![I(2 \ cm) = I_0 e^{- 1.2}](https://tex.z-dn.net/?f=I%282%20%5C%20cm%29%20%3D%20I_0%20e%5E%7B-%201.2%7D)
![I(2 \ cm) = 0.3012 \ I_0](https://tex.z-dn.net/?f=I%282%20%5C%20cm%29%20%3D%200.3012%20%5C%20I_0%20)
<h2>b</h2>
The optical density od is given by:
.
So, after travelling 1 cm:
![od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )](https://tex.z-dn.net/?f=od%28%201%5C%20cm%29%20%3D%20-%20log_%7B10%7D%20%28%20%5Cfrac%7B0.5488%20%5C%20I_0%7D%7BI_0%7D%20%29)
![od( 1\ cm) = - log_{10} ( 0.5488 )](https://tex.z-dn.net/?f=od%28%201%5C%20cm%29%20%3D%20-%20log_%7B10%7D%20%28%200.5488%20%29)
![od( 1\ cm) = - ( - 0.2606)](https://tex.z-dn.net/?f=od%28%201%5C%20cm%29%20%3D%20-%20%28%20%20-%200.2606%29)
![od( 1\ cm) = 0.2606](https://tex.z-dn.net/?f=od%28%201%5C%20cm%29%20%3D%20%200.2606)
After travelling 2 cm:
![od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )](https://tex.z-dn.net/?f=od%28%202%5C%20cm%29%20%3D%20-%20log_%7B10%7D%20%28%20%5Cfrac%7B0.3012%20%5C%20I_0%7D%7BI_0%7D%20%29)
![od( 2\ cm) = - log_{10} ( 0.3012 )](https://tex.z-dn.net/?f=od%28%202%5C%20cm%29%20%3D%20-%20log_%7B10%7D%20%28%200.3012%20%29)
![od( 2\ cm) = - ( - 0.5211)](https://tex.z-dn.net/?f=od%28%202%5C%20cm%29%20%3D%20-%20%28%20%20-%200.5211%29)
![od( 2\ cm) = 0.5211](https://tex.z-dn.net/?f=od%28%202%5C%20cm%29%20%3D%20%200.5211)
Answer:
Target Behavior would be the behavior known to be changed, it can be defined by function or topography
When two forces act in the same direction, they add together. ... Equal forces acting in opposite directions are called balanced forces. Balanced forces acting on an object will not change the object's motion. When you add equal forces in opposite direction, the net force is zero.
Answer:
A
Explanation:
you see red because it's reflected and other colors are absorbed, light transmitted through something is when it travels through something like glass or a gem