Question

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (with an incident intensity, lo) after it has traveled through 1 cm of the material? After it has traveled through 2 cm of the material? (b) What is the optical density of a 1 cm thick piece of such a material? Of a 2 cm thick piece of such a material?

Answer 1

Answer:

a.

• After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

• After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

b.

• After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

• After it has traveled through 2 cm :  $od( 2\ cm) = 0.5211$

Explanation:

a.

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

b

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$.

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$