Chapter 21: Problem 16

The fraction of nonreflected radiation that is transmitted through a 5 -mm thickness of a transparent material is $0.95 .$ If the thickness is increased to $12 \mathrm{mm}$, what fraction of light will be transmitted?

Short Answer

Expert verified

Based on the given information and our calculations, when the thickness of the transparent material is increased to 12 mm, approximately 88.1% of non-reflected radiation will be transmitted through the material.

Step by step solution

Calculate transmitted intensities $I_t_1$ and $I_t_2$

Given that the fraction of nonreflected radiation transmitted through the $5$ mm thickness of a transparent material is $0.95$, we can express the ratio of transmitted intensity $I_t_1$ to initial intensity $I_0$ as: $$ T_1 = \frac{I_{t_1}}{I_0} = 0.95 $$ We will find the transmitted intensity $I_t_2$ when the thickness is increased to $12$ mm.

Obtain an equation for absorption coefficient $\alpha$

Using the formula for transmitted intensity, we can write: $$ I_{t_1} = I_{0} e^{-\alpha d_1}, $$ where $d_1 = 5$ mm is the thickness at transmitted fraction $T_1$. We want to find the absorption coefficient $\alpha$.

Calculate the absorption coefficient $\alpha$

We can rewrite the equation in step 2 as: $$ \alpha = \frac{-\ln{\frac{I_{t_1}}{I_{0}}}}{d_1} $$ Substituting $T_1 = \frac{I_{t_1}}{I_0} = 0.95$ and $d_1 = 5$ mm: $$ \alpha = \frac{-\ln{0.95}}{5} $$ Then, calculate the absorption coefficient $\alpha$: $$ \alpha \approx 0.01035/mm $$

Find the intensity $I_t_2$

Now that we have the absorption coefficient $\alpha$, we can find the transmitted intensity $I_t_2$ when the thickness is increased to $12$ mm ($d_2 = 12$ mm) using the formula: $$ I_{t_2} = I_0 e^{-\alpha d_2} $$

Calculate the fraction of light transmitted $T_2$

Finally, we can find the fraction of light transmitted $T_2$ by expressing the ratio of transmitted intensity $I_t_2$ to initial intensity $I_0$ as: $$ T_2 = \frac{I_{t_2}}{I_0} = e^{-\alpha d_2} $$ Substitute $\alpha \approx 0.01035/mm$ and $d_2 = 12$ mm, then calculate the fraction: $$ T_2 \approx e^{-(0.01035)(12)} \approx 0.881 $$ So when the thickness is increased to $12$ mm, approximately $88.1 \%$ of light will be transmitted through the transparent material.

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The fraction of nonreflected radiation that is transmitted through a 5 -mm thickness of a transparent material is \(0.95 .\) If the thickness is increased to \(12 \mathrm{mm}\), what fraction of light will be transmitted?

Short Answer

Step by step solution

Calculate transmitted intensities \(I_t_1\) and \(I_t_2\)

Obtain an equation for absorption coefficient \(\alpha\)

Calculate the absorption coefficient \(\alpha\)

Find the intensity \(I_t_2\)

Calculate the fraction of light transmitted \(T_2\)

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