Chapter 31, Problem 82P

(a)

To determine

The proof that $I_{\text{T}}=I_0{\left[\frac{4n}{{\left(n+1\right)}^2}\right]}^2$ .

Answer to Problem 82P

It is proved that $I_0{\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^2$ .

Explanation of Solution

Formula used:

The expression for $I_{\text{T}}$ is given by,

  $I_{\text{T}}=I_{\text{glass}}-I_{\text{R,2}}$

The expression for $I_{\text{glass}}$ is given by,

  $I_{\text{glass}}=I_0\left[\frac{4n}{{\left(1+n\right)}^2}\right]$

The expression for $I_{\text{R,2}}$ is given by,

  $I_{\text{R,2}}={\left(\frac{1-n}{1+n}\right)}^2{I}_{\text{glass}}$

Calculation:

The expression for $I_{\text{R,2}}$ is calculated as,

  $\begin{array}{c}{I}_{\text{R,2}}={\left(\frac{1-n}{1+n}\right)}^2{I}_{\text{glass}}\\ ={\left(\frac{1-n}{1+n}\right)}^2\left[I_0\left[\frac{4n}{{\left(1+n\right)}^2}\right]\right]\end{array}$

The expression for $I_{\text{T}}$ is calculated as,

  $\begin{array}{c}{I}_{\text{T}}=I_{\text{glass}}-I_{\text{R,2}}\\ =I_0\left[\frac{4n}{{\left(1+n\right)}^2}\right]-{\left(\frac{1-n}{1+n}\right)}^2\left[I_0\left[\frac{4n}{{\left(1+n\right)}^2}\right]\right]\\ =I_0\left[1-{\left(\frac{1-n}{1+n}\right)}^2\right]\left[\frac{4n}{{\left(1+n\right)}^2}\right]\\ =I_0{\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^2\end{array}$

Conclusion:

Therefore, it is proved that $I_0{\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^2$ .

(b)

To determine

The ratio of transmitted intensity and incident intensity.

Answer to Problem 82P

The ratio of transmitted intensity and incident intensity is ${\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^{2N}$ .

Explanation of Solution

Formula used:

The expression for transmitted intensity for $N$ is given by,

  $I_{\text{T}}=I_0{\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^{2N}$

Calculation:

The ratio of transmitted intensity and incident intensity is calculated as,

  $\begin{array}{c}{I}_{\text{T}}=I_0{\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^{2N}\\ \frac{I_{\text{T}}}{I_0}={\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^{2N}\end{array}$

Conclusion:

Therefore, the ratio of transmitted intensity and incident intensity is ${\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^{2N}$ .

(c)

To determine

The number of slabs.

Answer to Problem 82P

The number of slabs is $28$ .

Explanation of Solution

Given:

The refraction index is $n=1.5$ .

Calculation:

The number of slabs is calculated as,

  $\begin{array}{c}\frac{I_{\text{T}}}{I_0}={\left[\frac{4n}{{\left(1+n\right)}^2}\right]}^{2N}\\ 0.1={\left[\frac{4\left(1.5\right)}{{\left(1+1.5\right)}^2}\right]}^{2N}\\ N=\frac{\log \left(0.1\right)}{2\log \left[\frac{4\left(1.5\right)}{{\left(1+1.5\right)}^2}\right]}\\ \approx 28\end{array}$

Conclusion:

Therefore, the number of slabs is $28$ .