In Fig. 35-33, two light pulses are sent through layers of plastic with thicknesses of either $\mathit{L}$or $\mathbf{2}\mathit{L}$as shown and indexes of refraction ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{55}$, ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}$, ${\mathit{n}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}$, ${\mathit{n}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{45}$,${\mathit{n}}_{\mathbf{5}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{59}$ , ${\mathit{n}}_{\mathbf{6}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{65}$ and ${\mathit{n}}_{\mathbf{7}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}$. (a) Which pulse travels through the plastic in less time? (b) What multiple of $\frac{\mathbf{L}}{\mathbf{c}}$gives the difference in the traversal times of the pulses?

Thickness is$L$ or $2L$.

The index of refractions are as below.

$$\begin{gathered} n_1=1.55 \\ {n}_2=1.70 \\ {n}_3=1.60 \\ {n}_4=1.45 \\ {n}_5=1.59 \\ {n}_6=1.65 \\ {n}_7=1.50 \end{gathered}$$

The expression to calculate the time is given as follows.

$\mathit{t}\mathbf{=}\frac{\mathbf{d}}{\mathbf{v}}$ ......(1)

Here, $\mathit{d}$ is the distance and role="math" localid="1663063320152" $\mathit{v}$is the speed.

The expression to calculate the difference in traversal time is given as follows.

$\mathit{\Delta }\mathit{t}\mathbf{=}{\mathit{t}}_{\mathbf{1}}\mathbf{-}{\mathit{t}}_{\mathbf{2}}$ ......(2)

Here,${\mathit{t}}_{\mathbf{1}}$ is the time for the pulse 1and${\mathit{t}}_{\mathbf{2}}$ is time for the pulse 2 .

The speed of light in the medium, $v=\frac{c}{n}$

Calculate the time of travelling pulse through medium,

Substitute $\frac{c}{n}$ for $v$into equation (1).

$\begin{array}{rcl}{t}_1& =& \frac{d}{\left(\frac{c}{n}\right)}\\ & =& \frac{d}{c}n\end{array}$

By using the above equation, calculate the time for pulse 1.

$\begin{array}{rcl}{t}_1& =& \frac{2L}{c}\times n_5+\frac{L}{c}\times n_6+\frac{L}{c}\times n_7\\ & =& \frac{L}{c}\left(2n_5+n_6+n_7\right)\end{array}$

Substitute $1.59$ for $n_5$, $1.65$for $n_6$ and localid="1663064157431" $1.50$ for localid="1663064176819" $n_7$ into above equation.

$\begin{array}{rcl}{t}_1& =& \frac{L}{c}\left(2\times 1.59+1.65+1.50\right)\\ & =& \frac{L}{c}\left(6.33\right)\\ & =& 6.33\frac{L}{c}\end{array}$

Calculate the time for the pulse 2.

$\begin{array}{rcl}{t}_2& =& \frac{L}{c}{n}_1+\frac{L}{c}{n}_2+\frac{L}{c}{n}_3+\frac{L}{c}{n}_4\\ & =& \frac{L}{c}\left(n_1+n_2+n_3+n_4\right)\end{array}$

Substitute $1.55$ for $n_1$, $1.70$for $n_2$, $1.60$ for $n_3$ , and $1.45$ for $n_4$ into above equation.

$\begin{array}{rcl}{t}_2& =& \frac{L}{c}\left(1.55+1.70+1.60+1.45\right)\\ & =& \frac{L}{c}\left(6.3\right)\\ & =& 6.3\frac{L}{c}\end{array}$

Hence, the pulse 2 take less than to travel through plastic.

Calculate the difference in the traversal of the time.

Substitute $6.33\frac{L}{c}$for $t_1$ and $6.3\frac{L}{c}$ for $t_2$ into equation (2)

$\begin{array}{rcl}\Delta t& =& 6.33\frac{L}{c}-6.3\frac{L}{c}\\ & =& 0.03\frac{L}{c}\end{array}$

Hence, the multiple of $\frac{L}{c}$ which gives the difference in the traversal of the pulses time is$0.03$ .