The figure shows the design of a Texas arcade game, Four laser pistols are pointed toward the center of an array of plastic layers where a clay armadillo is the target. The indexes of refraction of the layers are ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{55}\mathbf{,}{\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{,}{\mathit{n}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{45}\mathbf{,}{\mathit{n}}_{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{,}{\mathit{n}}_{\mathbf{5}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{45}\mathbf{,}{\mathit{n}}_{\mathbf{6}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{61}\mathbf{,}{\mathit{n}}_{\mathbf{7}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{59}\mathbf{,}{\mathit{n}}_{\mathbf{8}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}$and ${\mathit{n}}_{\mathbf{9}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}$. The layer thicknesses are either 2.00 mm or 4.00 mm, as drawn. What is the travel time through the layers for the laser burst from (a) pistol 1, (b) pistol 2, (c) pistol 3, and (d) pistol 4? (e) If the pistols are fired simultaneously, which laser burst hits the target first?

The indexes of refraction of the layers are:

The thicknesses are,

$$\begin{gathered} \Delta x_1=2\text{mm}=2\times {10}^{-3}\text{m} \\ \Delta x_2=4\text{mm}=4\times {10}^{-3}\text{m} \end{gathered}$$

The speed of light is, $c=3\times {10}^{8}m/s$

$$\begin{gathered} n_1=1.55, \\ {n}_2=1.70, \\ {n}_3=1.45, \\ {n}_4=1.60, \\ {n}_5=1.45, \\ {n}_6=1.61, \\ {n}_7=1.59, \\ {n}_8=1.70, \\ {n}_9=1.60 \end{gathered}$$

Use the formula for refractive index,

$n=\frac{c}{v},\text{and}v=\frac{\Delta x}{\Delta t}$

Here, $c$is speed in air, $v$is speed in medium

Straight forward application of Eq 35.3, $n=\frac{c}{v},\text{and}v=\frac{\Delta x}{\Delta t}$yields the result: pistol 1 with a time to

$$\begin{gathered} \Delta t=\frac{n\Delta x}{c} \\ \Delta t=\frac{\left(n_1+n_2+n_4+n_5\right)\Delta x}{c} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} ∆t=\frac{\left(1.55+1.70+1.60+1.45\right)\times 2\times {10}^{-3}m}{3\times {10}^{8}m/s} \\ =4.2\times {10}^{-11}s \\ =42\times {10}^{-12}s \\ ∆t=42ps \end{gathered}$$

Hence, the travel by laser burst from pistol 1 is $42ps$.

For pistol 2, use the formula and solve as follows:

$$\begin{gathered} \Delta t=\frac{n\Delta x}{c} \\ \Delta t=\frac{\left(n_2+n_3\right)\Delta x_1+n_4\Delta x_2}{c} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} ∆t=\frac{\left(1.70+1.45\right)\times 2\times {10}^{-3}m+1.60\times 4\times {10}^{-3}m}{3\times {10}^{8}m/s} \\ =4.233\times {10}^{-11}s \\ ∆t=42.3\times {10}^{-12}s \end{gathered}$$

Hence the travel time is equal to$42.3\times {10}^{-12}\text{s}$ .

For pistol 3, Use the formula and solve as follows:

$$\begin{gathered} \Delta t=\frac{n\Delta x}{c} \\ \Delta t=\frac{\left(n_8+n_9\right)\Delta x_1+n_7\Delta x_2}{c} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} ∆t=\frac{\left(1.70+1.60\right)\times 2\times {10}^{-3}m+1.59\times 4\times {10}^{-3}m}{3\times {10}^{8}m/s} \\ =4.32\times {10}^{-11}s \\ ∆t=43.2\times {10}^{-12}s \end{gathered}$$

Hence the travel time is equal to $43.2\times {10}^{-12}\text{s}$.

For pistol 4, use the formula and solve as follows:

$$\begin{gathered} \Delta t=\frac{n\Delta x}{c} \\ \Delta t=\frac{\left(n_5+n_9\right)\Delta x_1+n_6\Delta x_2}{c} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} ∆t=\frac{\left(1.45+1.60\right)\times 2\times {10}^{-3}m+1.61\times 4\times {10}^{-3}m}{3\times {10}^{8}m/s} \\ =4.18\times {10}^{-11}s \\ ∆t=41.8\times {10}^{-12}s \end{gathered}$$

The travel time is equal to$41.8\times {10}^{-12}\text{s}$ .

The time taken to travel by pistol 4 is less. So, observe that the blast from pistol 4 arrives first.

Hence, the laser burst hits the target first 4.