91Ó°ÊÓ

The indexes of refraction of the layers are:

The thicknesses are,

$$\begin{gathered} \mathrm{Δ}{x}_1=2\text{mm}=2×{10}^{-3}\text{m} \\ \mathrm{Δ}{x}_2=4\text{mm}=4×{10}^{-3}\text{m} \end{gathered}$$

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

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

Use the formula for refractive index,

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

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

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

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

Substitute all the value in the above equation.

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

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

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

$$\begin{gathered} \mathrm{Δ}t=\frac{n\mathrm{Δ}x}{c} \\ \mathrm{Δ}t=\frac{\left(n_2+n_3\right)\mathrm{Δ}{x}_1+n_4\mathrm{Δ}{x}_2}{c} \end{gathered}$$

Substitute all the value in the above equation.

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

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

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

$$\begin{gathered} \mathrm{Δ}t=\frac{n\mathrm{Δ}x}{c} \\ \mathrm{Δ}t=\frac{\left(n_8+n_9\right)\mathrm{Δ}{x}_1+n_7\mathrm{Δ}{x}_2}{c} \end{gathered}$$

Substitute all the value in the above equation.

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

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

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

$$\begin{gathered} \mathrm{Δ}t=\frac{n\mathrm{Δ}x}{c} \\ \mathrm{Δ}t=\frac{\left(n_5+n_9\right)\mathrm{Δ}{x}_1+n_6\mathrm{Δ}{x}_2}{c} \end{gathered}$$

Substitute all the value in the above equation.

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

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

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.