91Ó°ÊÓ

The angle between the radius R and the normal line on the interface between the gas and the water represents the critical angle${\mathrm{θ}}_a$. The very small difference between S and R could make this angle be calculated by

${\mathrm{θ}}_a=\frac{S}{R}$

Plug the values for S and R to get in radians then convert them into degrees

${\mathrm{θ}}_a=\frac{S}{R}=\frac{1.09m}{1.10m}=0.991rad$

From rad to degree we get the use the next formula

${\mathrm{θ}}_a=\left(0.991rad\right)\frac{180°}{\mathrm{Ï€°ù²¹»å}}=56.78°$

The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by the speed of light in the material. Snell's law is given by an equation in the form

$n_a\sin {\mathrm{θ}}_a=n_b\sin {\mathrm{θ}}_b$ (1)

Where$n_a$is the refractive index of the water equals 1.333. The total reflection occurs, therefore, the angle${\mathrm{θ}}_b=90°$. Now, we solve equation (1) for$n_b$and substitute the values of${\mathrm{θ}}_a$and$n_a$to get$n_b$

$$\begin{gathered} n_b=n_a\sin {\mathrm{θ}}_a \\ {n}_b=\left(1.333\right)\sin 56.78° \\ {n}_b=1.12 \end{gathered}$$

$t=\frac{d}{v}=\frac{2R}{\left(c/n_a\right)}=\frac{2R{n}_a}{c}$ (2)

Now, plug the values for R, c and into equation (2) to get

$$\begin{gathered} t=\frac{2R{n}_a}{c} \\ =\frac{2\left(1.10m\right)\left(1.333\right)}{3×{10}^{-9}s} \\ =9.78×{10}^{-9}s \\ =9.78ns \end{gathered}$$

$t={\left(\frac{2R{n}_a}{c}\right)}_{water}+{\left(\frac{2R{n}_a}{c}\right)}_{gas}$ (3)

Now, plug the values for R, c, and into equation (3) to get t

$$\begin{gathered} t={\left(\frac{2R{n}_a}{c}\right)}_{water}+{\left(\frac{2R{n}_b}{c}\right)}_{gas} \\ =\frac{\left(1.10m\right)\left(1.333\right)}{3×{10}^{8}m/s}+\frac{\left(1.10m\right)\left(1.12\right)}{3×{10}^{8}m/s} \\ =8.98×{10}^{-9}s \\ =8.98ns \end{gathered}$$