A horizontal cylindrical tank 2.20 m in diameter is half full of water. The space above the water is filled with a pressurized gas of unknown refractive index. A small laser can move along the curved bottom of the water and aims a light beam toward the center of the water surface (Fig.). You observe that when the laser has moved a distance S= 1.09 m or more (measured along the curved surface) from the lowest point in the water, no light enters the gas.

(a) What is the index of refraction of the gas?

(b) What minimum time does it take the light beam to travel from the laser to the rim of the tank when (i) S>1.09 m and (ii) S<1.09 m?

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

${\theta }_a=\frac{S}{R}$

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

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

From rad to degree we get the use the next formula

${\theta }_a=\left(0.991rad\right)\frac{180°}{\mathrm{\pi rad}}=56.78°$

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

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

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

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

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

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

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

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

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

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