Q52P

Question

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?

Step-by-Step Solution

Verified

Answer

a) (a) The refractive index of the gas is 1.12.

b) (i) t= 9.78 ns (ii) t = 8.98 ns.

1Step 1: calculate the refractive index.

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

$θ_{a} = \frac{S}{R}$

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

$θ_{a} = \frac{S}{R} = \frac{1.09 m}{1.10 m} = 0.991 r a d$

From rad to degree we get the use the next formula

$θ_{a} = (0.991 r a d) \frac{180 °}{πrad} = 56.78 °$

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

$n_{a} \sin θ_{a} = n_{b} \sin θ_{b}$ (1)

Where $n_{a}$ is the refractive index of the water equals 1.333. The total reflection occurs, therefore, the angle $θ_{b} = 90 °$ . Now, we solve equation (1) for $n_{b}$ and substitute the values of $θ_{a}$ and $n_{a}$ to get $n_{b}$

$n_{b} = n_{a} \sin θ_{a} n_{b} = (1.333) \sin 56.78 ° n_{b} = 1.12$

2Step 2: Calculate the travel time of the light beam.

(i) The ray travel a distance R to the surface of the gas, then another distance R to the rim, so, the total distance that the light travels is d = 2R with the speed v but not the speed of light c as it travels through water. So, from the equation, we could get this speed $v = \frac{c}{n_{a}}$ . Now. we can get the time by dividing the distance d by the speed v

$t = \frac{d}{v} = \frac{2 R}{(c / n_{a})} = \frac{2 R n_{a}}{c}$ (2)

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

$t = \frac{2 R n_{a}}{c} = \frac{2 (1.10 m) (1.333)}{3 \times 10^{- 9} s} = 9.78 \times 10^{- 9} s = 9.78 n s$

(ii) At distance S<1.09, the total reflection doesn't occur and refraction is occurring to the light and reach the rim at the side of the gas. So, it travels two distances, first through the water R and the second is R through the gas. Hence, we can get the time t by

$t = {(\frac{2 R n_{a}}{c})}_{w a t e r} + {(\frac{2 R n_{a}}{c})}_{g a s}$ (3)

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

$t = {(\frac{2 R n_{a}}{c})}_{w a t e r} + {(\frac{2 R n_{b}}{c})}_{g a s} = \frac{(1.10 m) (1.333)}{3 \times 10^{8} m / s} + \frac{(1.10 m) (1.12)}{3 \times 10^{8} m / s} = 8.98 \times 10^{- 9} s = 8.98 n s$

Q51P

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