A horizontal cylindrical tank 2.20 m in diameter is half...

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 ($\textbf{Fig. P33.52}$). 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?

Key Concepts

Speed of Light in Media

The speed at which light travels through a medium is given by the speed in vacuum divided by the medium's index of refraction. This relationship is crucial for calculating time-of-flight measurements and understanding the delay in optical paths within different materials.

Snell's Law

This fundamental principle of optics describes how light rays bend when passing from one medium into another with a different index of refraction. It relates the angles of incidence and refraction to the indices of refraction of the two media, forming the basis for understanding refraction phenomena at interfaces.

Total Internal Reflection and Critical Angle

When light passes from a medium with a higher index of refraction to one with a lower index, there is a specific incident angle—the critical angle—above which all light is reflected back into the original medium rather than refracted. This concept is essential in explaining scenarios where light fails to exit a medium.

Index of Refraction

The index of refraction is a measure of how much the speed of light is reduced in a medium compared to the speed of light in a vacuum. It not only determines the bending of light at boundaries but also influences the propagation speed of light through the medium.

Geometric Relationships in Circular Systems

Understanding the geometry of circles, particularly arc lengths and their relation to central angles, is essential in problems involving curved surfaces. This concept helps in determining distances traveled along curved paths and in setting up the correct spatial relationships for optical trajectories.

Transcript

00:01 Okay, so for the first question, we know sine theta c is equal to ng over nw.

00:06 So theta c here is the critical angle.

00:08 Ng is the index of refraction of gas, which is the one we are looking for to determine in this question.

00:14 And nw is the index of refraction in water.

00:17 So therefore, we have ng is equal nw sine theta c.

00:19 We know the critical angle can be equal to the move distance divided by the radius, which is 1 .09 meter over 2 .20 meter over 2, since the diameter is given as 2 .20 meter.

00:30 So therefore, radius is half diameter, which is 2 .20 meter divided by 2.

00:36 So eventually we have theta c is equal to 0 .9 radiance.

00:38 If we convert it to angles, it's about 56 .8 degree.

00:44 So we know the index of refraction in water is 1 .33.

00:49 So therefore, we can determine the index of refraction in the gas, which is ng is equal to 1 .33 times sine.

01:05 56 .8 degree, which is equal to 1 .11.

01:13 And for the next question, well, there are two sub questions.

01:16 So first sub question, so when the move distance is greater than 1 .09 meter, the time should be equal to 2t1, which is 2 times nw times r over c.

01:29 Okay? because when s is greater than 1 .09 meter, the angle of instant is greater than the critical angle...

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