A scuba diver training in a pool looks at his instructor as shown below. What angle does the ray from the instructor's face make with the perpendicular to the water at the point where the ray enters? The angle between the ray in the water and the perpendicular to the water is \(25.0^{\circ}\).

Short Answer

Expert verified
The angle the ray from the instructor's face makes with the perpendicular to the water at the point where the ray enters is approximately \(24.62^{\circ}\).

Step by step solution

01

Understand Snell's Law

Snell's Law states that the ratio of the sines of the angles of incidence (\(\theta_1\)) and refraction (\(\theta_2\)) is equal to the ratio of the refractive indices of the two media. It is given by the formula: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\) Here, \(n_1\) and \(n_2\) represent the refractive indices of the first and second media respectively, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction. In this problem, we have a ray traveling from water to air. The refractive index of water is approximately 1.33, and the refractive index of air is approximately 1.00.
02

Determine the given information

We know the angle between the ray in the water and the perpendicular to the water surface is \(25.0^{\circ}\). We can use this to find the angle of incidence, as the angle of incidence is equal to the complementary angle of the given angle. Thus, we have: \(\theta_1 = 90^{\circ} - 25.0^{\circ} = 65.0^{\circ}\).
03

Applying Snell's Law to solve for the angle of refraction

We will now use Snell's Law to solve for the angle of refraction (\(\theta_2\)). We have \(n_1 = 1.33\), \(n_2 = 1.00\), and \(\theta_1 = 65.0^{\circ}\). Plugging these values into Snell's Law equation, we get: \(1.33 \sin(65.0^{\circ}) = 1.00 \sin(\theta_2)\) Now we can solve for \(\theta_2\): \(\theta_2 = \sin^{-1} \left(\frac{1.33 \sin(65.0^{\circ})}{1.00}\right)\)
04

Calculate the angle of refraction

Using a calculator or mathematical software, compute the value of \(\theta_2\): \(\theta_2 \approx \sin^{-1} (1.0976) \approx 65.38^{\circ}\) Finally, since we want to find the angle the ray makes with the perpendicular to the water surface, we must calculate the complementary angle of \(\theta_2\):
05

Find the final angle

The final angle, which is the angle between the ray from the instructor's face and the perpendicular to the water at the point where the ray enters, can be found by: Final angle = \(90^{\circ} - \theta_2 = 90^{\circ} - 65.38^{\circ} \approx 24.62^{\circ}\) Thus, the angle the ray from the instructor's face makes with the perpendicular to the water at the point where the ray enters is approximately \(24.62^{\circ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Incident Ray and Snell's Law
The incident ray in optics is the ray that strikes a surface, making contact with it at a specific point and angle. In the context of Snell's Law, the incident ray is a crucial concept because it helps determine the path of light as it moves between different media with varying optical densities.

When light travels through different substances, it bends at the interface; this bending is due to the change in speed of light in different materials. The incident ray and the refracted ray are related to each other by Snell's Law, which is written as: \[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\].
Here, \(n_1\) and \(n_2\) are the refractive indices of the two media, and \(\theta_1\) is the angle of incidence—the angle between the incident ray and the perpendicular to the surface at the point of contact. This law can be employed to solve various optical phenomena such as the bending of light when moving from air to water, which is a common scenario for swimmers and divers.
Refractive Index
The refractive index is a dimensionless number that indicates how much the speed of light is reduced in a particular medium compared to the speed of light in a vacuum. When light moves from a medium with a low refractive index to a medium with a higher refractive index, it bends towards the normal, or perpendicular line, at the point of incidence. Conversely, it bends away from the normal when moving into a less dense medium.

In the given scuba diver problem, the refractive indices provided for water and air are essential in understanding the behavior of light and using Snell's Law effectively. Interestingly, temperature and wavelength can affect the refractive index, although these considerations are not part of this particular problem. An important point for students to remember is that the refractive index helps us predict how much an incident ray will bend upon entering a different medium.
Angle of Refraction
The angle of refraction is the angle between the refracted ray and the normal to the surface at the point where the light enters the new medium. In Snell's Law, this angle demonstrates how much the path of the incident ray has been altered due to entering a new medium.

In the exercise involving the scuba diver, we calculated this angle to understand the change in direction of the ray from the instructor's face as it leaves the water. It is important to grasp that the angle of refraction is dependent on both the angle of incidence and the ratio of the refractive indices of the two media. If the incident ray enters the medium at a larger angle with respect to the normal, the refracted ray will bend more significantly, and the angle of refraction will reflect this change.
When solving for the angle of refraction, there's usually a step involving inverse trigonometric functions, as demonstrated in the 'step by step solution'. Acknowledging that the angle of refraction can also give us information about the perceived position of objects under or above the water will help students understand phenomena like the 'bent pencil' illusion or why the pool's depth appears shallower than it really is.

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Most popular questions from this chapter

The light incident on polarizing sheet \(P_{1}\) is linearly polarized at an angle of \(30.0^{\circ}\) with respect to the transmission axis of \(\mathrm{P}_{1} .\) Sheet \(\mathrm{P}_{2}\) is placed so that its axis is parallel to the polarization axis of the incident light, that is, also at \(30.0^{\circ}\) with respect to \(\mathrm{P}_{1}\). (a) What fraction of the incident light passes through \(P_{1} ?\) (b) What fraction of the incident light is passed by the combination? (c) By rotating \(\mathrm{P}_{2},\) a maximum in transmitted intensity is obtained. What is the ratio of this maximum intensity to the intensity of transmitted light when \(P_{2}\) is at \(30.0^{\circ}\) with respect to \(\mathrm{P}_{1} ?\)

From his measurements, Roemer estimated that it took 22 min for light to travel a distance equal to the diameter of Earth's orbit around the Sun. (a) Use this estimate along with the known diameter of Earth's orbit to obtain a rough value of the speed of light. (b) Light actually takes \(16.5 \mathrm{min}\) to travel this distance. Use this time to calculate the speed of light.

At what minimum angle will you get total internal reflection of light traveling in water and reflected from ice?

A flat mirror is neither converging nor diverging. To prove this, consider two rays originating from the same point and diverging at an angle \(\theta\) (see below). Show that after striking a plane mirror, the angle between their directions remains \(\theta\). Unless otherwise specified, for problems 1 through \(10,\) the indices of refraction of glass and water should be taken to be 1.50 and \(1.333,\) respectively.

Does the fact that the light flash from lightning reaches you before its sound prove that the speed of light is extremely large or simply that it is greater than the speed of sound? Discuss how you could use this effect to get an estimate of the speed of light.

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