A submarine is 3.00 ×10^2 m horizontally from shore and 1.00 ×10^2 m beneath the surface of th

A submarine is 3.00 ×10^2 m horizontally from shore and...

A submarine is $3.00 \times 10^{2} \mathrm{~m}$ horizontally from shore and $1.00 \times 10^{2} \mathrm{~m}$ beneath the surface of the water. A lasenbeam is sent from the submarine so that the beam strikes the surface of the water $2.10 \times 10^{2} \mathrm{~m}$ from the shore. $\mathrm{A}$ building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level. (a) Draw a diagram of the siuation, identifying the two triangles that are important to finding the solution. (b) Find the angle of incidence of the beam striking the water-air interface. (c) Find the angle of refraction. (d) What angle does the refracted beam make with respect to the horizontal? (e) Find the height of the target above sea level.

A submarine is $3.00 \times 10^{2} \mathrm{~m}$ horizontally from shore and $1.00 \times 10^{2} \mathrm{~m}$ beneath the surface of the water. A lasenbeam is sent from the submarine so that the beam strikes the surface of the water $2.10 \times 10^{2} \mathrm{~m}$ from the shore. $\mathrm{A}$ building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level. (a) Draw a diagram of the siuation, identifying the two triangles that are important to finding the solution. (b) Find the angle of incidence of the beam striking the water-air interface.
(c) Find the angle of refraction.
(d) What angle does the refracted beam make with respect to the horizontal?
(e) Find the height of the target above sea level.

Key Concepts

Right Triangle Geometry

Understanding right triangle geometry is essential for solving problems involving distances and angles in optics. It allows one to model the situation using triangles where one leg represents the vertical distance (such as depth or height) and the other represents the horizontal distance. This concept is crucial because it provides a basis for using trigonometric ratios to find unknown lengths or angles, which is critical when analyzing the paths taken by light rays.

Trigonometric Functions

Trigonometric functions, including sine, cosine, and tangent, are used to relate the angles and side lengths of right triangles. In optical problems, these functions help calculate the angles of incidence and refraction by relating the known distances (such as depths and horizontal separations) to the angles formed by the light ray paths. Their use is fundamental in converting geometric measurements into angular information.

Snell's Law

Snell's Law is a fundamental principle in optics that describes how light bends, or refracts, when it passes from one medium into another. It relates the angle of incidence in the first medium to the angle of refraction in the second medium through the indices of refraction of the two media. This law is crucial for calculating the new direction of a light beam upon crossing the boundary between different refractive environments.

Interface Between Media

The interface between two media, such as water and air, is a critical concept in optical physics. At this boundary, properties such as the speed of light change, leading to refraction. Recognizing and properly analyzing the behavior of light at this interface using laws like Snell's Law is necessary for predicting the path of a light beam as it transitions between different media.

Geometric Diagramming

Drawing a clear geometric diagram is key to visualizing the problem scenario involving light paths. In many physics problems, especially those dealing with refraction, identifying important triangles and the relationships among the distances and angles helps in setting up the necessary equations and understanding the physical behavior of the system. A well-constructed diagram aids in applying both trigonometry and optics principles effectively.

Transcript

00:01 Hi there.

00:01 So for this problem, we are told that a submarine is 3 times 10 to the 2 horizontally from the shore.

00:10 And it is 1 times 10 to the 2 meters beneath the surface of the water.

00:20 So a laser beam is sent from the submarine so that the beam strides the surface of the water 2 .10 times 10 to the 2 meters from the shore.

00:34 And there is a building that stands on the shore, and the laser beam hits a target at the top of the building.

00:42 So the goal is to find the height of the target above the sea level.

00:47 So for part a of this problem, we are asked to draw a diagram of the situation, identify the two triangles that are important to find in the solution.

00:58 So the first thing that we are gonna draw in here is, well, this is for part a.

01:06 First, i'm going to draw a line in here.

01:12 And i'm going to draw the building in here.

01:19 So this is the building in the shore.

01:26 And let me just move this a little bit.

01:31 Ok, so now we're going to draw the submarine.

01:38 Well, let me just put that in here.

01:40 The submarine is in here.

01:44 This is the submarine.

01:45 The submarine is at a death that is equal to 1 times 10 to the 2 meters.

01:55 We're going to see that this in here, there is a laser beam that points to this position.

02:04 And this, the length of this is 90 meters.

02:12 We know that the earth, it has a, inters of refraption of 1.

02:17 So this is what we are going to call the angle theta 1.

02:23 That is going to be the same as this, theta 1.

02:28 And now we have some refraption because now the laser beam is passing from water to air and it's going to point in here.

02:38 So we're going to have this refruption right here.

02:44 And this is going to be what we are going to call, angle teta 2.

02:51 We're going to call this other angle in here fee.

02:55 And we have that this is the height of the building, h.

03:01 And this length in here corresponds to the length that we are given of 2 .10 times 10 to the 2 meters.

03:16 So we also know another important thing that we need to know is that the indesirruption of water is equal to 1 .333.

03:30 So this is the sketch of the situation that we have in here, submarine.

03:38 Now, for part b, as you can see, well, as you can see the two triangles that we will help us to solve this problem, is this one right here and this one right here.

03:52 So for part b, we are told to, to find the angle of incidence of the being striking the water or surface...

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