/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 A 4.00 -m-long pole stands verti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 4.00 -m-long pole stands vertically in a lake having a depth of \(2.00 \mathrm{m} .\) The Sun is \(40.0^{\circ}\) above the horizontal. Determine the length of the pole's shadow on the bottom of the lake. Take the index of refraction for water to be 1.33.

Short Answer

Expert verified
To find out the length of the shadow on the bottom of the lake, first use Snell's law to determine the refraction angle, then use trigonometry to find the shadow length. The steps involve applying Snell's law and simple trigonometry principles.

Step by step solution

01

Define Given Values

First, list out all the facts from the problem. This includes the length of the pole (4.00 m), the depth of the lake (2.00 m), the index of refraction for water (1.33), and the angle the Sun is above the horizontal (40.0°). The angle of incidence (\( θi\)) is 40.0° and the index of refraction of air is 1 (since light travels in a straight line in air).
02

Calculate the Angle of Refraction

Use Snell's Law to calculate the angle of refraction (\( θr\)), which is the angle the light rays make with the normal, when they enter the water. Snell's law states that the ratio of the sine of the angle of incidence (\( θi\)) to the sine of the angle of refraction (\( θr\)) is equal to the ratio of the velocity of light in the first medium to the velocity of light in the second medium. This can also be written as: \( n1 * sin(θi) = n2 * sin(θr) \). Here, \( n1\) is the refractive index of air and \( n2\) is the refractive index of the water. By substituting the given values, we get \( 1 * sin(40°) = 1.33 * sin(θr) \). Solve for \( θr\) to get the angle of refraction.
03

Find the Length of the Shadow

Now we need to find the length of the shadow at the bottom of the lake. As the light hits the water it bends or refracts, which causes the shadow to extend. We can use simple trigonometry here as we have the depth of the lake and we know the angle of refraction of light. The tangent of the \( θr\) is equal to the opposite side (which is the shadow) divided by the adjacent side (which is the depth of the lake). This implies that: Shadow = Depth of the lake * tan(θr). Substitute the known values and solve for the shadow length.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Snell's Law
Snell's Law is a principle in physics that describes the way light or other waves change direction, or refract, when they pass through different mediums. Its primary concern is the relationship between the angles of incidence and refraction, which are crucial to understanding how light behaves as it enters substances like water.
To delve deeper, imagine a beam of light traveling from air into water. The speed of light is different in air compared to water, which causes the light to bend at the surface. According to Snell's Law, the ratio of the sine of the angle of incidence (\(\theta_i\)) to the sine of the angle of refraction (\(\theta_r\)) is equal to the ratio of the refractive indices of the two media:
  • \(n_1 \times \sin(\theta_i) = n_2 \times \sin(\theta_r)\)
Here, \(n_1\) and \(n_2\) are the refractive indices of the air and water, respectively. This equation helps to predict how much light will bend and is widely used in various scientific and engineering applications including optics and photography. In our exercise, we use Snell's Law to determine the angle of refraction as sunlight transitions from air into water.
Refraction
Refraction is the bending of light as it passes from one medium into another with a different refractive index. This phenomenon occurs because light travels at different speeds in different media.
When the speed of light changes, its wavelength also changes, which results in a change in direction. The extent of this bending depends on how drastically the light speed changes—from air to water, light slows down, causing it to bend towards the normal.
  • Refraction is responsible for a variety of optical illusions, such as the apparent "bending" of objects partially submerged in water.
  • It plays a crucial role in lenses, allowing us to focus light and produce clear images.
  • In natural settings, refraction manifests when you view the bottom of a pool appearing shallower than it is.
In our problem, refraction causes the sunlight to change direction as it hits the water, making the shadow of the pole longer on the lake's bottom than it would be on a flat surface without water.
Trigonometry in Physics
Trigonometry is a branch of mathematics that deals with the relationships between angles and sides of triangles. In physics, it provides a powerful tool to solve problems involving angles and various geometric configurations.
When applying trigonometry to physics exercises, one often makes use of the basic trigonometric functions: sine, cosine, and tangent. These functions relate one angle to ratios of the sides of a right triangle.
  • Sine relates the opposite side to the hypotenuse.
  • Cosine relates the adjacent side to the hypotenuse.
  • Tangent relates the opposite side to the adjacent side.
In the context of the exercise, once the angle of refraction is calculated using Snell’s Law, trigonometry allows us to determine the length of the shadow. By knowing the depth of the water and the angle of refraction, using the tangent function gives us the relationship:
\[\text{Shadow Length} = \text{Depth} \times \tan(\theta_r)\]This relationship illustrates how trigonometry integrates with physics principles to solve real-world problems, such as determining shadow lengths created by refraction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Apollo 11 astronauts set up a panel of efficient corner cube retro- reflectors on the Moon's surface. The speed of light can be found by measuring the time interval required for a laser beam to travel from Earth, reflect from the panel, and return to Earth. If this interval is measured to be \(2.51 \mathrm{s}\) what is the measured speed of light? Take the center-to-center distance from Earth to Moon to be \(3.84 \times 10^{8} \mathrm{m},\) and do not ignore the sizes of the Earth and Moon.

A room contains air in which the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) The walls of the room are made of concrete, in which the speed of sound is \(1850 \mathrm{m} / \mathrm{s} .\) (a) Find the critical angle for total internal reflection of sound at the concrete-air boundary. (b) In which medium must the sound be traveling in order to undergo total internal reflection? (c) "A bare concrete wall is a highly efficient mirror for sound." Give evidence for or against this statement.

The distance of a light bulb from a large plane mirror is twice the distance of a person from the plane mirror. Light from the bulb reaches the person by two paths. It travels to the mirror at an angle of incidence \(\theta,\) and reflects from the mirror to the person. It also travels directly to the person without reflecting off the mirror. The total distance traveled by the light in the first case is twice the distance traveled by the light in the second case. Find the value of the angle \(\theta\).

A dance hall is built without pillars and with a horizontal ceiling \(7.20 \mathrm{m}\) above the floor. A mirror is fastened flat against one section of the ceiling. Following an earthquake, the mirror is in place and unbroken. An engineer makes a quick check of whether the ceiling is sagging by directing a vertical beam of laser light up at the mirror and observing its reflection on the floor. (a) Show that if the mirror has rotated to make an angle \(\phi\) with the horizontal, the normal to the mirror makes an angle \(\phi\) with the vertical. (b) Show that the reflected laser light makes an angle \(2 \phi\) with the vertical. (c) If the reflected laser light makes a spot on the floor \(1.40 \mathrm{cm}\) away from the point vertically below the laser, find the angle \(\phi\).

Builders use a leveling instrument with the beam from a fixed helium-neon laser reflecting in a horizontal plane from a small flat mirror mounted on an accurately vertical rotating shaft. The light is sufficiently bright and the rotation rate is sufficiently high that the reflected light appears as a horizontal line wherever it falls on a wall. (a) Assume the mirror is at the center of a circular grain elevator of radius \(R .\) The mirror spins with constant angular velocity \(\omega_{m} .\) Find the speed of the spot of laser light on the wall. (b) What If? Assume the spinning mirror is at a perpendicular distance \(d\) from point \(O\) on a flat vertical wall. When the spot of laser light on the wall is at distance \(x\) from point \(O,\) what is its speed?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.