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Chapter 26: Problem 113

A scuba diver, submerged under water, looks up and sees sunlight at an angle of \(28.0^{\circ}\) from the vertical. At what angle, measured from the vertical, does this sunlight strike the surface of the water?

Short Answer

Expert verified
The sunlight strikes the water surface at an angle of approximately 38.6 degrees from the vertical.

Step by step solution

01

Understand the Problem

The problem involves light refraction from water to air. The light changes direction when it enters the air from the water due to different refractive indices. We need to find the angle of incidence for the light as it leaves water and enters air.
02

Snell's Law

Snell's Law describes how light rays bend when entering a new medium, given by:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]where \( n_1 = 1.33 \) (refractive index of water), \( \theta_1 = 28.0^{\circ} \) (angle of light from vertical within water), \( n_2 = 1.00 \) (refractive index of air), and \( \theta_2 \) is the angle of incidence in air from the vertical.
03

Rearrange Snell's Law

We need \( \theta_2 \), so rearrange the equation to solve for \( \theta_2 \):\[ \sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1) \]Plug in the known values: \( n_1 = 1.33 \), \( n_2 = 1.00 \), and \( \theta_1 = 28.0^{\circ} \).
04

Calculate \(\theta_2\)

Calculate the sine of \( \theta_2 \):\[ \sin(\theta_2) = \frac{1.33}{1.00} \sin(28.0^{\circ}) \]\[ \sin(\theta_2) = 1.33 \times 0.4695 \]\[ \sin(\theta_2) = 0.624 \]Now find \( \theta_2 \) using the inverse sine function: \( \theta_2 = \arcsin(0.624) \).
05

Find Angle in Degrees

Compute the angle using a calculator for \( \theta_2 \):\[ \theta_2 \approx 38.6^{\circ} \]
06

Conclusion: Angle of Incidence

The angle at which the sunlight strikes the surface of the water, measured from the vertical, is approximately 38.6 degrees.

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Key Concepts

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

Light Refraction
When light passes from one medium to another, such as from water into air, it bends. This bending of light is called refraction. It's a result of the change in speed that light experiences when moving between media with different optical densities.
The light refraction happens at the boundary separating the two media.
For example, as light exits the water and enters the air, it travels faster and bends away from the normal (perpendicular to the surface). The study of light refraction helps explain phenomena like the apparent shift in position of objects underwater, and is crucial in understanding how lenses function.
  • Less dense medium to denser medium: Light bends towards the normal.
  • Denser medium to less dense medium: Light bends away from the normal.
Refractive Index
The refractive index is a measure that indicates how fast light travels through a medium. It鈥檚 denoted by the symbol "n." Water has a refractive index of approximately 1.33, while air's refractive index is about 1.00. These values mean light travels approximately 33% slower in water than in air.
Refractive indices help determine the bending angle when light enters a new medium, playing a pivotal role in Snell's Law. Snell's Law uses the refractive indices of the respective media to calculate how much the light will bend.
  • Higher refractive index: Light slows down, bends more.
  • Lower refractive index: Light speeds up, bends less.
Angle of Incidence
The angle of incidence refers to the angle at which a light ray hits a surface relative to a line perpendicular to the surface, known as the normal. In our exercise, the diver perceives light rays at a 28.0掳 angle from vertical within water. This is the angle of incidence from the perspective of the water-to-air boundary.
When light transitions between different media, the angle of incidence determines how much the light ray will refract.
Snell's Law connects this angle with the angle of refraction, which is the angle of light as it leaves one medium to enter another. Understanding angles of incidence is important for predicting light behavior at surfaces like lenses and air-water interfaces.
Inverse Sine Function
The inverse sine function, also called arcsin, helps find the angle of a triangle knowing its sine value. In problems like ours, we use it to determine the angle of refraction once the calculation gives us a sine value.
For example, after computing \( \sin(\theta_2) = 0.624 \), we use the inverse sine function to discover \( \theta_2 \), which is roughly \( 38.6^\circ \).
Using a calculator, the arcsin function reverses the sine function to output the actual angle in degrees. This is essential in practical applications like finding light angles in refraction problems.
  • Provides angle from known sine.
  • Key tool in trigonometry and physics.

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Most popular questions from this chapter

A farsighted man uses eyeglasses with a refractive power of 3.80 diopters. Wearing the glasses 0.025 m from his eyes, he is able to read books held no closer than 0.280 m from his eyes. He would like a prescription for contact lenses to serve the same purpose. What is the correct contact lens prescription, in diopters?

A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is \(1.39 \times 10^{9} \mathrm{m},\) and its mean distance from the earth is \(1.50 \times 10^{11} \mathrm{m} .\) The camper is using a converging lens whose focal length is 10.0 \(\mathrm{cm} .\) (a) What is the area of the sun's image on the paper? (b) If 0.530 \(\mathrm{W}\) of sunlight passes through the lens, what is the intensity of the sunlight at the paper?

A paperweight consists of a 9.00-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposite faces of the cube. On each side of the paper is printed a different joke that can be read by looking perpendicularly straight into the cube. When read from one side (the top), the apparent depth of the paper in the plastic is 4.00 cm. When read from the opposite side (the bottom), the apparent depth of the paper in the plastic is 1.63 cm. What is the index of refraction of the plastic?

A stargazer has an astronomical telescope with an objective whose focal length is 180 cm and an eyepiece whose focal length is 1.20 cm. He wants to increase the angular magnification of a galaxy under view by replacing the telescope鈥檚 eyepiece. Once the eyepiece is replaced, the barrel of the telescope must be adjusted to bring the galaxy back into focus. If the barrel can only be shortened by 0.50 cm from its current length, what is the best angular magnification the stargazer will be able to achieve?

mmh A glass block \((n=1.56)\) is immersed in a liquid. A ray of light within the glass hits a glass-liquid surface at a \(75.0^{\circ}\) angle of incidence. Some of the light enters the liquid. What is the smallest possible refractive index for the liquid?
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