/*! 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 11 Sunlight strikes the surface of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 11

Sunlight strikes the surface of a lake. A diver sees the Sun at an angle of \(42.0^{\circ}\) with respect to the vertical. What angle do the Sun's rays in air make with the vertical?

Short Answer

Expert verified
Answer: The angle between the Sun's rays in air and the vertical is approximately 30.77掳.

Step by step solution

01

Identify the given information

We are given the angle between the Sun's rays and the vertical in water, which is \(\theta_{2}=42.0^{\circ}\).
02

Find the index of refraction for air and water

The index of refraction for air (n1) is approximately 1.0003, and for water (n2) is approximately 1.333. These values are well-known and can be found in tables or literature.
03

Write down Snell's law of refraction

Snell's law states that: $$ n_1\sin(\theta_1)=n_2\sin(\theta_2) $$ Where \(n_1\) and \(n_2\) are the indices of refraction for air and water, respectively, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction in the media, respectively.
04

Solve for the angle of incidence, \(\theta_1\)

We are given \(n_1, n_2\), and \(\theta_2\), and we need to find \(\theta_1\). We can rearrange Snell's Law to solve for \(\theta_1\): $$ \theta_1 = \arcsin(\frac{n_2\sin(\theta_2)}{n_1}) $$ Now plug in the values we know: $$ \theta_1 = \arcsin(\frac{1.333\sin(42.0^{\circ})}{1.0003}) $$ Calculate the value of \(\theta_1\): $$ \theta_1 \approx 59.23^{\circ} $$
05

Calculate the angle between the Sun's rays in air and the vertical

We found the angle of incidence \(\theta_1\), which is the angle between the Sun's rays in air and the normal (perpendicular line) to the surface of the water. To find the angle between the Sun's rays and vertical, we need to calculate the complementary angle: $$ \text{Angle with the vertical} = 90^{\circ} - 59.23^{\circ} = 30.77^{\circ} $$ So, the angle between the Sun's rays in air and the vertical is approximately \(30.77^{\circ}\).

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 91影视!

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

Samantha puts her face \(32.0 \mathrm{cm}\) from a makeup mirror and notices that her image is magnified by 1.80 times. (a) What kind of mirror is this? (b) Where is her face relative to the radius of curvature or focal length? (c) What is the radius of curvature of the mirror?
Light rays from the Sun, which is at an angle of \(35^{\circ}\) above the western horizon, strike the still surface of a pond. (a) What is the angle of incidence of the Sun's rays on the pond? (b) What is the angle of reflection of the rays that leave the pond surface? (c) In what direction and at what angle from the pond surface are the reflected rays traveling?
A 5.0 -cm-tall object is placed \(50.0 \mathrm{cm}\) from a lens with focal length \(-20.0 \mathrm{cm} .\) (a) How tall is the image? (b) Is the image upright or inverted?
Show that the deviation angle \(\delta\) for a ray striking a thin converging lens at a distance \(d\) from the principal axis is given by \(\delta=d l f .\) Therefore, a ray is bent through an angle \(\delta\) that is proportional to \(d\) and does not depend on the angle of the incident ray (as long as it is paraxial). [Hint: Look at the figure and use the small-angle approximation \(\sin \theta=\tan \theta \approx \theta(\text { in radians) } .]\)

A glass prism bends a ray of blue light more than a ray of red light since its index of refraction is slightly higher for blue than for red. Does a diverging glass lens have the same focal point for blue light and for red light? If not, for which color is the focal point closer to the lens?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Waves Physics

Read Explanation

Medical Physics

Read Explanation

Radiation

Read Explanation

Dynamics

Read Explanation

Solid State Physics

Read Explanation
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.