You are under water in a pond and look up at the smooth surface of the water, noticing the sun in the sky. Is the sun in fact higher in the sky than it appears to you while under water, or is it lower?

Short Answer

Expert verified
Answer: The sun appears higher in the sky than its actual position to an observer underwater due to refraction.

Step by step solution

01

Understanding refraction

Refraction is the change in direction of a wave due to a change in its speed as it passes from one medium to another. In this case, when the sun is in the sky, its light travels through the air and then enters the water. The speed of light changes at the air-water boundary, causing the light to refract. This bending of light occurs due to the difference in the index of refraction of the two mediums (air and water).
02

Identifying Snell's Law

Snell's Law is used to determine the angle of refraction when light passes from one medium to another. It is derived from the relationship between the angles of incidence and refraction with the indices of refraction of the incident and transmitted media. Mathematically, Snell's Law is expressed as: n_1 \times \sin{\theta}_1 = n_2 \times \sin{\theta}_2 Here, n_1 and n_2 are the indices of refraction of the two media (air and water, respectively), while \theta_1 and \theta_2 are the angles of incidence and refraction, respectively.
03

Comparing the indices of refraction

In this scenario, the index of refraction for air (n_1) is approximately 1, while the index of refraction for water (n_2) is approximately 1.33. Since n_2 > n_1, when light passes from air to water, it bends towards the normal (a line perpendicular to the surface at the point of incidence).
04

Analyzing the apparent position of the sun

As the light enters the water and bends towards the normal, the angle \theta_2 becomes smaller than the angle \theta_1. This bending of light creates the illusion that the sun is at a higher position in the sky than its actual location. Therefore, to an observer underwater, the sun appears higher in the sky than it actually is.
05

Conclusion

Based on the analysis of refraction and the application of Snell's Law, it can be concluded that the sun appears higher in the sky than it actually is to an observer underwater.

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 Vaia!

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

A person sits \(1.0 \mathrm{~m}\) in front of a plane mirror. What is the location of the image?

A physics student is eying a steel drum, the top part of which has the approximate shape of a concave spherical surface. The surface is sufficiently polished that she can just barely make out the reflection of her finger when she places it above the drum. As she slowly moves her finger toward the surface and then away from it, you ask her what she is doing. She replies that she is estimating the radius of curvature of the drum. How can she do that?

For specular reflection of a light ray, the angle of incidence a) must be equal to the angle of reflection. b) is always less than the angle of reflection. c) is always greater than the angle of reflection. d) is equal to \(90^{\circ}\) - the angle of reflection. e) may be greater than, less than, or equal to the angle of reflection.

Reflection and refraction, like all classical features of light and other electromagnetic waves, are governed by the Maxwell equations. The Maxwell equations are time-reversal invariant, which means that any solution of the equations reversed in time is also a solution. a) Suppose some configuration of electric charge density \(\rho,\) current density \(\vec{j},\) electric field \(\vec{E},\) and magnetic field \(\vec{B}\) is a solution of the Maxwell equations. What is the corresponding time-reversed solution? b) How, then, do "one-way mirrors" work?

A \(45^{\circ}-45^{\circ}-90^{\circ}\) triangular prism can be used to reverse a light beam: The light enters perpendicular to the hypotenuse of the prism, reflects off each leg, and emerges perpendicular to the hypotenuse again. The surfaces of the prism are not silvered. If the prism is made of glass with in dex of refraction \(n_{\text {glass }}=1.520\) and the prism is surrounded by air, the light beam will be reflected with a minimum loss of intensity (there are reflection losses as the light enters and leaves the prism). a) Will this work if the prism is under water, which has index of refraction \(n_{\mathrm{H}_{2} \mathrm{O}}=1.333 ?\) b) Such prisms are used, in preference to mirrors, to bend the optical path in quality binoculars. Why?

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.

Sign-up for free