Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 2236

A Sound wave travels from air to water. The angle of incidence is \(\alpha_{1}\) and the angle of reflection is \(\alpha_{2}\) If the snell's Law is valid then, (A) \(\alpha_{1} \geq \alpha_{2}\) (B) \(\alpha_{1}=\alpha_{2}\) (C) \(\alpha_{1}>\alpha_{2}\) (D) \(\alpha_{1}<\alpha_{2}\)

Short Answer

Expert verified
(C) \(\alpha_{1} > \alpha_{2}\)

Step by step solution

01

Write down Snell's Law formula

Recall that Snell's Law states the relationship between the angles of incidence, refraction, and the refractive indices of the two mediums. The formula for Snell's Law is: \[ n_1 \sin\alpha_{1} = n_2 \sin\alpha_{2} \]
02

Relate the refractive indices to the speed of sound in each medium

Refractive index is the ratio of the speed of light in a vacuum to the speed of light in a given medium. For sound waves, we can use a similar ratio by defining the index of refraction as the ratio of the speed of sound in the reference medium (air) to the speed of sound in the given medium. Using this definition, we can write the refractive indices of air and water as \(n_1 = \frac{v_{air}}{v_{air}}\) and \(n_2 = \frac{v_{air}}{v_{water}}\), where \(v_{air}\) and \(v_{water}\) are the speeds of sound in air and water, respectively.
03

Insert the refractive indices into Snell's Law formula

Now, substitute the expressions for refractive indices into Snell's Law formula: \[ \frac{v_{air}}{v_{air}} \sin\alpha_{1} = \frac{v_{air}}{v_{water}} \sin\alpha_{2} \]
04

Simplify the equation

Notice that \(v_{air}\) cancels out on both sides of the equation, leaving: \[ \sin\alpha_{1} = \frac{v_{water}}{v_{air}} \sin\alpha_{2} \]
05

Compare the speeds of sound in air and water

In general, the speed of sound is faster in water than in air (\(v_{water} > v_{air}\)). Therefore, the fraction \(\frac{v_{water}}{v_{air}}\) is greater than 1.
06

Analyze the relationship between angles of incidence and refraction

With the information from the previous step, we can rewrite the equation as: \[ \sin\alpha_{1} = k \sin\alpha_{2}\ ,\quad k > 1 \] Since \(k\) is greater than 1, for the same values of \(\alpha_{1}\) and \(\alpha_{2}\), the sine of the angle of incidence will always be greater than the sine of the angle of refraction.
07

Determine the correct answer choice

Based on our analysis in Steps 5 and 6, we can conclude that when a sound wave travels from air to water, the angle of incidence \(\alpha_{1}\) will always be greater than the angle of refraction \(\alpha_{2}\). Therefore, among the given answer choices, the correct one is: (C) \(\alpha_{1} > \alpha_{2}\)

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.

Start your free trial

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

An observer look at a tree of height 10 meters away with a telescope of magnifying power \(10 .\) To him, the tree appears (A) 10 times taller (B) 10 times smaller (C) 10 times nearer (D) 20 times nearer

Relation between critical angle of water \(C_{w}\) and that of the glass \(\mathrm{C}_{\mathrm{g}}\) is (given, $\left.\mathrm{n}_{\mathrm{w}}=(4 / 3), \mathrm{n}_{\mathrm{g}}=1.5\right)$ (A) \(\mathrm{C}_{\mathrm{w}}<\mathrm{C}_{\mathrm{g}}\) (B) \(\mathrm{C}_{\mathrm{w}}=\mathrm{C}_{\mathrm{g}}\) (C) \(\mathrm{C}_{\mathrm{w}}>\mathrm{C}_{\mathrm{g}}\) (D) \(\mathrm{C}_{\mathrm{w}}=\mathrm{C}_{\mathrm{g}}=\mathrm{O}\)

Light of wave-length \(\lambda\) is incident on a slit of width \(\mathrm{d}\). The resulting diffraction pattern is observed on a screen placed at a distance \(\mathrm{D}\). The linear width of the principal maximum is equal to the width of the slit, then \(\mathrm{D}=\) (A) \(\left(\mathrm{d}^{2} / 2 \lambda\right)\) (B) \(\left(2 \lambda^{2} / \mathrm{d}\right)\) (C) \((\mathrm{d} / \lambda)\) (D) \((2 \lambda / \mathrm{d})\)

A Slit of width \(12 \times 10^{-7} \mathrm{~m}\) is illuminated by light of wavelength \(6000 \AA\). The angular width of the central maxima is approximately (A) \(30^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(0^{\circ}\)

The phenomenon of polarization of electromagnetic waves proves that the electromagnetic waves are (A) mechanical (B) longitudinal (C) transverse (D) none of these
See all solutions

Recommended explanations on English Textbooks

Blog

Read Explanation

Essay Prompts

Read Explanation

Phonetics

Read Explanation

Email

Read Explanation

Linguistic Terms

Read Explanation

English Grammar

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.

Sign-up for free