/*! 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} Q37P Physicians use high-frequency 1f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Physicians use high-frequency 1f = 195 MHz2 sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 m>s in muscle and 344 m/s in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell’s law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of 9.73° from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 m/s.) (b) What is the critical angle for sound waves in air incident on muscle?

Short Answer

Expert verified

(a) The angle at which the beam enters the heart is 47.01°

(b) The critical angle at which the beam enters the air is 13.44°

Step by step solution

01

Snell’s law

Snell’s law;

The refractive index of medium a isna

The incident angle in the medium a isθa

The refractive index of medium b isnb

The incident angle in the medium b isθb

02

The angle at which the beam enters the heart

A substance's index of refraction is defined as the ratio of the speed of sound in air to the speed of sound in the material. As a result, the muscle's refractive index will be;

nmuscle=3441480=0.2324

The speed of waves in air is 344m/s

The speed of the waves in muscles is 1480m/s

  1. Incident angle is

So, the angle is;

Hence, the angle at which the beam enters the heart is47.01°

03

The critical angle at which the beam enters the air

(b) For the critical angle in the air, the wave should leave at an angle of 90°

Thus, using Snell’s law;

Hence, the critical angle at which the beam enters the air is13.44°

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!

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 lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object.

(a) What is the focal length of the lens? Is the lens converging or diverging?

(b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted?

(c) Draw a principal-ray diagram.

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible light ranges in wavelength from 380 nm (violet) to 750 nm (red), as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

The Galilean Telescope. Figure P34.100 is a diagram of a Galilean telescope, or opera glass, with both the object and its final image at infinity. The image serves as a virtual object for the eyepiece. The final image is virtual and erect. (a) Prove that the angular magnification is M=-f1/f2. (b) A Galilean telescope is to be constructed with the same objective lens as in Exercise 34.61. What focal length should the eyepiece have if this telescope is to have the same magnitude of angular magnification as the one in Exercise 34.61? (c) Compare the lengths of the telescopes. Figure P34.100

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 μ³¾apart, and in line with an observer, so that one source is2.04 μ³¾farther from the observer than the other. (a) For what visible wavelengths (380 to 750 nm) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is2.04 μ³¾farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

The professor once again returns the apparatus to its original setting, but now she adjusts the oscillator to produce sound waves of half the original frequency. What happens? (a) The students who originally heard a loud tone again hear a loud tone, and the students who originally heard nothing still hear nothing.(b) The students who originally heard a loud tone now hear nothing, and the students who originally heard nothing now hear a loud tone. (c) Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing. (d) Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone.

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.