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\(\bullet$$\bullet\) Heart sonogram. Physicians use high-frequency \((f=\) 1 MHz to 5 MHz) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 \(\mathrm{m} / \mathrm{s}\) in muscle and 344 \(\mathrm{m} / \mathrm{s}\) 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 \(9.73^{\circ}\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 \(\mathrm{m} / \mathrm{s} .\) ) (b) What is the critical angle for sound waves in air incident on muscle?

Short Answer

Expert verified
(a) Ultrasound enters the heart at approximately 42.18° from normal. (b) The critical angle for sound waves in air incident on muscle is approximately 13.46°.

Step by step solution

01

Calculate the Index of Refraction of Lungs

The index of refraction \( n \) is defined as the ratio of the speed of sound in air to the speed of sound in the material (lungs, in this case). Given that the speed of sound in air \( v_{air} = 344 \, \mathrm{m/s} \) and in lungs \( v_{lungs} = 344 \, \mathrm{m/s} \), we can calculate the index of refraction as follows:\[ n_{lungs} = \frac{v_{air}}{v_{lungs}} = \frac{344}{344} = 1 \]
02

Calculate the Index of Refraction of Muscle

Similarly, calculate the index of refraction of muscle. The speed of sound in muscle \( v_{muscle} = 1480 \, \mathrm{m/s} \). Therefore, the index of refraction is:\[ n_{muscle} = \frac{v_{air}}{v_{muscle}} = \frac{344}{1480} \approx 0.2324 \]
03

Apply Snell's Law for Refraction

According to Snell's Law, \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). Here, \( \theta_1 = 9.73^{\circ} \), \( n_1 = n_{lungs} = 1 \), and \( n_2 = n_{muscle} \approx 0.2324 \). Substitute the values into Snell's Law:\[ \sin 9.73^{\circ} = 0.2324 \sin \theta_2 \]Calculate \( \sin \theta_2 \) and then \( \theta_2 \):\[ \theta_2 = \sin^{-1}\left( \frac{\sin 9.73^{\circ}}{0.2324} \right) \]
04

Calculate the Critical Angle

The critical angle \( \theta_c \) occurs when the angle of refraction is \( 90^{\circ} \). Using Snell's Law, \( n_1 \sin \theta_{c} = n_2 \sin 90^{\circ} \). Here, \( n_1 = n_{air} = 1 \) and \( n_2 = n_{muscle} = 0.2324 \). Thus:\[ \sin \theta_c = \frac{0.2324}{1} = 0.2324 \]\[ \theta_c = \sin^{-1}(0.2324) \]
05

Solve for Angles

Perform the calculations using a calculator.- From Step 3: Calculate \( \theta_2 \).- From Step 4: Calculate \( \theta_c \).

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Key Concepts

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

Snell's Law
Snell's Law is a principle that describes how waves, such as sound or light, bend when they pass from one medium into another. This bending occurs due to the change in wave speed from one medium to the other. The law is expressed mathematically as:\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]where:
  • \( n_1 \) and \( n_2 \) are the indices of refraction for each medium
  • \( \theta_1 \) is the angle of incidence
  • \( \theta_2 \) is the angle of refraction
Snell's Law is crucial in technologies like ultrasound imaging, where understanding the angle of wave entry and exit helps create accurate internal body scans. When sound waves travel from lungs to muscle, their speed alters, necessitating the use of Snell's Law to predict the change in the wave's direction.
Index of Refraction
The index of refraction is a dimensionless number that gives insight into how much a wave, such as sound, slows down in a particular medium compared to air. It is calculated as the ratio of the speed of sound in air to the speed in the medium:\[ n = \frac{v_{\text{air}}}{v_{\text{medium}}} \]In our scenario:
  • Lungs have an index of refraction of 1, since the speed of sound in lungs is equal to that in air.
  • Muscle has an index of refraction of approximately 0.2324, meaning sound travels slower in muscle than in air.
This difference affects how sound waves move through the body, a critical factor in ultrasound imaging.
Critical Angle
The critical angle is a concept related to total internal reflection. It is the angle of incidence at which a wave, such as a sound wave, strikes the boundary between two media and refracts along the boundary, making an angle of 90 degrees with the normal. Above this angle, the wave is completely reflected back.To find the critical angle \( \theta_c \) for sound waves moving from air to muscle, we use Snell's Law with the understanding that the refracted angle is \( 90^{\circ} \):\[ \sin \theta_c = \frac{n_{\text{muscle}}}{n_{\text{air}}} \]Given \( n_{\text{muscle}} = 0.2324 \) and \( n_{\text{air}} = 1 \), we can calculate:\[ \theta_c = \sin^{-1}\left( 0.2324 \right) \]This calculation is crucial in ultrasound as it helps define the limits within which waves will successfully penetrate tissues without reflecting completely back into the air.
Speed of Sound
The speed of sound is how fast sound waves travel through a medium. In ultrasound imaging, understanding the speed of sound in various tissues helps us interpret images accurately. Different materials affect sound speed:
  • Air: 344 m/s
  • Lungs: 344 m/s (same as air, hence no initial refraction)
  • Muscle: 1480 m/s
The vast difference in speed when sound moves between media is why we see bending due to refraction. Calculations involving speed of sound allow practitioners to adjust ultrasound equipment settings to better capture images based on the body's internal characteristics.
Ultrasound Imaging
Ultrasound imaging is a powerful, non-invasive technique used by physicians to visualize internal organs. It utilizes high-frequency sound waves, generally in the 1 MHz to 5 MHz range, which create images by bouncing off tissues and returning echoes to the sensor. The speed at which these waves travel through different types of tissue impacts the image's clarity and accuracy.
  • High-resolution images: Achieved by using different frequencies suited for specific tissues.
  • Minimal risk: Unlike X-rays, ultrasound does not use ionizing radiation.
  • Dynamic assessment: Real-time imaging allows for the observation of moving structures, like the heart.
Understanding the physics behind ultrasound, particularly wave speed and refraction, helps technicians obtain and interpret these images correctly. This technique is vital in diagnostics, allowing for deeper insights into the body's function and health.

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Most popular questions from this chapter

\(\bullet\) Laboratory lasers. He-Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 \(\mathrm{nm}\) and a power of 0.500 \(\mathrm{mW}\) spread over a cylindrical beam 1.00 \(\mathrm{mm}\) in diameter (although these quantities can vary). (a) What is the intensity of this laser beam? (b) What are the maximum values of the electric and magnetic fields? (c) What is the average energy density in the laser beam?

\(\bullet$$\bullet\) Plane-polarized light passes through two polarizers whose axes are oriented at \(35.0^{\circ}\) to each other. If the intensity of the original beam is reduced to \(15.0 \%,\) what was the polarization direction of the original beam, relative to the first polarizer?

\(\bullet\) Ultraviolet radiation. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 \(\mathrm{nm}\) to 400 nm. It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 \(\mathrm{nm}\) and \(320 \mathrm{nm},\) is much more dangerous, because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

\(\bullet$$\bullet\) A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 \(\mathrm{cm}\) and an electric field amplitude of \(5.40 \times 10^{-2} \mathrm{V} / \mathrm{m}\) at a distance of 250 \(\mathrm{m}\) from the antenna. Calculate: (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

\(\bullet\) Unpolarized light with intensity \(I_{0}\) is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at \(41.0^{\circ}\) to that of the first. Determine (a) the intensity of the beam after it has passed through the second polarizer and (b) its state of polarization.

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