A narrow beam of ultrasonic waves reflects off the liver...

A narrow beam of ultrasonic waves reflects off the liver tumor illustrated in Figure P35.20. The
speed of the wave is 10.0$\%$ less in the liver than in the surrounding medium. Determine the depth of the tumor.

Key Concepts

Ultrasonic Wave Propagation

Ultrasonic waves are high-frequency sound waves used primarily in medical imaging. Their propagation through various media depends on the medium’s properties, such as density and elasticity, which affect both speed and attenuation.

Time-of-Flight Measurement

This concept involves calculating the distance to an object by measuring the time taken for an ultrasound pulse to travel to a target and back. The depth is determined by the product of the speed of sound in the medium and half the round-trip travel time.

Speed of Sound Variation in Media

The speed of sound can differ between various tissues or media due to differences in their acoustic properties. Changes in speed, such as a 10% reduction in one medium compared to another, must be accounted for in calculations of distance and depth.

Reflection of Waves

When an ultrasonic wave encounters a boundary between two media with different acoustic properties, part of it is reflected. The analysis of these echoes is crucial in identifying and localizing structures within the body.

Transcript

00:01 So in this problem, we are told, this is a tough one, actually, so i'll show you.

00:08 We have our layer of skin, then we have, or rather, this is the liver, and we have a tumor at this depth.

00:26 So our tumor is right here, and we have an incident light or incident high frequency coming in here and then leaving there.

00:40 So it's coming in at this angle.

00:46 So to the vertical, that is a 50 degree angle.

00:52 So we are also told that this gap is 12 centimeters, right? so another thing we're told is that the speed in the medium, the outside medium, the speed in the outside medium is 10 % faster than the speed in the liver.

01:32 So that means that v2, which is the speed inside the liver, is 0 .9 of v1.

01:39 So now we can use this to solve for the depth of this tumor.

01:45 So the depth will be the distance from the start of the liver to the tumor.

01:53 So we're going to call this depth.

01:58 Now to start, we need to determine what n is.

02:03 What is the refractive index of the liver? so the good thing is that to determine this, all we have to do is find the speed of, is divide the speed in the outside medium by the speed in the liver.

02:24 So that will give us 0 .9 to the negative 1 power, which is just basically a rearrangement of what you see up here.

02:39 So that would be 1 .11.

02:44 The next equation we're going to use is n1 of sine of theta of 1 is equal to n2 sine theta of 2.

02:56 So for simplicity's sake, we're going to assume that n1 is 1.

03:02 And we have just established that the n inside the liver is 1 .11.

03:07 So we're going to solve for this angle theta.

03:13 We're going to call this theta 2 and this theta 1...