A narrow beam of ultrasonic waves reflects off the liver tum

A narrow beam of ultrasonic waves reflects off the liver tumor illustrated in the figure below. The speed of the wave is 9.1% less in the liver than in the surrounding medium. Determine the depth of the tumor.

Solution

Let v1 be the speed of sound in the surrounding medium: then the speed of the sound in the liver is

v2 = 1 - 0.107 = 0.893*v1.

Since the refraction index of a wave is inversely proportional to the speed of the wave, if n1 is the refraction index in the surrounding medium the refraction index in the liver is n2 = n1/0.909

Let be the refraction angle. According to Snell\'s law:

sin / sin(50°) = n1/n2 = 0.909

sin = 0.909* sin(50°) = 0.684

= 44.13°

The tumor depth is therefore

d = 6.00/tan(44.13°) = 6.3 cm