Question

Sound waves undergo reflection and refraction, much as electromagnetic waves do. Here is one practical application of reflection and refraction in the field of health care: determining the location of a liver tumor.

Suppose that a narrow beam of ultrasonic waves travels through surrounding tissue and enters the liver with an incidence angle of 48.0°. These inaudible sound waves travel 14.5% more slowly through the liver than through the medium that lies above.

Suppose that the beam reflects off the tumor and emerges from the liver at a distance 12.5 cm from its entry point. Calculate the depth of the tumor (in cm) below the surface of the liver.\

*An illustration shows a beam entering a liver and being reflected by a tumor. The liver and tumor are represented by two rectangles with the liver on top of the tumor. The depth of the liver is larger than the depth of the tumor. A beam is moving down and to the right and is incident on the surface of the liver at an angle of 48.0° with the vertical. Inside the liver, it continues to move down and to the right but at a steeper slope than the incident beam. The beam then reflects off the tumor and moves up and to the right and exits the liver, where the beam continues to travel up and to the right but at a shallower slope than within the liver. The distance between the position where the beam enters the liver and the position where the beam exits the liver is 12.5 cm.*

Answer 1

This question can be solved by using concept of refraction and reflection of light through parallel glass slab as shown in figure below:

We know that refrective index is given by

So, ​​​​​V​​​1 = V, then V2= V - 14.5/100 V= 85.5/100 V

So refractive index = 100/85.5= 1.17

At point A angle of refraction is given by Shell's law

So, r = 39.42°

Now, at point Depth of tumor below the surface of liver is 4.04 cm