Question

A narrow beam of ultrasonic waves reflects off the liver tumor in the figure below. If the speed of the wave is 24.5% less in the live Your response differs from the correct answer by more than 10%. Double check your calculations. cm

Answer 1

Question, you provided, is not completely visible. But what I suppose is that it is asking for the depth of the liver. So I am solving for that. If there is some other term asking for then please follow the procedure. You'll get hint for that too(in case only)

Angle of incidence, $i= 50 ^{\circ}$

Let speed of ray in air be "v" then

speed of ray in liver will be $v_2= v-\frac{24.5}{100}v= v(1-0.245)= 0.755v$

then if angle of refraction is "r" then by Using Snell's Law,

$\frac{sin(i)}{sin(r)}= \frac{v_1}{v_2}$

using all given values,

$\frac{sin(50^{\circ})}{sin(r)}= \frac{v}{0.755v}$

$sin(r)= \frac{0.766 \times 0.755}{1}= 0.5783$

   $r= sin^{-1}(0.5783)= 35.33^{\circ}$

ABove is the angle which the ray makes with the normal. Thus the angle which the ray makes with the upper horizontal surface of the liver inside it is

$\theta= 90^{\circ}-r= 90^{\circ}-35.33^{\circ}= 54.67^{\circ}$

Since reflection is totally symmetric thus we can assume that area covered by the ray inside the liver to be consisting of 2 right angle triangles with the angle $\theta$ and base equal to

   $b= \frac{12}{2}= 6cm$

then depth of the liver will be $d= btan\theta= 6(tan54.67^{\circ})= 8.464cm$ (ANS)

Also the total distance travelled by ray inside the liver will be

   $D_t= 2(\frac{b}{cos(54.67^{\circ})})=2(\frac{6}{0.578})= 20.761cm$