Question

An ultrasound beam is incident on liver tissue at an angle of 50 degrees. This beam undergoes refraction at the interface between the fluid above and the liver. It then reflects off the interface of the liver and tumor and is detected as exiting the liver 12.0 cm displaced from where it entered. Use this information to determine the depth of the tumor in the liver. Assume that the index of fraction for ultrasound waves in the liver is 1.2 times that of the fluid above (n_"liver"

Answer 1

Step-by-step explanation:

Let the angle of refraction be θ .

sin 50 / sinθ = μ₂ /μ₁ , μ₂ is index of refraction of liver which is 1.2 times μ₁ .

sin 50 / sinθ = 1.2

sinθ = sin 50 / 1.2

= .6384

θ = 40°

If we take d is depth of tumer in the lever

d is base of the triangle within tumer , 1.2 / 2 = .6 cm is the perpendicular with respect to angle of refraction of 40° .

.6 / d = tan 40

d = .6 / tan40

= .715 cm .