The fraction of incident ultrasound intensity reflected at an interface with Z1 = 1.60 MRayls and Z2 = 1.70 MRayls is?

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Multiple Choice

The fraction of incident ultrasound intensity reflected at an interface with Z1 = 1.60 MRayls and Z2 = 1.70 MRayls is?

Explanation:
When ultrasound encounters an interface at normal incidence, the reflected fraction of the incident intensity depends on the mismatch of the acoustic impedances Z1 and Z2. The intensity reflection coefficient is R = [(Z2 − Z1)/(Z2 + Z1)]^2. Plugging in Z1 = 1.60 and Z2 = 1.70 MRayls gives: Z2 − Z1 = 0.10, Z2 + Z1 = 3.30, so the ratio is 0.10/3.30 ≈ 0.0303. Squaring this gives R ≈ (0.0303)^2 ≈ 0.000918, which is 0.0918% when expressed as a percentage. So the fraction reflected is about 0.092%.

When ultrasound encounters an interface at normal incidence, the reflected fraction of the incident intensity depends on the mismatch of the acoustic impedances Z1 and Z2. The intensity reflection coefficient is R = [(Z2 − Z1)/(Z2 + Z1)]^2.

Plugging in Z1 = 1.60 and Z2 = 1.70 MRayls gives: Z2 − Z1 = 0.10, Z2 + Z1 = 3.30, so the ratio is 0.10/3.30 ≈ 0.0303. Squaring this gives R ≈ (0.0303)^2 ≈ 0.000918, which is 0.0918% when expressed as a percentage.

So the fraction reflected is about 0.092%.

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