BR Physics Ultrasound

[starbaduk]

The last passage in TBR Section 4:

Ir/Ii = ((rho1 * v1 - rho2 * v2) / (rho1* v1 + rho2 * v2)^2)

where Ir and Ii represent reflected and incident intensity,

65. The fraction of sound transmitted through a bodily interface is small, if:

I picked B. density difference between media is small.

The answer they picked is A. the density difference between media is great.

Looking at the term on the right side of the equation, higher density difference would increase the numerator therefore increase the reflected intensity, no? So, wouldn't smaller fraction of sound transmitted through the interface would result from smaller density difference?

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[Fort]

I don't have BR, but I'll try to help.

If the difference in density between the interface is great, then v2 will much slower than v1. Under this condition, the numerator will be relatively big, ultimately leading to a higher Ir/Ii (a high reflective intensity).

If the reflective intensity is high, then the transmitted intensity must be low. So there explanation makes sense to me.

Hope this helps.

 

[starbaduk]

I don't have BR, but I'll try to help.

If the difference in density between the interface is great, then v2 will much slower than v1. Under this condition, the numerator will be relatively big, ultimately leading to a higher Ir/Ii (a high reflective intensity).

If the reflective intensity is high, then the transmitted intensity must be low. So there explanation makes sense to me.

Hope this helps.

Thanks. You are right. I confused transmitted intensity with reflected.