Arrival time from two US transducers

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Hi,

Please refer to the attached image.

T1 and T2 are ultrasound transducers. The focal point of the transducers is 2cm into the liver tissue.

Q) A pulse is emitted from both transducers at the same time. Calculate the difference in arrival time at the focal point.

I seem to be getting the wrong answer. This is what I'm doing:

The arrival time from T2 is: [tex]t_{T2} = 1cm/1450 + 2cm/1540 = 19.9us[/tex]

From this we can calculate an equivalent average speed across both tissue:
[tex]c_{equiv} = 3cm/19.9us = 1508.8m/s[/tex]

Using this we'll calculate the arrival time for T1
[tex]t_{T1} = \sqrt{(9.6mm)^2 + (3cm)^2}/c_{equiv} = 20.9us[/tex]

Thus the difference in arrival time is [tex]1us[/tex]
However the answer sheet indicates [tex]0.4us[/tex]

Thanks for your input.
 
Attachments
  • Diagram of two ultrasound transducers T1 and T2 with a red focal point at the right; left region labelled Fat tissue, right region Liver tissue; T1 to T2 distance 9.6mm, horizontal 1cm and 2cm, C fat = 1450m/s, C liver = 1540m/s.
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Discussion

Hint: Snell's law

EDIT: On reflection (pun!) I can see that refraction of the sound wave at the fat/liver interface is not going to make a great deal of difference in the result... So perhaps your answer is correct and the given answer is not.
 
Last edited:
Hi,

Thanks. We aren't expected to consider to reflection in this question.

So my approach of finding the equivalent sound speed is right?
 
enc08 said:
Hi,

Thanks. We aren't expected to consider to reflection in this question.

So my approach of finding the equivalent sound speed is right?

Sure. The method works because, not considering refraction, the ratio of the times spent in each media is the same for both paths.

If refraction was taken into account and resulted in a different ratio, then this would not be the case.