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The speed of sound in human tissue is of the order of `1500 m//s`. A `3.5 MHz` probe is used for an ultrasonic procedure. (a) If the effective physical depth of the ultrasound is `250` wavelengths, what is the physical depth in metres ? (b) What is the time lapse for the ultrasound to make a round trip if reflected from an object at the effective depth ? (c ) The smallest detail capable of being detected is of the order of one wavelength of the ultrasound. What would this be ? |
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Answer» (a) `lamda = (v)/(v) = (1500 m//s)/(3.5 xx 10^6 m) = 428.6 xx 10^6 m` Physical depth of ultrasound in metres `= 200 lamda = 250 (428.6 xx 10^6 m) = 0.10715 m` (b) `t = (s)/(v) = (2 xx 0.10715 m)/(1500 m//s) = 0.000143 s = 1.43 xx 10^-4 s` ( c) The smallest possible detail capable of being detected `= lamda = 428.6 xx 10^-6 m = 4.29 xx 10^-4 m`. |
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