Ultrasonic waves, like all other waves, exhibit diffraction. Recall that the min

Question

Ultrasonic waves, like all other waves, exhibit diffraction. Recall that the minimum half angle of a beam is given by:
alpha = (1.22*lamda_m)/d
The wavelengths used in ultrasound imaging techniques are generally in the range from 0.3 mm to 0.75 mm [longer waves (lower frequencies) penetrate more deeply but provide less resolution of detail than the short waves]. In order for an ultrasonic beam to spread by less than a centimeter after it has traveled across a human torso (20 cm), the source of the waves must be about 10 or 20 times larger than the wavelength. The source of the waves is called a transducer.

Question:

A transducer is 18.5 mm in diameter and emits a wave of wavelength 0.50 mm. The beam travels a total distance of 37 cm through the patient and back to the transducer. How much has the beam spread after returning to the transducer? (Hint: Start with the formula, alpha = (1.22*lamda_m)/d , and recall that this is the minimum HALF angle of the beam.)

__________________ mm  

Explanation / Answer

I think that the answer may be 24.4mm. How about it. I got alpha=0.033. It Is minimum half angle so the angle of diffraction should be 0.066. so spread length = 0.066*370mm=24.4(mm)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.