Question:medium

In an ultrasonic transducer with radius \( R \) and ultrasound wavelength \( \lambda \), the near field extends to a distance given by:

Show Hint

In clinical practice where the wavelength \( \lambda \) is extremely tiny compared to the transducer radius \( R \), the second term \( \frac{\lambda}{4} \) becomes so negligible that the formula is frequently abbreviated as simply \( x \approx \frac{R^2}{\lambda} \) or \( \frac{D^2}{4\lambda} \). However, the mathematically precise boundary expression includes the exact correction factor: \( \left(\frac{R^2}{\lambda}\right) - \left(\frac{\lambda}{4}\right) \).
Updated On: Jun 23, 2026
  • \( \left(\frac{R^2}{\lambda}\right) - \left(\frac{\lambda}{4}\right) \)
  • \( \left(\frac{R^2}{\lambda}\right) - \left(\frac{\lambda}{16}\right) \)
  • \( \left(\frac{R}{\lambda}\right) - \left(\frac{\lambda}{2}\right) \)
  • \( \left(\frac{R^2}{2\lambda}\right) - \left(\frac{\lambda}{4}\right) \)
Show Solution

The Correct Option is A

Solution and Explanation

Was this answer helpful?
0