Near field length formula for a circular-aperture transducer with diameter D and wavelength λ?

• A. NF ≈ D^2 / (4λ)
• B. NF ≈ 2λ / D
• C. NF ≈ D/λ
• D. NF ≈ πD^2/(4λ)

Answer: (A)

The main idea is how far the finite circular aperture influences the beam before it settles into a stable, predictable pattern. In the near field, the wavefronts from different parts of the transducer interfere strongly, and the field isn’t yet a clean, uniform beam. As you move farther away, the contributions from all parts of the aperture begin to combine into a smoother, quasi-plane wave in the far field.

For a circular aperture of diameter D, look along the axis at a distance z. The path difference between the center of the aperture and the edge is roughly Δ ≈ sqrt(z^2 + (D/2)^2) − z. When z is not much larger than D, this difference is significant; using a small-angle approximation gives Δ ≈ D^2/(8z). The near-field region is considered to extend until this phase difference becomes about a half-wavelength (onset of substantial destructive interference across the aperture), so set Δ ≈ λ/2. Solving D^2/(8z) ≈ λ/2 yields z ≈ D^2/(4λ).

Thus, the near field length for a circular-aperture transducer is approximately D^2/(4λ). This means a larger aperture or a shorter wavelength (higher frequency) increases the extent of the near field.