Question

36.3 diffraction by a circular aperture
a circular diaphragm 59.45 cm in diameter oscillates at a frequency of 16.84 khz as an underwater source of sound used for submarine detection. far from the source, the sound intensity is distributed as the diffraction pattern of a circular hole whose diameter equals that of the diaphragm. take the speed of sound in water to be 1450. m/s, and find the angle (in degrees) between the normal to the diaphragm and a line from the diaphragm to the first minimum.

Explanation:

Step1: Find the wavelength of the sound wave

The formula for the wavelength \(\lambda\) is \(\lambda=\frac{v}{f}\), where \(v = 1450\space m/s\) is the speed of sound in water and \(f=16.84\times 10^{3}\space Hz\) is the frequency.
\(\lambda=\frac{1450}{16.84\times 10^{3}}\approx0.0861\space m\)

Step2: Use the circular aperture diffraction formula for the first minimum

The formula for the angle \(\theta\) of the first minimum in circular aperture diffraction is \( \sin\theta = 1.22\frac{\lambda}{D}\), where \(D = 59.45\space cm=0.5945\space m\) is the diameter of the diaphragm.
Substitute \(\lambda = 0.0861\space m\) and \(D = 0.5945\space m\) into the formula:
\(\sin\theta=1.22\times\frac{0.0861}{0.5945}\)
\(\sin\theta=1.22\times0.1448\)
\(\sin\theta\approx0.1767\)

Step3: Find the angle \(\theta\)

Take the inverse sine of \(0.1767\) to get \(\theta\):
\(\theta=\arcsin(0.1767)\approx10.2^{\circ}\)

Answer:

\(10.2\) (The answer may vary slightly due to rounding during calculation)