Chapter 36: Q32P (page 1111)

(a) A circular diaphragm 60 cm in diameter oscillates at a frequency of 25 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 between the normal to the diaphragm and a line from the diaphragm to the first minimum. (b) Is there such a minimum for a source having an (audible) frequency of 1.0 kHz?

Expert verified

Step by step solution

For the first minima in diffraction pattern, the diffraction angle is given by,

$s i n θ = 1.22 \frac{λ}{d} θ = s i n^{- 1} (1.22 \frac{λ}{d}) . . . . . . (1)$

Find the wavelength as follows.

$λ = \frac{v}{f} = \frac{1450}{25 \times 10^{3}} = 0.058 m$

Substitute $0.6 m$ for $d$ and $0.058 m$ for $λ$ in equation (1).

$θ = \sin^{- 1} (\frac{1.22 (0.058 m)}{0.6 m}) θ = \sin^{- 1} (0.1179) θ = 6.77 °$

Therefore, the required angle is $θ = 6.77 °$ .

Find the wavelength as follows.

$λ = \frac{v}{f} = \frac{1450}{1 \times 10^{3}} = 1.45 m$

Substitute $0.6 m$ for $d$ and $1.45 m$ for $λ$ in equation (1).

$\sin θ = (\frac{1.22 (1.45 m)}{0.6 m}) \sin θ = 2.95 > 1$

It is known that $\sin θ > 1$ is not possible. So, no minima are possible.

Therefore, there is no minimum for a source having a frequency of $1 kHz$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.