Question

Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.
At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.

Answer 1

Concepts and reason

The concept of diffraction is used to solve this problem.

Use the condition of the single slit diffraction to find the minimum angle expression. Substitute the value and find the angle. Find the angle for $m = 1,2,3$ .

Fundamentals

The condition of the single slit diffraction is given as follows:

$\sin \theta = \frac{{m\lambda }}{W}$

Here, $\lambda$ is the wavelength of the source, W is the width of the slit and m is the integer.

The expression to evaluate the wavelength is given as follows:

$\lambda = \frac{c}{f}$

Here, c is the speed of the source and f is the frequency of the source.

Substitute $344{\rm{ m/s}}$ for c and 1250 Hz for f in the expression $\lambda = \frac{c}{f}$ .

$\begin{array}{c}\\\lambda = \frac{{344{\rm{ m/s}}}}{{1250{\rm{ Hz}}}}\\\\ = 0.2752{\rm{ m}}\\\end{array}$

Re-arrange the expression $\sin \theta = \frac{{m\lambda }}{W}$ for $\theta$ .

$\theta = {\sin ^{ - 1}}\left( {\frac{{m\lambda }}{W}} \right)$

Substitute 0.2752 m for $\lambda$ , 1.05 m for W and 1 for m in the expression $\theta = {\sin ^{ - 1}}\left( {\frac{{m\lambda }}{W}} \right)$ .

$\begin{array}{c}\\\theta = {\sin ^{ - 1}}\left( {\frac{{\left( 1 \right)\left( {0.2752{\rm{ m}}} \right)}}{{1.05{\rm{ m}}}}} \right)\\\\ = 15.19^\circ \\\end{array}$

Substitute 0.2752 m for $\lambda$ , 1.05 m for W and 2 for m in the expression $\theta = {\sin ^{ - 1}}\left( {\frac{{m\lambda }}{W}} \right)$ .

$\begin{array}{c}\\\theta = {\sin ^{ - 1}}\left( {\frac{{\left( 2 \right)\left( {0.2752{\rm{ m}}} \right)}}{{1.05{\rm{ m}}}}} \right)\\\\ = 31.61^\circ \\\end{array}$

Substitute 0.2752 m for $\lambda$ , 1.05 m for W and 3 for m in the expression $\theta = {\sin ^{ - 1}}\left( {\frac{{m\lambda }}{W}} \right)$ .

$\begin{array}{c}\\\theta = {\sin ^{ - 1}}\left( {\frac{{\left( 3 \right)\left( {0.2752{\rm{ m}}} \right)}}{{1.05{\rm{ m}}}}} \right)\\\\ = 51.84^\circ \\\end{array}$

Ans:

The first three minimum value of angle $\theta$ is $15.19^\circ ,31.61^\circ {\rm{ and 51}}{\rm{.84}}^\circ$ .