Question

Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 8.90 cm passes through a narrow slit 11.5 cm wide. A microphone is placed 44.0 cm directly in front of the center of the slit, corresponding to point  O.The microphone is then moved in a direction perpendicular to the line from the center of the slit to point  O.

At what minimal distance from O will the intensity detected by the microphone be zero?

Answer 1

Concepts and reason

The concept required to solve this problem is single slit diffraction.

Initially, draw a diagram showing the slit and the microphone’s position. Then, determine the minimal distance from the point where the intensity detected by the microphone be zero by using the condition for the minima in a single slit diffraction.

Fundamentals

The expression of the diffraction minima in a single slit diffraction can be given as follows:

$d\sin \theta = m\lambda$

Here, d is the slit width, $\theta$ is the diffraction angle, m is the order of diffraction, and $\lambda$ is the wavelength of the sound.

The figure 1 shows the single slit diffraction pattern. The slit is of width ‘d’ and the microphone is placed at point O. the new position of the microphone is at point A, where the intensity of sound becomes zero.

The angle $\theta$ can be calculated by using the trigonometry as follows:

$\tan \theta = \frac{{\;{\rm{perpendicular}}}}{{{\rm{base}}}}$

Here, perpendicular of the triangle is y and base is L.

Substitute y for perpendicular and L for base in the above expression.

$\begin{array}{c}\\\tan \theta = \frac{y}{L}\\\\y = L\tan \theta \\\end{array}$

The expression of the diffraction minima in a single slit diffraction can be given as follows:

$d\sin \theta = m\lambda$

Rearrange the above expression for $\theta$ .

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

Substitute 1 for m, 8.90 cm for $\lambda$ , and 11.5 cm for d in the above expression.

$\begin{array}{c}\\\theta = {\sin ^{ - 1}}\left( {\frac{{\left( 1 \right)\left( {8.90{\rm{ cm}}} \right)}}{{11.5{\rm{ cm}}}}} \right)\\\\ = 0.88499693{\rm{ rad}}\\\end{array}$

Now, substitute 0.88499693 rad for $\theta$ and 44.0 cm for L in the equation

$\begin{array}{c}\\y = \left( {44.0{\rm{ cm}}} \right)\tan \left( {0.8849969{\rm{ rad}}} \right)\\\\ = 53.77{\rm{ cm}}\\\\ = 53.8{\rm{ cm}}\\\end{array}$

Ans:

The minimal distance from point O is 53.8 cm.