How Is Sound Intensity Affected by Diffraction and Slit Width?

[6021023]

Homework Statement

Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 9.20 cm passes through a narrow slit 11.5 cm wide. A microphone is placed 35.0 cm directly in front of the center of the slit. The microphone is then moved in a direction perpendicular to the line from the center of the slit to point O, the center of the diffraction pattern.

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

Homework Equations

sin \theta = \frac{m\lambda}{a}

tan \theta = \frac{y}{L}

\lambda = 9.20 cm, a = 11.5 cm, L = 35.0 cm, m = 1

The Attempt at a Solution

tan \theta = sin \theta

\frac{y}{L} = \frac{m\lambda}{a}

y = \frac{m\lambda L} {a}

y = \frac{1 \times 9.20 cm \times 35.0 cm}{11.5 cm}

y = 28 cm

However, this is not the correct answer. The correct answer is given as y = 46.7 cm. I suspect that my solution didn't work because the angles are not small enough so that the sine and tangent functions can not be set equal to each other. How do you get the correct answer?

 

[hage567]

I suspect that my solution didn't work because the angles are not small enough so that the sine and tangent functions can not be set equal to each other. How do you get the correct answer?

So try it without this approximation. Based on the geometry of the problem how would you express sinθ? Figure that out and put it in your equation. The math gets a bit messier but it is relatively a straightforward calculation.