Q5E - University Physics with Modern Physics

Question

Question: Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 9.00 cm  passes through a slit 12.0 cm wide. A microphone is placed  8.00 m 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. At what distances from O will the intensity detected by the microphone be zero?

Step-by-Step Solution

Answer

the intensity of the microphone will be zero at  $\pm 9.07 m$.

Step 1: Given Data

It is given that,

Wavelength of light source,  $\lambda =9cm=9x{10}^{-2}\hspace{0.33em}m$

Width of the slit, $a=12cm=12\times {10}^{-2}\hspace{0.33em}m$ 

Distance between screen and slit, R = 8.0 m  

Step 2: Concept

When a propagating sound wave moves around the edges of an object of size smaller or equal to the wavelength of sound, producing interference effects, it is known as the diffraction of sound.

Step 3: Total number of dark fringes

 

 

Position of dark fringe in one slit experiment is given by:

$y_m=R tan {\theta }_m$ 

  

Here, $y_m$ is the distance from the central maxima, R is the distance between screen and slits and  ${\theta }_m$  is the diffraction angle.

 

For a light of wavelength  $\lambda$ and slit width a , the angle is given by-

 

 $$\begin{gathered} \sin {\theta }_1=\frac{m\lambda }{a} \\ {\theta }_1={\sin }^{-1}\left(\frac{m\lambda }{a}\right) \end{gathered}$$ 

 

For first dark fringe, m = 1:

 

$$\begin{gathered} y_1=R \tan \left[{\sin }^{-1}\left(\frac{\lambda }{a}\right)\right] \\ =\left(8.0 m\right)\times \tan \left[{\sin }^{-1}\left(\frac{9.0\times {10}^{-2} m}{12.0\times {10}^{-2} m}\right)\right] \\ =\pm 9.07 m \end{gathered}$$ 

 

For second dark fringe, m = 2:

 

$$\begin{gathered} y_2=R \tan \left[{\sin }^{-1}\left(\frac{2\lambda }{a}\right)\right] \\ =\left(8.0 m\right)\times \tan \left[{\sin }^{-1}\left(\frac{2\times 9.0\times {10}^{-2} m}{12.0\times {10}^{-2} m}\right)\right] \\ =Not defined \end{gathered}$$ 

 

Since $\frac{2\lambda }{a}>1,$ 

$\sin \left(\frac{2\lambda }{a}\right)$ is not defined for that point.

 

Therefore, the only answer is $\pm 9.07 m$ at which intensity detected by microphone will be zero.