91Ó°ÊÓ

It is given that,

Wavelength of light source, $\mathrm{λ}=9cm=9x{10}^{-2} m$

Width of the slit, $a=12cm=12×{10}^{-2} m$

Distance between screen and slit, R = 8.0 m

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.

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

$y_m=Rtan{\mathrm{θ}}_m$

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

For a light of wavelength$\mathrm{λ}$and slit width a , the angle is given by-

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

For first dark fringe, m = 1:

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

For second dark fringe, m = 2:

$$\begin{gathered} y_2=R\tan \left[{\sin }^{-1}\left(\frac{2\mathrm{λ}}{a}\right)\right] \\ =\left(8.0m\right)×\tan \left[{\sin }^{-1}\left(\frac{2×9.0×{10}^{-2}m}{12.0×{10}^{-2}m}\right)\right] \\ =Notdefined \end{gathered}$$

Since $\frac{2\mathrm{λ}}{a}>1,$

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

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