91Ó°ÊÓ

Acoustic ceiling tiles lay into a suspended or dropped ceiling grid and are used alter the acoustics in the room. These tiles can be made of various acoustical materials like fiber glass, foam, wood, polyester, and other substrates.

According to Rayleigh’s criterion, the angular separation $\mathrm{θ}$ between two point sources is given as follows

$\mathrm{θ}=\frac{1.22\mathrm{λ}}{d}$

Here,$\mathrm{λ}$ is wave length, and d is circular aperture diameter.

Since the regular separation $\mathrm{θ}$ is very small, the value of $\sin \mathrm{θ}$ is nearby equal to$\mathrm{θ}$

$\mathrm{θ}=\frac{D}{L}$

Here, D is distance between two point sources centre to centre, and L is distance between screen and source.

Substitute$\frac{D}{L}$ for $\mathrm{θ}$ in the above equation

$\mathrm{θ}=\frac{1.22\mathrm{λ}}{d}$ to obtain the expression for L as follows:

$\frac{D}{L}=\frac{1.22\mathrm{λ}}{d}$

Rearrange the above equation for L as follows:

$L=\frac{Dd}{d1.22\mathrm{λ}}$

Converts the units of D from millimetre to metre as follows:

$$\begin{gathered} 5.0\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right) \\ =0.005\text{m} \end{gathered}$$

Convert the units of diameter d from millimetre to metre as follows:

$$\begin{gathered} d=4.0\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right) \\ =0.004\text{m} \end{gathered}$$

Converts the units of wave length $\mathrm{λ}$from nanometre to metre as follows:

$$\begin{gathered} \mathrm{λ}=550\text{mm}\left(\frac{1\text{m}}{{10}^9\text{nm}}\right) \\ =550×{10}^{-9}\text{m} \end{gathered}$$

Substitute $0.005\text{m}$for D,$0.004m$for d , and

$550×{10}^{-9}\text{m}$ for $\mathrm{λ}$ in the above equation

$L=\frac{Dd}{1.22\mathrm{λ}}$ to obtain the value of L as follows:

$$\begin{gathered} L=\frac{\left(0.005\text{m}\right)\left(0.004\text{m}\right)}{1.22\left(550×{10}^{-9}\text{m}\right)} \\ =29.8\text{m} \end{gathered}$$

Thus, the distance that the person from the tile distinguishes the individual holes

(Rounding off to two significant figures) is $30m$.