Question

The wall of a large room is covered with acoustic tile in which small holes are drilled $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{mm}$from centre to centre. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer’s eye to be $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{mm}$, and the wavelength of the room light to be $\mathbf{550}\mathbf{nm}$?

Short answer

The distance that the person from the tile distinguishes the individual holes

(Rounding off to two significant figures) is$30\text{m}$ .

Step-by-step solution

Introduction

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.

Concept

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

$\theta =\frac{1.22\lambda }{d}$

Here,$\lambda$ is wave length, and d is circular aperture diameter.

Since the regular separation $\theta$ is very small, the value of $\sin \theta$ is nearby equal to$\theta$

$\theta =\frac{D}{L}$

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

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

$\theta =\frac{1.22\lambda }{d}$ to obtain the expression for L as follows:

$\frac{D}{L}=\frac{1.22\lambda }{d}$

Rearrange the above equation for L as follows:

$L=\frac{Dd}{d1.22\lambda }$

Calculation

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 $\lambda$from nanometre to metre as follows:

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

Find the distance

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

$550\times {10}^{-9}\text{m}$ for $\lambda$ in the above equation

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

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

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

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