Answer:
The distance between their eye and the ceiling is 33.06 m.
Explanation:
Given that,
The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center, D = 4.35 mm
Diameter of the eye of pupil, d = 5.1 mm
Wavelength, [tex]\lambda=550\ nm[/tex]
We need to find the distance between their eye and the ceiling. Using Rayleigh criteria, we get the distance as follows :
[tex]L=\dfrac{Dd}{1.22\lambda}\\\\L=\dfrac{4.35\times 10^{-3}\times 5.1\times 10^{-3}}{1.22\times 550\times 10^{-9}}\\\\L=33.06\ m[/tex]
So, the distance between their eye and the ceiling is 33.06 m.
