Question

The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center. If a person with a pupil of diameter 5.10 mm walks into the hall when it is illuminated by 550 nm wavelength light and they can just distinguish the individual holes, what is the distance between their eye and the ceiling?

Answer 1

33.0 m

Step-by-step explanation:

We are given that

Wavelength,
`\lambda=550nm=550* 10^(-9) m`

`1nm=10^(-9) m`

Diameter of pupil=d=5.1 mm=
`5.1* 10^(-3) m`

`1mm=10^(-3) m`

`y=4.35 mm=4.35* 10^(-3)m`

By Rayleigh criteria

`sin\theta=1.22(\lambda)/(d)`

`(y)/(L)=1.22(\lambda)/(d)`

`L=(yd)/(1.22\lambda)`

Substitute the values

`L=(4.35* 10^(-3)* 5.1* 10^(-3))/(1.22* 550* 10^(-9))`

`L=33.0 m`