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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

User Xarcell
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Answer:

The distance between their eye and the ceiling is 33.06 m.

Step-by-step 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,
\lambda=550\ nm

We need to find the distance between their eye and the ceiling. Using Rayleigh criteria, we get the distance as follows :


L=(Dd)/(1.22\lambda)\\\\L=(4.35* 10^(-3)* 5.1* 10^(-3))/(1.22* 550* 10^(-9))\\\\L=33.06\ m

So, the distance between their eye and the ceiling is 33.06 m.

User Sobin Augustine
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