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Hoochie [10]
3 years ago
8

While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish

two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.
Physics
1 answer:
Bingel [31]3 years ago
6 0

Answer:

the smallest separation between two objects is 0.8067 m

Explanation:

Given the data in the question;

Altitude h = 5.75 km = 5750 m

Diameter D = 4.0 mm = 0.004 m

λ = 460 nm = 4.6 × 10⁻⁷ m

Now, Using Rayleigh criterion for Airy disks resolution.

we know that, Minimum angular separation for resolving two points is;

θ = 1.22λ / D

so we substitute

θ = (1.22 × 4.6 × 10⁻⁷)  / 0.004

θ = 5.612 × 10⁻⁷ / 0.004

θ = 1.403 × 10⁻⁴ rad  

so minimum separation d_{min =  θh

so we substitute

d_{min = (1.403 × 10⁻⁴) × 5750 m

d_{min = 0.8067 m

Therefore, the smallest separation between two objects is 0.8067 m

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