Question

A pilot at an altitude of h=7300 m looks down at the ground. His eye has an aperture of D=2.2 mm and index of refraction of n=1.35. His eye can see wavelengths up to λ=721 nm. Randomized Variables h=7300 m

D=2.2 mm
n=1.35
λ=721 nm
​
\& 50% Part (a) Write an expression, in terms of h,D, and n, for the minimum separation d two objects on the ground can have and still be distinguishable at the wavelength λ d=nhD

Answer 1

The minimum separation, d, between two objects on the ground that can still be distinguishable at a wavelength, λ, can be determined by the formula d = n×h×D.. Substituting the given values into this formula, the minimum separation is approximately 21.726 meters.

Step-by-step explanation:

The minimum separation, d, between two objects on the ground that can still be distinguishable at a wavelength, λ, can be determined by the formula d = n×h×D.

Using the given values:

• h = 7300 m
• D = 2.2 mm = 0.0022 m
• n = 1.35
• λ = 721 nm = 0.000721 m

Substituting these values into the formula, we have: d = 1.35 × 7300 × 0.0022 = 21.726 m

Therefore, the minimum separation, d, between the two objects on the ground is approximately 21.726 meters.