Question

You are in an airplane at an altitude of 10,000 m. The pupil of your eye is about 3.0 mm in diameter and the wavelength of light $\lambda$ = 550 nm. If you look down at the ground, the minimum separation s between objects that you could distinguish is :

• A. 1.24 m
• B. 2.24 m
• C. 9.92 m
• D. 4.48 m

Answer 1

Answer: (B)

2.24 m

Answer 2

The minimum separation $s$ between two objects that can be distinguished is given by the formula based on the Rayleigh criterion for a circular aperture:

$s = R \theta$

where $R$ is the distance to the objects (altitude of the airplane) and $\theta$ is the angular resolution of the eye. The angular resolution is given by:

$\theta = 1.22 \frac{\lambda}{d}$

where $\lambda$ is the wavelength of light and $d$ is the diameter of the aperture (pupil).

Given values:

• Altitude $R = 10,000$ m
• Pupil diameter $d = 3.0$ mm $= 3.0 \times 10^{-3}$ m
• Wavelength $\lambda = 550$ nm $= 550 \times 10^{-9}$ m

First, calculate the angular resolution $\theta$:

$\theta = 1.22 \times \frac{550 \times 10^{-9} \text{ m}}{3.0 \times 10^{-3} \text{ m}}$

$\theta = 1.22 \times \frac{550}{3.0} \times 10^{-6} \text{ radians}$

Now, calculate the minimum separation $s$:

$s = R \theta$

$s = 10000 \text{ m} \times \left( 1.22 \times \frac{550}{3.0} \times 10^{-6} \right) \text{ radians}$

$s = 10^4 \times 1.22 \times \frac{550}{3.0} \times 10^{-6} \text{ m}$

$s = 1.22 \times \frac{550}{3.0} \times 10^{4-6} \text{ m}$

$s = 1.22 \times \frac{550}{3.0} \times 10^{-2} \text{ m}$

$s = \frac{1.22 \times 550}{300} \text{ m}$

$s = \frac{671}{300} \text{ m}$

$s \approx 2.2367 \text{ m}$

Rounding to two decimal places, the minimum separation is approximately 2.24 m.