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You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headlights that are 0.641 m apart. At what distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source? Take the wavelength of the light to be 549 nm and your pupil diameter to be 4.89 mm.

User AtoMerz
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2 Answers

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Final answer:

Using the Rayleigh criterion for resolution, we find that the human eye can marginally distinguish two separate headlights that are 0.641 m apart at a distance of approximately 468 meters, which is 0.468 kilometers.

Step-by-step explanation:

To determine the distance at which a person can distinguish two separate headlights, we use the Rayleigh criterion for resolution, which states that the minimum angular resolution, θ, that the eye can discern is given by:

θ = 1.22 λ / D

where λ is the wavelength of light and D is the diameter of the pupil. The distance, d, at which the two points of light (the headlights) become resolvable is then calculated using the small angle approximation:

d = l / θ

where l is the distance between the two headlights.

Given λ = 549 nm = 549 × 10⁻⁹ m, D = 4.89 mm = 4.89 × 10⁻3 m, and l = 0.641 m, we can now solve for θ and consequently for d:

θ = 1.22 × 549 × 10⁻⁹ m / 4.89 × 10⁻3 m = 1.37 × 10⁻B radians

d = 0.641 m / (1.37 × 10⁻B radians) = 468 meters

Therefore, the distance in kilometers is 0.468 km, which is when you would be marginally able to discern that there are two separate headlights.

User Ronnette
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Answer:


d_e=4.679km

Step-by-step explanation:

x' = 2 * 1.22 * lambda * ( f / D )

lambda = wavelength of light in mm

f = focal length in mm

D = aperture diameter in mm


sin(\beta)=1.22*(L)/(d)


sin(\beta)=1.22*(549x10^(-9)m)/(4.98x10^(-3)m)


sin(\beta)=1.36x10^(-4)m

sinβ=light separation/ distance eye

distance eye=light /sinβ


d_e=(0.641m)/(1.369x10^(-4)m)=4679.87m


d_e=4.679km

User Pluckyglen
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