You are driving at night on a long, straight highway in the desert as another ca

Question

You are driving at night on a long, straight highway in the desert as another car approaches. What is the maximum distance at which you can tell that the car is not a motorcycle by seeing its two headlights, which are 1.3 m apart?
(a) Assume that your visual acuity is limited only by diffraction. Use 525 nm for the vacuum wavelength, and assume that your pupils are dilated to a 6.2 mm diameter. (The refractive index for eye is 1.34.)


(b) Assume a more realistic, typical visual acuity with ?min = 5.0 x10-4 rad.

Explanation / Answer

The resolving power of an eye is     ? = 1.22 ?/ D here ? is the wavelegnth of light in a medium , D is diameter of the pupil
           ? = ? ' / n     =525nm/1.34     = 391.8nm for small angle  ? = y / L here y is the distance between the lights, L is distance from car to eye    y / L = 1.22 ?/D L = y D / 1.22 ?    = 1.3m(6.2x10^-3)/1.22(391.8x10^-9) =  16862 m b) given ? = 5x10^-4 rad    ? = y /L L = y/? =1.3 / 5x10^-4     = 2600m
The resolving power of an eye is     ? = 1.22 ?/ D here ? is the wavelegnth of light in a medium , D is diameter of the pupil
           ? = ? ' / n     =525nm/1.34     = 391.8nm for small angle  ? = y / L here y is the distance between the lights, L is distance from car to eye    y / L = 1.22 ?/D L = y D / 1.22 ?    = 1.3m(6.2x10^-3)/1.22(391.8x10^-9) =  16862 m b) given ? = 5x10^-4 rad    ? = y /L L = y/? =1.3 / 5x10^-4     = 2600m
The resolving power of an eye is     ? = 1.22 ?/ D here ? is the wavelegnth of light in a medium , D is diameter of the pupil
           ? = ? ' / n     =525nm/1.34     = 391.8nm for small angle  ? = y / L here y is the distance between the lights, L is distance from car to eye    y / L = 1.22 ?/D L = y D / 1.22 ?    = 1.3m(6.2x10^-3)/1.22(391.8x10^-9) =  16862 m b) given ? = 5x10^-4 rad    ? = y /L L = y/? =1.3 / 5x10^-4     = 2600m

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