Question

36.3 diffraction by a circular aperture
the two headlights on an on-coming car are 1.47 m apart. at what distance (in km) will an eye with pupil diameter of 4.88 mm be able to resolve the two headlights? assume a wavelength of 584 nm for the light.
km
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Explanation:

Step1: Recall Rayleigh Criterion

The Rayleigh criterion for angular resolution is \(\theta = 1.22\frac{\lambda}{D}\), where \(\lambda\) is wavelength, \(D\) is aperture diameter.
\(\lambda = 584\,\text{nm} = 584\times10^{-9}\,\text{m}\), \(D = 4.88\,\text{mm} = 4.88\times10^{-3}\,\text{m}\).

Step2: Relate Angular Resolution to Distance

For small angles, \(\theta\approx\frac{d}{L}\), where \(d = 1.47\,\text{m}\) (separation of headlights), \(L\) is distance.
Set \(1.22\frac{\lambda}{D}=\frac{d}{L}\), solve for \(L\): \(L=\frac{dD}{1.22\lambda}\).

Step3: Substitute Values

\(L=\frac{1.47\times4.88\times10^{-3}}{1.22\times584\times10^{-9}}\)
Calculate numerator: \(1.47\times4.88\times10^{-3}\approx7.1736\times10^{-3}\)
Denominator: \(1.22\times584\times10^{-9}\approx7.1248\times10^{-7}\)
\(L\approx\frac{7.1736\times10^{-3}}{7.1248\times10^{-7}}\approx10068.5\,\text{m}\approx10.1\,\text{km}\) (rounded appropriately).

Answer:

\(10.1\) (or more precise calculation: ~10.1 km, exact value may vary slightly with calculation precision)