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The head lights of a jeep are $1.2 \mathrm{~m}$ apart. If the pupil of the eye of an observer has a diameter of $2 \mathrm{~mm}$ and light of wavelength $5896 Å$ is used, what should be the maximum distance of the jeep from the observer if the two head lights are just separated?
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The correct answer is:
$3.33 \mathrm{~km}$
Distance of jeep, $x=\frac{D \times d}{1.22 \times \lambda}$
$D=$ diameter of lens, $d=$ separation between sources.
or $x=\frac{\left(2 \times 10^{-3}\right) \times 1.2}{1.22 \times 5896 \times 10^{-10}} \mathrm{~m}$
or $x=3336 \mathrm{~m}$ or $x=3.33 \mathrm{~km}$.
$D=$ diameter of lens, $d=$ separation between sources.
or $x=\frac{\left(2 \times 10^{-3}\right) \times 1.2}{1.22 \times 5896 \times 10^{-10}} \mathrm{~m}$
or $x=3336 \mathrm{~m}$ or $x=3.33 \mathrm{~km}$.
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