The headlights of a jeep are 1.2 ,m apart. If | vedclass

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(C) The condition for the two headlights to be just resolved is given by the Rayleigh criterion: $	heta = 1.22 \frac{\lambda}{D}$, where $	heta$ is

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$3.34 \,km$

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(C) The condition for the two headlights to be just resolved is given by the Rayleigh criterion: $	heta = 1.22 \frac{\lambda}{D}$, where $	heta$ is the angular resolution, $\lambda$ is the wavelength of light, and $D$ is the diameter of the pupil.Also, $	heta = \frac{d}{x}$, where $d$ is the separation between the headlights and $x$ is the distance of the jeep from the observer.Equating the two expressions: $\frac{d}{x} = 1.22 \frac{\lambda}{D} \Rightarrow x = \frac{d 	imes D}{1.22 	imes \lambda}$.Given: $d = 1.2 \,m$, $D = 2 \,mm = 2 	imes 10^{-3} \,m$, $\lambda = 5896 \,	ext{Å} = 5896 	imes 10^{-10} \,m$.Substituting the values: $x = \frac{1.2 	imes 2 	imes 10^{-3}}{1.22 	imes 5896 	imes 10^{-10}} \approx 3336 \,m$.Converting to kilometers: $x \approx 3.34 \,km$.

The headlights of a jeep are $1.2 \,m$ apart. If the pupil of the eye of an observer has a diameter of $2 \,mm$ and light of wavelength $5896 \,\text{Å}$ is used, what should be the maximum distance of the jeep from the observer if the two headlights are just separated?

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