Once dark adapter, the pupil of your eye is approximately $7mm$diameter. The headlights of an oncoming car are $120cm$apart. if the lens of your eye is diffraction-limited, at what distance are the two headlights marginally resolved? assume a wavelength of $600nm$and that the index of refraction inside the eye is $1.33$

(Your eye is not really good enough to resolve headlight at this distance, due both to aberrations in the lens and to the size of the receptors in your retina, but it comes reasonably close.

We have given that:

pupil's eye $7mmdiameter$.

headlights of car $120cm$.

assume wavelength $600nm$.

refraction inside the eye $1.33$.

Distance between oncoming carslocalid="1649144003539" $∆x=1.2m$,

diameter of eye$D=7x{10}^{-3}m$,

wavelength of air${\lambda }_{air}=600nm$

and refractive index$n=1.33$.

We need to find the value of $\lambda$.

Let us find the value of $\lambda$,

$\lambda ={\lambda }_{air}/n$

substituting the value in equation,

$\lambda =\frac{600nm}{1.33}$

$\lambda =450nm$

By using the formula we need to find the value of $L$,

localid="1649143988920" $∆x=\alpha L$

where $\alpha =\frac{1.22\lambda }{D}$

localid="1649143979018" $∆x=\frac{1.22\lambda }{D}L$

localid="1649143956866" $\frac{1.22\lambda }{D}=\frac{∆x}{L}$

$$\begin{gathered} L=\frac{1}{1.22\lambda } \\ D \end{gathered}$$

localid="1649143936404" $L=\frac{∆x\times D}{1.22\lambda }$

By substituting the values in the equation we get,

$L=\frac{1.20m(7.0x{10}^{-3}m)}{1.229(450x{10}^{-9}m)}$

$L=15300.54m\to 15.30km.$

Here, $L$is the distance between two headlights marginally resolved.