91Ó°ÊÓ

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${\mathrm{λ}}_{air}=600nm$

and refractive index$n=1.33$.

We need to find the value of $\mathrm{λ}$.

Let us find the value of $\mathrm{λ}$,

$\mathrm{λ}={\mathrm{λ}}_{air}/n$

substituting the value in equation,

$\mathrm{λ}=\frac{600nm}{1.33}$

$\mathrm{λ}=450nm$

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

localid="1649143988920" $∆x=\mathrm{Î\pm }L$

where $\mathrm{Î\pm }=\frac{1.22\mathrm{λ}}{D}$

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

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

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

localid="1649143936404" $L=\frac{∆x×D}{1.22\mathrm{λ}}$

By substituting the values in the equation we get,

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

$L=15300.54m→15.30km.$

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