The two headlights of an approaching automobile are $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}$ apart. At what (a) angular separation and (b) maximum distance will the eye resolve them? Assume that the pupil diameter is $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{mm}$, and use a wavelength of $\mathbf{550}\mathbf{nm}$ for the light. Also assume that diffraction effects alone limit the resolution so that Rayleigh’s criterion can be applied.

Distance between two head lights$D=1.4\text{m}$

Diameter of the pupil $d=5.0\text{mm}$

Wavelength of light $\lambda =550 \text{nm}$

The Rayleigh criteria specify the minimal distance between two light sources that must exist in order to resolve them into separate objects.

The angular separation is defined by,

${\mathit{\theta }}_{\mathbf{R}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}}$ ...(i)

Here, d is the pupil diameter.

The maximum distance at which the eye will resolve is:

$\mathit{L}\mathbf{=}\frac{\mathbf{D}\mathbf{d}}{\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{\lambda }}$ ...(ii)

Where D is the distance between the two headlights

The arrangement of separation D between two head lights their angular separation and the viewing distance L .

Distance between two head lights$D=1.4\text{m}$

Diameter of the pupil $d=5.0\text{mm}$

$$\begin{gathered} d=5.0\text{mm} \\ \text{=}\left(5.0\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right) \\ =5.0\times {10}^{-3}\text{m} \end{gathered}$$

Wavelength of the light$\lambda =550\text{nm}$

role="math" localid="1663024691134" $$\begin{gathered} d\text{=}\left(550\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right) \\ =550\times {10}^{-9}\text{m} \end{gathered}$$

The angular separation between the headlights as shown in the above figure is given by an equation (i)

$$\begin{gathered} {\theta }_R=1.22\frac{\lambda }{d} \\ =1.22\left(\frac{550\times {10}^{-9}\text{m}}{5\times {10}^{--3}\text{m}}\right) \\ =1.34\times {10}^{-4}\text{rad} \end{gathered}$$

Therefore, the angular separation is$1.34\times {10}^{-4}\text{rad}$

From equation (ii), the maximum distance at which the eye can resolve the head lights as separate is

$$\begin{gathered} \text{L =}\frac{Dd}{1.22\lambda } \\ =\frac{\left(1.4\text{m}\right)\left(5\times {10}^{-3}\text{m}\right)}{1.22\left(550\times {10}^{-9}\text{m}\right)} \\ =10.43\times {10}^3\text{m} \\ \text{=}\left(10.43\times {10}^3\text{m}\right)\left(\frac{1\text{km}}{{10}^3\text{m}}\right) \\ =10.43\text{km} \end{gathered}$$

Therefore, the maximum distance at which the eye will resolve the headlights is$10.43 \text{km}$ .