Chapter 19, Problem 66A

To determine

TO-DETERMINE: the distance upto which the human eye can distinguish the two headlights at night.

Answer to Problem 66A

The distance upto which the eye can distinguish the two objects is $L_{\text{obj}}=22\text{km}$ .

Explanation of Solution

Given info:

The wavelength of the light is $\lambda =525\text{nm}$ .

The diameter of the aperture of the eye is $D=8.0\text{mm}$ .

The distance between the headlight is $x_{\text{obj}}=1.8\text{m}$ .

Consider the distance upto which the human eye can distinguish the two headlights at night is denoted by $L_{\text{obj}}$ .

Formula used:

Consider the eye distinguish the two headlights placed at a distance $L_{\text{obj}}$ .

Show the expression for the resolution of the eye as follows:

  $x_{\text{obj}}=\frac{1.22\lambda L_{\text{obj}}}{D}$

Here, $D$ is the diameter of the aperture, $\lambda$ is the wavelength of the light, and $L_{\text{obj}}$ is the distance between the eye and the object.

Calculation:

Consider the eye distinguish the two headlights placed at a distance $L_{\text{obj}}$ .

Find the distance $\left(L_{\text{obj}}\right)$ as follows:

  $\begin{array}{l}{L}_{\text{obj}}=\frac{x_{\text{obj}}D}{1.22\lambda }\\ L_{\text{obj}}=\frac{1.8\text{m}\times \left(8\text{mm}\times \frac{1\text{m}}{1000\text{mm }}\right)}{1.22\times \left(525\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ L_{\text{obj}}=\left(\frac{1.8\text{m}\times 0.008\times {10}^{-3}\text{m}}{1.22\times 525\times {10}^{-9}\text{m}}\right)\\ L_{\text{obj}}=22\text{km}\end{array}$

Hence, the distance upto which the eye can distinguish the two objects is $L_{\text{obj}}=22\text{km}$ .

Conclusion:

Thus, the distance upto which the eye can distinguish the two objects is $L_{\text{obj}}=22\text{km}$ .