Question

The head light of a jeep are \(1.2 \mathrm{~m}\) apart. If the pupil of the eye of an observer has a diameter of \(2 \mathrm{~mm}\) and light of wavelength \(5896 \AA\) is used what should be the maximum distance of the jeep from the observer if two head lights are just seem to be separated apart? (A) \(30.9 \mathrm{~km}\) (B) \(33.4 \mathrm{~km}\) (C) \(3.34 \mathrm{~km}\) (D) \(30.9 \mathrm{~km}\)

Short answer

(C) \(3.34 \mathrm{~km}\)

Step-by-step solution

Convert Given Parameters to Compatible Units

Since the calculations will involve trigonometric functions which often expect radian inputs, it would be good to convert the given units into SI units. The diameter of the pupil is given in millimeters (mm). Convert this into meters (m) by multiplying by \(1 \times 10^{-3}\). So, \(D = 2 \times 10^{-3} m\). The wavelength of light is given in Angstroms (\( \AA\)). Convert this to meters by multiplying by \(1 \times 10^{-10}\). So, \( \lambda = 5896 \times 10^{-10} m\).

Calculate Minimum Resolvable Angle

Using the formula for the Rayleigh criterion \(\theta = 1.22 \cdot \frac{\lambda}{D}\), plug in the values for \(D\) and \( \lambda\), as calculated in the previous step. This gives \(\theta = 1.22 \cdot \frac{5896 \times 10^{-10}}{2 \times 10^{-3}}\) which simplifies to \(\theta = 3.58 \times 10^{-4} rad\). This is the minimum resolvable (or distinguishable) angle, in radians, for the two headlights.

Calculate the Maximum Distance of the Jeep

From the calculated angle, we can determine the maximum distance of the Jeep. The position of the headlights can be considered as a small angle triangle, where the headlights' separation (\(1.2 m\)) is the opposite side, and the distance to the Jeep is the adjacent side. So, the tangential value of the angle \( \theta\) would be the ratio of these two. Therefore, we can use the relation \(\tan(\theta) = \frac{opposite}{adjacent}\) to calculate the maximum distance of the Jeep from the observer. Rearrange for 'adjacent' to get \(d = \frac{1.2}{\tan(\theta)}\). Plug the values to get \(d = \frac{1.2}{\tan(3.58 \times 10^{-4})}\), which comes out to be approximately \(3340\ m\) or \(3.34 km\). Thus, the observer will be able to distinguish the two headlights as long as the Jeep is within approximately \(3.34 km\). So the answer is: (C) \(3.34 \mathrm{~km}\)

Key concepts

Resolvability of Headlights

The ability to distinguish between two close points of light, such as a vehicle's headlights, depends on several optical factors. In our case, the headlight resolvability is determined by the separation between the lights, the diameter of the observing pupil, and the wavelength of the light. When two headlights are too close together, they can appear as one single light source to an observer. This limit of resolution is crucial in optics and defined by the Rayleigh Criterion. The separation must be sufficiently large relative to the observer's eye and light wavelength for the headlights to appear distinct.

• Headlights need to be sufficiently distant for effective segregation.
• Smaller diameters can resolve finer details, similar to using a larger camera lens.

Minimum Resolvable Angle

To find the minimum angle at which two points of light are resolvable, we use the Rayleigh Criterion. This formula gives us the smallest angular separation between two light sources that can be perceived distinctly:
\[\theta = 1.22 \frac{\lambda}{D}\]Here, \(\theta\) is the minimum resolvable angle, \(\lambda\) is the light's wavelength, and \(D\) is the diameter of the observer's aperture—in this case, the pupil size. For headlights, this angle becomes the deciding factor in how far apart they can appear distinct at a certain distance.

• A smaller \(D\) or larger \(\lambda\) means a larger \(\theta\), so less detail is resolvable.
• Narrower angles help determine longer distances over which the headlights are noticeable.

Trigonometric Relations in Optics

Trigonometry plays an essential role in optimizing optical systems, especially when calculating distances or angles. In this exercise, we employed the approach of angles to find the maximum viewing distance of the Jeep. By considering the formation of a triangle:

• The headlights' separation is the triangle's opposite side.
• The distance to the observer forms the adjacent side.

The tangent function becomes a tool that facilitates this calculation:
\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]By using this relation and knowing \(\theta\), we can rearrange the equation to solve for the Jeep's maximum distance:
\[d = \frac{1.2}{\tan(3.58 \times 10^{-4})}\]This reveals that, for the given scenario, the Jeep can be discerned from a maximum distance of approximately \(3.34 \) km.