Problem 51 - Physics

Question

Under these conditions the pupils of your eyes have diameters of about \(7.0 \mathrm{~mm}\). The taillights of this car are separated by a distance of \(1.2 \mathrm{~m}\) and emit red light (wavelength \(=660 \mathrm{~nm}\) in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

Step-by-Step Solution

Answer

The car is approximately 10.4 km away when the taillights merge into a single spot due to diffraction.

Step 1: Understand the Problem

The problem involves determining the distance at which two separate light sources (taillights) appear as one due to diffraction. This involves using the concept of the resolution limit calculated based on the diffraction limit for a circular aperture (like an eye) and wavelength of light.

Step 2: Determine the Resolution Limit

Using Rayleigh's criterion, the angular resolution \( \theta \) for a circular aperture is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light (660 nm) and \( D \) is the diameter of the pupil (7 mm). Convert these to meters: \( \lambda = 660 \times 10^{-9} \mathrm{~m} \), \( D = 7 \times 10^{-3} \mathrm{~m} \).

Step 3: Calculate the Angular Resolution \( \theta \)

Substitute the values into Rayleigh's formula: \( \theta = 1.22 \cdot \frac{660 \times 10^{-9}}{7 \times 10^{-3}} \). This calculates \( \theta \) in radians. Perform the calculation: \( \theta = 1.22 \times 94.286 \times 10^{-6} = 1.15 \times 10^{-4} \mathrm{~radians} \).

Step 4: Relate Angular Resolution to Distance

The small angle approximation states that \( \theta \approx \frac{d}{L} \), where \( d = 1.2 \mathrm{~m} \) is the separation between the taillights and \( L \) is the distance we wish to find. Rearrange to find \( L \): \( L = \frac{d}{\theta} = \frac{1.2}{1.15 \times 10^{-4}} \).

Step 5: Calculate the Distance \( L \)

Substitute and calculate the distance: \( L = \frac{1.2}{1.15 \times 10^{-4}} = 10434.78 \mathrm{~m} \). This means the car is around 10,435 m (or approximately 10.4 km) away when its taillights appear merged into one spot.

Key Concepts

Rayleigh's criterion

Rayleigh's criterion helps us understand the limit at which two close but distinct light sources appear as a single light source due to the phenomenon of diffraction. This criterion is particularly useful for optical systems like telescopes or even human eyes in determining resolution limits.
Rayleigh's criterion states that two light sources are just resolvable when the principal diffraction maximum of one image coincides with the first minimum of the other. This is mathematically expressed for a circular aperture as:

• \( \theta = 1.22 \frac{\lambda}{D} \)

where \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture, like the pupil of an eye. The formula helps calculate the smallest angular separation \( \theta \) at which two points of light can be distinguished. In our context, it allows us to determine how far away a car's taillights can be before they blend together in our vision.

Angular resolution

Angular resolution is the ability of an optical system to distinguish between two points that are close together. It's a measure of the system's sharpness and clarity. When we talk about angular resolution in the context of Rayleigh's criterion, it directly relates to how fine the details in the visual field can be discerned.
In practical terms, angular resolution is calculated using Rayleigh's criterion equation, providing a value in radians. This value tells us the smallest angle at which the human eye can differentiate between two sources of light. A smaller angular resolution means better clarity and separation of points. In the given problem, the angular resolution \( \theta = 1.15 \times 10^{-4} \mathrm{~radians} \) gives us a benchmark for assessing how much detail the eye can resolve between the taillights as two distinct sources.

Small angle approximation

The small angle approximation simplifies calculations in optics by equating angles to their tangent when angles are expressed in radians and are significantly small. This is especially useful when dealing with diffraction and resolution problems.
In our scenario, the approximation \( \theta \approx \frac{d}{L} \) assists us in linking the angular resolution with physical separation \( d \) and the distance \( L \).
Here,

• \( \theta \) is the angular resolution,
• \( d \) is the actual separation between the light sources (taillights),
• \( L \) represents how far the observer (like you) is from the sources.

This approximation lets us determine how far away the car is based on the angular resolution calculated from Rayleigh's criterion. It acts as a bridge between measurable physical distances and the abstract concept of angular resolution. In the problem, using the small angle approximation, we calculate the distance \( L \approx 10435 \mathrm{~m} \), translating this abstract concept into something tangible.