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Chapter 27: Problem 51

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?

Short Answer

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

Step by step solution

01

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.
02

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: \( \lambda = 660 \times 10^{-9} \mathrm{~m} \), \( D = 7 \times 10^{-3} \mathrm{~m} \).
03

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: \( \theta = 1.22 \times 94.286 \times 10^{-6} = 1.15 \times 10^{-4} \mathrm{~radians} \).
04

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 \( L \): \( L = \frac{d}{\theta} = \frac{1.2}{1.15 \times 10^{-4}} \).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength \(625 \mathrm{~nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{~m} ?\)

Light shines on a diffraction grating, and a diffraction pattern is produced on a viewing screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) From trigonometry, how is the distance \(y\) from the central bright fringe to the second-order bright fringe related to the diffraction angle \(\theta\) and the distance \(L\) between the grating and the screen? (b) From physics, how is \(\theta\) related to the order \(m\) of the bright fringe, the wavelength \(\lambda\) of the light, and the separation \(d\) between the slits? (c) In this problem, the angle \(\theta\) is small (less than a few degrees). When the angle is small, \(\tan \theta\) is approximately equal to \(\sin \theta\), or \(\tan \theta \approx \sin \theta\). Using this approximation, obtain an expression for \(y\) in terms of \(L, m, \lambda\), and \(d\). (d) If the entire apparatus in the drawing is submerged in water, would you expect the distance \(y\) to increase, decrease, or remain unchanged? Why? Light of wavelength \(480 \mathrm{~nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(5.0 \times 10^{-7} \mathrm{~m}\). The distance between the grating and the viewing screen is \(0.15 \mathrm{~m}\). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right)\), what is the distance \(y ?\) Be sure your answer is consistent with part (d) of the Concept Questions.

A nonreflective coating of magnesium fluoride \((n=1.38)\) covers the glass \((n=1.52)\) of a camera lens. Assuming that the coating prevents reflection of yellow-green light (wavelength in vacuum \(=565 \mathrm{~nm}\) ), determine the minimum nonzero thickness that the coating can have.

The same diffraction grating is used with two different wavelengths of light, \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\). The fourth-order principal maximum of light A exactly overlaps the third-order principal maximum of light \(\mathrm{B}\). Find the ratio \(\lambda_{\mathrm{A}} / \lambda_{\mathrm{B}}\).

Consult Interactive Solution \(\underline{27} .17\) at to review a model for solving this problem. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength \(=640.0 \mathrm{nm}\) in vacuum). Assuming that the visible spectrum extends from 380 to \(750 \mathrm{nm}\), for which visible wavelength(s) (in vacuum) will the film appear bright due to constructive interference?
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