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A converging lens 7.20 \(\mathrm{cm}\) in diameter has a focal length of 300 \(\mathrm{mm}\) . If the resolution is diffraction limited, how far away can an object be if points on it 4.00 \(\mathrm{mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{nm} .\)

Short Answer

Expert verified
The object can be 431 meters away.

Step by step solution

01

Determine the formula to use

We use Rayleigh's criterion which states that two points can be resolved if the central maximum of the diffraction pattern of one point coincides with the first minimum of the other point. The minimum resolvable angle \( \theta \) is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( D \) is the diameter of the lens and \( \lambda \) is the wavelength of the light.
02

Convert units to consistent system

It is essential to work in the same units. Convert all given dimensions to meters: \( D = 7.20 \ \text{cm} = 0.072 \ \text{m} \), \( \lambda = 550 \ \text{nm} = 550 \times 10^{-9} \ \text{m} \). Keep the focal length in meters as \( 300 \ \text{mm} = 0.3 \ \text{m} \).
03

Calculate the minimum resolvable angle

Plug the values into Rayleigh's formula: \[ \theta = 1.22 \frac{550 \times 10^{-9} \ \text{m}}{0.072 \ \text{m}} \approx 9.28 \times 10^{-6} \text{ rad} \].
04

Use the small angle approximation

Using small angle approximation, \( \tan \theta \approx \theta \). So, \( \theta = \frac{s}{d} \), where \( s = 4.00 \ \text{mm} = 0.004 \ \text{m} \) is the separation we want to resolve, and \( d \) is the distance to the object. Rearrange \( d = \frac{s}{\theta} \).
05

Calculate the distance to the object

Using the angle computed, \[ d = \frac{0.004 \ \text{m}}{9.28 \times 10^{-6} \text{ rad}} \approx 431 \ \text{m} \].

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Key Concepts

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

Converging Lens
A converging lens, also known as a convex lens, is essential in focusing light. It brings parallel rays of light toward a single focal point. This makes it an important tool in optics for capturing clear images. The focal length, a key lens parameter, measures the distance from the lens to the focal point where light converges. In optical systems, converging lenses are frequently used in cameras, microscopes, and glasses, helping to create a sharp image by adjusting the focus of light.
Key characteristics of converging lenses include:
  • They are thicker in the center than at the edges.
  • Capable of forming real images that can be projected onto a screen.
  • Their ability to magnify objects when the object is placed within one focal length.
Understanding how converging lenses work is crucial for solving problems in optics, such as determining how far an object can be resolved given a specific lens diameter and light wavelength.
Diffraction Limit
The diffraction limit is a fundamental principle in optics that describes the smallest angle or detail that can be distinguished by an imaging system. It is dictated by the wave nature of light and the size of the aperture through which light passes. In simpler terms, the diffraction limit is the barrier at which an optical system stops distinguishing between two closely spaced objects.
  • The diffraction limit arises because light waves spread out, or diffract, when they encounter obstacles, like a lens edge.
  • When observing or imaging objects at this limit, details blur together, limiting resolution.
To calculate the diffraction limit, we often use Rayleigh's criterion, which states that two points are resolvable if the main peak of one point's diffraction pattern lies at the minimum of the other's. This criterion helps determine the system's resolving power, depending on lens specifications and light wavelength.
Minimum Resolvable Angle
The minimum resolvable angle is a measure of how fine a detail an optical system can resolve. It is the smallest angular separation between two light sources or points that the system can distinguish as separate.
  • Rayleigh's criterion gives the formula for calculating this angle: \( \theta = 1.22 \frac{\lambda}{D} \). This considers both the wavelength \( \lambda \) of light and the diameter \( D \) of the lens.
  • A smaller angle indicates a higher resolution of the optical system, meaning it can distinguish finer details.
By knowing the minimum resolvable angle, one can determine how detailed an image a particular lens system can produce. It's crucial in designing optics, ensuring that devices meet the need for precision in various applications, like microscopy and astronomy.
Wavelength of Light
The wavelength of light is a fundamental parameter in optics, representing the distance between consecutive peaks of a light wave. It is significant in processes like diffraction and interference, influencing how light interacts with materials.
  • Different wavelengths of light correspond to different colors. For instance, blue light has a shorter wavelength than red light.
  • The wavelength affects the resolving power of optical systems. Shorter wavelengths generally provide better resolution.
In the context of Rayleigh's criterion, knowing the wavelength is essential for calculating the minimum resolvable angle and evaluating an optical system's performance. For example, using a wavelength of 550 nm (a typical wavelength for green light) is common in resolution calculations, as it falls within the visible range and forms a basis for many optical designs.

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Most popular questions from this chapter

Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 \(\mathrm{km}\) . If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

A diffraction grating has 650 slits \(/ \mathrm{mm}\) . What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(400-700 \mathrm{nm} . )\)

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 \(\mathrm{cm} / \mathrm{s}\) on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 \(\mathrm{m}\) away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm 61.3 \mathrm{cm}\) from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

A slit 0.360 \(\mathrm{mm}\) wide is illuminated by parallel rays of light that have a wavelength of 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 1.20 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\) . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?(\text { See Problem } 36.53 \text { , part (a), for a hint about how to }\) solve for the phase angle \(\beta . )\)

Intensity Pattern of \(N\) Slits. (a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\) . Show that at a time \(t,\) the electric field at a distant point \(P\) is $$ \begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned} $$ where \(E_{0}\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching \(P(\text { as measured from the perpendicular bisector of the slit arrangement }\)), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d .\) (b) To carry out the sum in part \((a),\) it is convenient to use the complex-number relationship $$ e^{i k}=\cos z+i \sin z $$ where \(i=\sqrt{-1}\) . In this expression, \(\cos z\) is the real part of the complex number \(e^{i k},\) and \(\sin z\) is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity $$ \sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega+n \phi)} $$ (c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A},\) show that the sum in part \((b)\) can be written as $$ E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)}=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i(k R-\omega t+(N-1) \phi / 2]} $$ Then, using the relationship \(e^{i k}=\cos z+i \sin z,\) show that the (real) electric field at point \(P\) is $$ E_{p}(t)=\left[E_{0} \frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right] \cos [k R-\omega t+(N-1) \phi / 2] $$ The quantity in the first square brackets in this expression is the amphitude of the electric field at \(P .\) (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$ I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2} $$ where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N=2\) . It will help to recall that \(\sin 2 A=2 \sin A \cos A\) . Explain why your result differs from Eq. \((35.10),\) the expression for the intensity in two-source interference, by a factor of \(4 .\) (Hint: Is \(I_{0}\) defined in the same way in both expressions?)

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