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Chapter 36: Problem 47

Observing Jupiter. You are asked to design a space telescope for earth orbit. When Jupiter is \(5.93 \times 10^{8} \mathrm{km}\) away (its closest approach to the earth), the telescope is to resolve, by Rayleigh's criterion, features on Jupiter that are 250 \(\mathrm{km}\) apart. What minimum- diameter mirror is required? Assume a wavelength of 500 \(\mathrm{nm} .\)

Short Answer

Expert verified
The minimum diameter required is approximately 1.45 meters.

Step by step solution

01

Understand Rayleigh's Criterion

Rayleigh's criterion states that the minimum angular resolution \( \theta \) that can be resolved by a telescope is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the telescope's aperture.
02

Convert Units Appropriately

Convert the given wavelength of light from nanometers to meters. Since \( 1 \text{ nm} = 10^{-9} \text{ m} \), the wavelength \( \lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m} \).
03

Calculate the Angular Resolution

Use the small angle approximation for angular resolution \( \theta \) using the resolved separation and the distance to Jupiter: \( \theta \approx \frac{\text{separation}}{\text{distance}} = \frac{250 \text{ km}}{5.93 \times 10^8 \text{ km}} \). Convert 250 km to meters: \( 250 \text{ km} = 250,000 \text{ m} \). Then \( \theta \approx \frac{250,000 \text{ m}}{5.93 \times 10^{11} \text{ m}} \).
04

Rearrange Rayleigh's Formula

With \( \theta = \frac{250,000}{5.93 \times 10^{11}} \), plug this value back into Rayleigh's formula: \( \theta = 1.22 \frac{\lambda}{D} \). Solve for the diameter \( D \): \( D = 1.22 \frac{\lambda}{\theta} \).
05

Calculate Minimum Diameter

Substitute the known values into the rearranged formula: \( D = 1.22 \frac{5 \times 10^{-7}}{\theta} \). Using the value of \( \theta \approx 4.21 \times 10^{-7} \), solve for \( D \): \( D = 1.22 \times 5 \times 10^{-7} / 4.21 \times 10^{-7} \approx 1.45 \text{ meters} \).

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Key Concepts

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

Angular Resolution
Angular resolution is the ability of a telescope to distinguish between two closely spaced objects. This is crucial for observing celestial objects that are far away, such as planets and distant stars. The better the angular resolution, the clearer and more detailed the image. The angular resolution of a telescope is often described using Rayleigh's criterion, which provides a formula to calculate it. This formula is:
  • \( \theta = 1.22 \frac{\lambda}{D} \)
Here \( \theta \) is the angular resolution in radians, \( \lambda \) is the wavelength of the light used (in meters), and \( D \) is the diameter of the telescope's aperture (also in meters). By measuring how finely a telescope can resolve details, astronomers can determine the telescope's capability. This is especially beneficial when observing planets where distinguishing surface features is essential.
Minimum Diameter Calculation
To determine the minimum diameter of a telescope necessary to observe specific features on a distant object, we use the Rayleigh's criterion formula. By rearranging the formula, we can solve for the diameter \( D \):
  • \( D = 1.22 \frac{\lambda}{\theta} \)
In the context of observing Jupiter, if the distance at closest approach is \(5.93 \times 10^{8}\) km and the details to be resolved are 250 km apart, the separation in meters is \(250,000\) m. Calculating the necessary angular resolution \( \theta \) using:
  • \( \theta \approx \frac{250,000}{5.93 \times 10^{11}} \approx 4.21 \times 10^{-7} \) radians
With a light wavelength \( \lambda \) of 500 nm (or \(5 \times 10^{-7}\) m), the minimum diameter is:
  • \( D \approx 1.22 \times \frac{5 \times 10^{-7}}{4.21 \times 10^{-7}} \approx 1.45 \) meters
This value informs us about the required size of the telescope's mirror to achieve the desired resolution.
Telescope Design
Designing a telescope involves several key considerations, including the aperture, the focal length, and the mechanical structure. The aperture, or diameter of the telescope's main lens or mirror, is crucial for collecting light and determining resolution. As seen in the minimum diameter calculation, a larger aperture allows for observing finer details at greater distances. Practical telescope design integrates:
  • Light Collection: A wide aperture collects more light, improving the observation of faint objects.
  • Resolution Enhancement: A larger diameter directly impacts resolving power, vital for detailed planetary observations.
  • Structural Stability: The telescope must maintain alignment and rigidity, especially in space where maintenance is not feasible.
One must also consider the materials for constructing the mirror or lens, as these affect the telescope's sensitivity to temperature changes and potential distortions. Modern telescopes often employ adaptive optics and advanced mirror coatings to further enhance image quality and resolution.

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Most popular questions from this chapter

Monochromatic \(\mathrm{x}\) rays are incident on a crystal for which the spacing of the atomic planes is 0.440 \(\mathrm{nm} .\) The first-order maximum in the Bragg reflection occurs when the incident and reflected \(x\) rays make an angle of \(39.4^{\circ}\) with the crystal planes. What is the wavelength of the \(\mathrm{x}\) rays?

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit 0.450 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slit. In terms of the intensity \(I_{0}\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) \(1.00 \mathrm{mm} ;\) (b) 3.00 \(\mathrm{mm}\) ; (c) 5.00 \(\mathrm{mm}\) ?

The intensity of light in the Fraunhofer diffraction pattern of a single slit is $$I=I_{0}\left(\frac{\sin \gamma}{\gamma}\right)^{2}$$ where $$\gamma=\frac{\pi a \sin \theta}{\lambda}$$ (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma .(\) b) Determine the three smallest positive values of \(\gamma\) that are solutions of this equation. (Hint: You can use a trial-and-error procedure. Guess a value of \(\gamma\) and adjust your guess to bring \(\tan \gamma\) closer to \(\gamma .\) A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.)

Phased-Array Radar. In one common type of radar installation, a rotating antenna sweeps a radio beam around the sky. But in a phased-array radar system, the antennas remain stationary and the beam is swept electronically. To see how this is done, consider an array of \(N\) antennas that are arranged along the horizontal \(x\) -axis at \(x=0, \pm d, \pm 2 d, \ldots, \pm(N-1) d / 2 .\) (The number \(N\) is odd.) Each antenna emits radiation uniformly in all directions in the horizontal \(x y\) -plane. The antennas all emit radiation coherently, with the same amplitude \(E_{0}\) and the save- length \(\lambda\) . The relative phase \(\delta\) of the emission from adjacent antennas can be varied, however. If the antenna at \(x=0\) emits a signal that is given by \(E_{0} \cos \omega t,\) as measured at a point next to the antenna, the antenna at \(x=d\) emits a signal given by \(E_{0} \cos (\omega t+\delta),\) as measured at a point next to that antenna. The corresponding quantity for the antenna at \(x=-d\) is \(E_{0} \cos (\omega t-\delta) ;\) for the antennas at \(x=\pm 2 d,\) it is \(E_{0} \cos (\omega t \pm 2 \delta) ;\) and so on. (a) If \(\delta=0,\) the interference pattern at a distance from the antennas is large compared to \(d\) and has a principal maximum at \(\theta=0\) (that is, in the \(+y\) -direction, perpendicular to the line of the antennas. Show that if \(d<\lambda,\) this is the only principal interference maximum in the angular range \(-90^{\circ}<\theta<90^{\circ}\) . Hence this principal maximum describes a beam emitted in the direction \(\theta=0 .\) As described in Section 36.4 , if \(N\) is large, the beam will have a large intensity and be quite narrow. (b) If \(\delta \neq 0,\) show that the principal intensity maximum described in part (a) is located at $$\theta=\arcsin \left(\frac{\delta \lambda}{2 \pi d}\right)$$ where \(\delta\) is measured in radians. Thus, by varying \(\delta\) from positive to negative values and back again, which can easily be done electronically, the beam can be made to sweep back and forth around \(\theta=0 .(\mathrm{c})\) A weather radar unit to be installed on an airplane emits radio waves at 8800 \(\mathrm{MHz}\) . The unit uses 15 antennas in an array 28.0 \(\mathrm{cm}\) long (from the antenna at one end of the array to the antenna at the other end. What must the maximum and minimum values of \(\delta\) be (that is, the most positive and most negative values) if the radar beam is to sweep \(45^{\circ}\) to the left or right of the airplane's direction of flight? Give your answer in radians.

What is the longest wavelength that can be observed in the third order for a transmission grating having 9200 slits/cm? Assume normal incidence.
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