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The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (380-750 nm). The Arecibo radio telescope in Puerto Rico is 305 m (1000 ft) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength 75 cm. (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

Short Answer

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(a) Hubble: 110 m, Arecibo: 1150 m crater on moon. (b) Max orbit for Hubble surveillance is 1045 km.

Step by step solution

01

Understanding the Diffraction Limit

The diffraction limit defines the smallest angular separation (resolution) that a telescope can distinguish, given by the formula \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution in radians, \( \lambda \) is the wavelength of light or radio wave, and \( D \) is the diameter of the telescope's aperture.
02

Calculating the Smallest Crater with Hubble

First, convert the wavelength of visible light to meters, choosing an average value \( \lambda = 565 \text{ nm} = 565 \times 10^{-9} \text{ m} \). For Hubble, with an aperture of 2.4 meters, calculate the angular resolution: \( \theta = 1.22 \frac{565 \times 10^{-9}}{2.4} \approx 2.87 \times 10^{-7} \text{ radians} \). The moon is about 384,400 km away, so the smallest crater size is: \( \Delta s = \theta \times 384,400 \times 10^{3} \approx 110 \text{ meters}\).
03

Calculating the Smallest Crater with Arecibo

For Arecibo with a diameter of 305 meters and a wavelength \( \lambda = 0.75 \text{ m} \), the angular resolution is: \( \theta = 1.22 \frac{0.75}{305} \approx 3.00 \times 10^{-3} \text{ radians} \). The smallest crater size this resolution can detect on the moon is: \( \Delta s = \theta \times 384,400 \times 10^{3} \approx 1150 \text{ meters} \).
04

Calculating Maximum Orbit for Hubble's Surveillance Use

For resolving a license plate (about 0.3 meters long), calculate the maximum orbit distance using \( 0.3 = 2.87 \times 10^{-7} \times H \), where \( H \) is the height above Earth's surface. Solving for \( H \): \( H = \frac{0.3}{2.87 \times 10^{-7}} \approx 1,045,300 \text{ meters} = 1045 \text{ km}\).

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Key Concepts

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

Hubble Space Telescope
The Hubble Space Telescope (HST) is a marvel of astronomical engineering, launched into low Earth orbit in 1990. It allows us to peer deep into the cosmos, capturing stunning images and shedding light on countless astronomical phenomena. One of the Hubble’s most crucial features is its large aperture, measuring 2.4 meters across. This size is key, as it allows Hubble to focus visible light, ranging from 380 nm to 750 nm. With such an aperture, the telescope can achieve a diffraction-limited resolution, which depends on both the wavelength of light it observes and the size of its aperture. The larger the aperture, the finer the details it can resolve. Specifically for the Hubble, this impressive capability allows it to see small features as tiny as 110 meters across on the moon's surface, even when observing from hundreds of thousands of kilometers away.
Angular Resolution
Angular resolution is a measure of a telescope's ability to distinguish between two closely spaced objects. It is defined by the smallest angle between two points that a telescope can effectively distinguish as separate. The angular resolution is determined using the formula \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution in radians, \( \lambda \) is the wavelength, and \( D \) is the telescope's aperture diameter. A smaller angular resolution value implies a sharper, more detailed image, as the telescope can resolve finer details. For Hubble, this equates to being able to discern features as small as fractions of a meter when at close distances, such as observing objects in Earth's orbit. For distant celestial bodies like the moon, this still enables high precision, resolving features down to 110 meters.
Telescopic Resolution
Telescopic resolution refers to the ability of a telescope to distinguish fine details and separate objects that are close together. While it might seem similar to angular resolution, telescopic resolution specifically considers the practical application of these calculations in observational astronomy. Factors such as weather conditions, atmosphere, and the quality of the telescope's optics can all affect telescopic resolution. The Hubble Space Telescope, with its significant aperture and operation above Earth's atmosphere, circumvents many of these challenges. This allows it to achieve a resolution limited mainly by diffraction, meaning it can observe significantly smaller features compared to many ground-based telescopes, whose resolutions are often degraded by atmospheric interference.
Radio Telescope Wavelength
Radio telescopes operate at much longer wavelengths than their optical counterparts, which presents unique challenges and opportunities. The Arecibo radio telescope, before its closure, was one of the most noteworthy, boasting a massive 305-meter diameter. It focused on radio wavelengths around 75 cm. Due to its large diameter, Arecibo had a decent angular resolution for its wavelength, though not as fine as those achieved by optical telescopes for visible light. This is a fundamental characteristic of radio astronomy; the longer the wavelength, the larger the telescope needed to achieve high resolution. Consequently, Arecibo could only resolve features around 1150 meters on the lunar surface, much larger than the capabilities of an optical telescope like Hubble. The construction of radio telescopes often involves creating massive dishes or arrays, underscoring their important role in capturing cosmic radio waves from deep within the universe.

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Most popular questions from this chapter

If the planes of a crystal are 3.50 \(\AA\) (1 \(\AA\) = 10\(^{-10}\) m = 1 \(\AA\)ngstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0\(^\circ\), and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

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(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 $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ 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 \(P\) (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 \(e^{iz} = cos z + i \space sin \space z\), where \(i = \sqrt{-1}\). In this expression, cos \(z\) is the \(real\) \(part\) of the complex number \(e^{iz}\), 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 $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that \(e^Ae^B = e^{(A+B)}\) and \((e^A)^n = e^{nA}\), show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship \(e^{iz}\) = cos \(z\) + \(i\) sin \(z\), show that the (real) electric field at point \(P\) is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude 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 $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ 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 I0 defined in the same way in both expressions?)

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