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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 of the mirror is calculated to be approximately 1.25 meters, assuming the calculation has been made correctly and all given conditions are met.

Step by step solution

01

Convert all distances to the same units

Our given distances are in kilometers (km) and nanometers (nm). To make the computations simpler, all distances have to be in the same units. 1 km equals \( 1 \times 10^{9} \) nm, so convert \(5.93 \times 10^{8}\) km to \(5.93 \times 10^{17}\) nm and \(250\) km to \(250 \times 10^{9}\) nm.
02

Calculate the angular size

Calculate the angular size of the smallest resolvable feature. The angular size \( \theta \) is given by \(\theta = d / r\) where \(d = 250 \times 10^{9} nm\) is the actual size of the feature and \( r = 5.93 \times 10^{17} nm \) is the distance from the feature. Plugging in these values yields \( \theta = (250 \times 10^{9})/(5.93 \times 10^{17}) radians \).
03

Calculate the minimum diameter of the mirror

Using the angular resolution formula, rewrite it to find \(D\): \(D = 1.22 \lambda / \theta\). Inserting the provided wavelength \(\lambda = 500 nm\) and the calculated \(\theta\), solve for \(D\).
04

Convert mirror diameter to more convenient units

The resulting mirror size will be very small and in nanometers. Convert it to a more convenient unit like meters by using the conversion \(1 nm = 1 \times 10^{-9} m\).

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Key Concepts

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

Understanding Rayleigh's Criterion
One of the fundamental aspects of telescope design is the instrument's ability to distinguish two closely spaced objects, a feature known as its resolution. A key principle governing this is Rayleigh's criterion. Simply put, this criterion states that two light sources are considered resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other.In more accessible terms, imagine the light from each space object spreading out and creating patterns of light and dark bands (like ripples from raindrops in a pond). For us to say that our telescope can 'see' both objects separately, the center of one set of ripples should fall on the first gap in the other set. This determines the finest detail a telescope can resolve and is particularly valuable in astronomy where we aim to see detailed structures of distant celestial bodies like planets and stars.
Rayleigh's criterion mathematically defines the minimum angular separation at which a telescope can distinguish between two points of light. This is directly related to the light's wavelength and the telescope's aperture, leading us into how technologists calculate the angular resolution and, thus, the required mirror diameter of a telescope.
Deciphering Angular Resolution
The angular resolution of a telescope describes its ability to distinguish small details of an object, which allows us to 'zoom in' on objects like the craters on the moon or storms on Jupiter. More precisely, it tells us the smallest angle over which we can tell that two dots (or stars) are distinct rather than a single blurred point.The finer (smaller) the angular resolution, the better the detail that can be resolved. The limit to this resolution isn't just about the quality of the telescope; it's governed by the laws of physics - specifically by diffraction. This effect is a result of light's wave-like characteristics. When light waves pass through an aperture, such as a telescope mirror, they spread out or 'diffract', limiting the sharpness of the image produced.
In the context of the exercise, to determine the angular size \( \theta \) of the smallest detail observable on Jupiter, we use the formula \( \theta = \frac{d}{r} \), where \( d \) is the size of the feature and \( r \) is the distance to the feature. Crucially, this angular resolution must meet or exceed what is defined by Rayleigh's criterion, for the telescope to resolve features on distant planets.
The Math Behind Mirror Diameter Calculation
To apply Rayleigh's criterion and calculate the angular resolution necessary for a telescope's design, we must connect these concepts to the size of the telescope's mirror. A larger mirror generally allows better resolution due to its ability to collect more light and spread it over a larger image; hence, diffraction has a smaller relative impact.When we look at the math, to find the minimum diameter \( D \) of a telescope's mirror necessary to resolve a certain detail, we use the formula \( D = 1.22 \frac{\lambda}{\theta} \) where \( \lambda \) is the wavelength of the light and \( \theta \) is the angular size of the smallest detail to be resolved, as determined by the earlier mentioned process. The numerical factor 1.22 is derived from the physics of diffraction through a circular aperture and is a constant in calculations like these.
In practice, after computing the angular size \( \theta \), we can solve for \( D \) to determine the minimum required diameter for our telescope's mirror. If calculated in nanometers, it's helpful to convert the result into meters or centimeters for a practical physical understanding, as the exercise illustrates in its final step. Knowing this not only guides the design of space telescopes but also deepens our appreciation of the intricate links between physics and astronomy.

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Most popular questions from this chapter

X rays of wavelength \(0.0850 \mathrm{nm}\) are scattered from the atoms of a crystal. The second-order maximum in the Bragg reflection occurs when the angle \(\theta\) in Fig. 36.22 is \(21.5^{\circ} .\) What is the spacing between adjacent atomic planes in the crystal?

If a diffraction grating produces a third-order bright spot for red light (of wavelength \(700 \mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength \(400 \mathrm{nm}\) )?

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longer-wavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are \(1.60 \mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only \(0.740 \mu \mathrm{m}\) apart. Repeat the calculation of part (a) for the DVD.

A wildlife photographer uses a moderate telephoto lens of focal length \(135 \mathrm{~mm}\) and maximum aperture \(f / 4.00\) to photograph a bear that is \(11.5 \mathrm{~m}\) away. Assume the wavelength is \(550 \mathrm{nm}\). (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0\), what would be the width of the smallest resolvable feature on the bear?

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