Skip to main content
Ch 35: Optical Instruments
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 35, Problem 46b

Alpha Centauri, the nearest star to our solar system, is 4.3 light years away. Assume that Alpha Centauri has a planet with an advanced civilization. Professor Dhg, at the planet’s Astronomical Institute, wants to build a telescope with which he can find out whether any planets are orbiting our sun. Building a telescope of the necessary size does not appear to be a major problem. What practical difficulties might prevent Professor Dhg’s experiment from succeeding?

Verified step by step guidance
1
Consider the concept of angular resolution, which is the ability of a telescope to distinguish between two closely spaced objects. The angular resolution depends on the wavelength of light used and the diameter of the telescope's aperture. The formula for angular resolution is given by: θ=λD, where θ is the angular resolution, λ is the wavelength of light, and D is the diameter of the telescope.
Realize that the distance between Alpha Centauri and our solar system (4.3 light years) is vast. Even with a large telescope, the angular separation between the Sun and its planets as seen from Alpha Centauri would be extremely small, making it challenging to resolve the planets from the Sun's glare.
Consider the issue of light interference. The Sun is a very bright source of light, and its brightness could overwhelm the faint light reflected by the planets. This would make it difficult to detect the planets even with a highly sensitive telescope.
Account for the interstellar medium, which consists of gas and dust particles. These particles can scatter and absorb light, potentially distorting or dimming the light from the Sun and its planets as it travels to Alpha Centauri.
Finally, think about the technological and observational challenges. Building a telescope with a sufficiently large aperture to achieve the required angular resolution might be feasible, but maintaining its stability, precision, and alignment over such a long distance would be a significant engineering challenge. Additionally, atmospheric effects on Alpha Centauri's planet could further complicate observations.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Light Year

A light year is a unit of distance that represents how far light travels in one year, approximately 5.88 trillion miles (9.46 trillion kilometers). This concept is crucial for understanding astronomical distances, such as the 4.3 light years to Alpha Centauri, which highlights the vastness of space and the challenges of interstellar observation.
Recommended video:
10:30
Ray Nature of Light

Telescope Design and Limitations

The design of a telescope involves considerations such as aperture size, lens or mirror quality, and sensitivity to light. Larger telescopes can gather more light and resolve finer details, but practical difficulties include material limitations, atmospheric interference, and the need for precise alignment, all of which can hinder the ability to detect distant planets.
Recommended video:
Guided course
03:50
Designing a Solenoid (Total Length of Wire)

Interstellar Communication and Observation

Interstellar communication and observation face significant challenges due to the immense distances involved. Signals weaken over vast distances, and the time delay in receiving information can span years. Additionally, detecting exoplanets requires advanced techniques like transit photometry or direct imaging, which may be limited by current technology and the faintness of distant stars.
Recommended video:
Guided course
06:03
The Doppler Effect (Light)
Related Practice
Textbook Question

FIGURE CP35.50 shows a lens combination in which the lens separation is less than the focal length of the converging lens. The procedure for combination lenses is to let the image of the first lens be the object for the second lens, but in this case the image of the first lens—shown as a dot—is on the far side of the second lens. This is called a virtual object, a point that light rays are converging toward but never reach. The top half of Figure CP35.50 shows that the converging rays are refracted again by the diverging lens and come to a focus farther to the right. The procedure for combination lenses will continue to work if we use a negative object distance for a virtual object. Equation 35.1 defined the effective focal length feff of a lens combination, but we didn't discuss how it is used. Although an actual ray refracts twice, once at each lens, we can extend the output rays leftward to where they need to bend only once in a plane called the principal plane. The principal plane is similar to the lens plane of a single lens, where a single bend occurs, but the principal plane generally does not coincide with the physical lens; it's just a mathematical plane in space. The effective focal length is measured from the principal plane, so parallel input rays are focused at distance feff beyond the principal plane. Find the positions of the principal planes for lens separations of 5 cm and 10 cm. Give your answers as distances to the left of the diverging lens.

1727
views
Textbook Question

A beam of white light enters a transparent material. Wavelengths for which the index of refraction is n are refracted at angle θ₂. Wavelengths for which the index of refraction is n + δn, where δn << n, are refracted at angle θ₂ + δθ. A beam of white light is incident on a piece of glass at 30°. Deep violet light is refracted 0.28° more than deep red light. The index of refraction for deep red light is known to be 1.552. What is the index of refraction for deep violet light?

1642
views
Textbook Question

The resolution of a digital camera is limited by two factors: diffraction by the lens, a limit of any optical system, and the fact that the sensor is divided into discrete pixels. Consider a typical point-and-shoot camera that has a 20-mm-focal-length lens and a sensor with 2.5μm x 2.5 μm pixels. What is the f-number of the lens for the diameter you found in part b? Your answer is a quite realistic value of the f-number at which a camera transitions from being pixel limited to being diffraction limited. For f-numbers smaller than this (larger-diameter apertures), the resolution is limited by the pixel size and does not change as you change the aperture. For f-numbers larger than this (smaller-diameter apertures), the resolution is limited by diffraction, and it gets worse as you “stop down” to smaller apertures.

1539
views
Textbook Question

High-power lasers are used to cut and weld materials by focusing the laser beam to a very small spot. This is like using a magnifying lens to focus the sun's light to a small spot that can burn things. As an engineer, you have designed a laser cutting device in which the material to be cut is placed 5.0 cm behind the lens. You have selected a high-power laser with a wavelength of 1.06 μm. Your calculations indicate that the laser must be focused to a 5.0-μm-diameter spot in order to have sufficient power to make the cut. What is the minimum diameter of the lens you must install?

1444
views
Textbook Question

The lens shown in FIGURE CP35.49 is called an achromatic doublet, meaning that it has no chromatic aberration. The left side is flat, and all other surfaces have radii of curvature R. Because of dispersion, either lens alone would focus red rays and blue rays at different points. Define ∆n1 and ∆n2 as nblue - nred for the two lenses. What value of the ratio ∆n1 / ∆n2 makes fblue = fred for the two-lens system? That is, the two-lens system does not exhibit chromatic aberration.

120
views
Textbook Question

The Hubble Space Telescope has a mirror diameter of 2.4 m. Suppose the telescope is used to photograph stars near the center of our galaxy, 30,000 light years away, using red light with a wavelength of 650 nm. For comparison, what is this distance as a multiple of the distance of Jupiter from the sun?

197
views