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Chapter 4: Problem 3

Compare the collecting areas of 5 and \(8 \mathrm{m}\) diameter reflectors. Comment on the significance of this comparison.

Short Answer

Expert verified
The 8-meter diameter reflector collects 2.56 times more light than the 5-meter diameter reflector, leading to significantly improved observational capabilities.

Step by step solution

01

- Area of a Circle Formula

The formula to compute the area of a circle is given by: \[ A = \pi r^2 \] where \(A\) is the area, \(\pi\) is a constant (approximately 3.14159), and \(r\) is the radius of the circle.
02

- Calculate Radius for Each Reflector

For the 5-meter diameter reflector, the radius \(r\) is: \[ r = \frac{5}{2} = 2.5 \ \, \mathrm{meters} \] For the 8-meter diameter reflector, the radius \(r\) is: \[ r = \frac{8}{2} = 4 \ \, \mathrm{meters} \]
03

- Calculate Area for Each Reflector

Using the formula \(A = \pi r^2\): For the 5-meter diameter reflector: \[ A_{1} = \pi(2.5)^{2} = 6.25 \pi \, \mathrm{square\ meters} \] For the 8-meter diameter reflector: \[ A_{2} = \pi(4)^{2} = 16 \pi \, \mathrm{square\ meters} \]
04

- Compare the Areas

Now that we have the areas for both reflectors, compare them: \[ A_{2} = 16 \pi \, \mathrm{square\ meters} \] \[ A_{1} = 6.25 \pi \, \mathrm{square\ meters} \] The 8-meter reflector has a significantly larger area. To quantify: \[ \frac{A_{2}}{A_{1}} = \frac{16 \pi}{6.25 \pi} = \frac{16}{6.25} = 2.56 \]
05

- Significance of Comparison

A larger collecting area implies a greater capability to gather light or electromagnetic signals. Compared to the 5-meter reflector, the 8-meter diameter reflector can collect approximately 2.56 times more light, which can lead to more detailed observations and greater sensitivity in various scientific applications.

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Key Concepts

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

Reflector Diameter
When discussing the diameter of a reflector, we refer to the straight-line distance passing through the center of the reflector from one edge to the other. This is crucial in telescopes, as a larger diameter increases the ability to capture more light.

  • For instance, a reflector with a diameter of 5 meters has a radius of 2.5 meters (since the radius is half the diameter).
  • Similarly, a reflector with an 8-meter diameter has a radius of 4 meters.

The diameter plays a pivotal role in determining the overall light-gathering power of the telescope. To contextualize, the diameter directly influences the area of the reflector, which brings us to our next concept.
Area Calculation
Calculating the area of a circular reflector uses the formula for the area of a circle: equation: \[ A = \pi r^{2} \] here, \pi is a mathematical constant (approximately 3.14159) and \ r is the radius. For our reflectors:
  • The 5-meter reflector with a radius of 2.5 meters has an area: \[ A_{1} = \pi (2.5)^{2} = 6.25\pi \approx 19.63 \mathrm{\,square meters} \]
  • The 8-meter reflector with a radius of 4 meters has an area: \[ A_{2} = \pi (4)^{2} = 16\pi \approx 50.27 \mathrm{\,square meters} \]
The larger the area, the more light the reflector can capture. This means the 8-meter reflector, with its larger area, captures much more light than the 5-meter reflector..
To quantify, we can compare these areas:
  • .\[ \frac{A_{2}}{A_{1}} = \frac{16\pi}{6.25\pi} = \frac{16}{6.25} = 2.56 \]
    • In essence, the 8-meter reflector captures 2.56 times more light. This brings us to its implications in astronomical observation.
Astronomical Observation
In astronomy, gathering light is crucial. Observational instruments, such as telescopes, need more light to observe faint and distant objects. The collectable light impacts how much detail and range they can capture.

By having a larger collecting area:
  • The 8-meter reflector can gather 2.56 times more light than the 5-meter one.
  • This translates to better-quality images, the ability to see fainter objects, and improved observational data.
Imagine looking at the night sky through a small window versus a large window. The bigger window lets you see more stars, more deeply, and more clearly.
This principle applies directly to astronomical observations.

Thus, the 8-meter reflector signifies an advancement in observational capability, allowing astronomers to push the boundaries of their discovery.

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Most popular questions from this chapter

The focal length of the objective on your telescope is \(0.8 \mathrm{m} .\) You are using a \(25 \mathrm{cm}\) focal length eyepiece. In the image you find that the angular separation between two stars is 10 arc sec. What is the actual angular separation on the sky between the two stars?

The sodium D lines in the Sun's spectrum are at wavelengths of 589.594 and \(588.997 \mathrm{nm}\) (a) If a grating has \(10^{4}\) lines/cm, how wide must the grating be to resolve the two lines in first order? (b) Under these conditions what is the angular separation between the two lines? (c) How would the results in (a) and (b) change for second order?

If a radio source emits a solar luminosity \(\left(4 \times 10^{33} \operatorname{erg} | \mathrm{s}\right)\) in radio waves, (a) what would be the power per surface area reaching us if we were (i) \(1 \mathrm{AU}\) away, (ii) \(1 \mathrm{pc}\) away? (b) If that power is uniformly spread out over a frequency range of \(10^{11} \mathrm{Hz}\), what is the flux density (power/surface area/Hz) in each case? (c) How does this compare with the power you receive from a \(50 \mathrm{kW}\) radio station \(10 \mathrm{km}\) from you?

If we have two objects \(\theta(")\) apart on the sky, how far apart, \(x,\) are their images on the film of a camera with a focal length \(f\). (Assume that we wish to express \(x\) and \(f\) in the same units.)

(a) Using the fact that the limiting magnitude of the eye is \(6,\) derive an expression for the limiting magnitude for direct viewing with a telescope of diameter \(D\). (Ignore the effects of sky brightness.) (b) Use this result to derive an expression for the farthest distance at which a telescope of diameter \(D\) can be used to see an object of absolute magnitude \(M\).
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