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?

Short Answer

Expert verified
3.125 arc sec

Step by step solution

01

Convert the focal lengths to the same units

Convert the eyepiece focal length from centimeters to meters. Since 1 meter = 100 centimeters, then the 25 cm eyepiece focal length is \[ \text{Focal length of eyepiece} = 25 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 0.25 \text{ m} \]
02

Calculate the magnification of the telescope

Use the formula for magnification, which is \[ \text{Magnification} = \frac{\text{Focal length of objective}}{\text{Focal length of eyepiece}} \] Substitute the given values into this formula:\[ \text{Magnification} = \frac{0.8 \text{ m}}{0.25 \text{ m}} = 3.2 \]
03

Calculate the actual angular separation

The formula to find the actual angular separation is \[ \text{Actual Angular Separation} = \frac{\text{Observed Angular Separation}}{\text{Magnification}} \] Given that the observed angular separation is 10 arc sec, we substitute in the values:\[ \text{Actual Angular Separation} = \frac{10 \text{ arc sec}}{3.2} \ = 3.125 \text{ arc sec} \]

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Key Concepts

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

focal length conversion
Converting focal lengths to the same units is essential for accurate astronomical calculations. Focal length refers to the distance between the lens and the point where it brings light into focus. Usually, measurements might be in different units, such as centimeters and meters. To convert the eyepiece focal length from centimeters to meters, you use the conversion factor where 1 meter equals 100 centimeters. For instance, a 25 cm focal length eyepiece is converted as follows: \[ \text{Focal length of eyepiece} = 25 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 0.25 \text{ m} \] This conversion helps in maintaining uniformity and simplifies further calculations like magnification.
angular separation
Angular separation measures the apparent distance between two celestial objects, such as stars, as seen from Earth. The observed angular separation is often given in arc seconds – where 1 degree equals 3600 arc seconds. To find the actual angular separation in the sky, you need to factor in the telescope's magnification: \[ \text{Actual Angular Separation} = \frac{\text{Observed Angular Separation}}{\text{Magnification}} \] For example, if the observed angular separation is 10 arc seconds and the telescope magnification is 3.2, then: \[ \text{Actual Angular Separation} = \frac{10 \text{ arc sec}}{3.2} = 3.125 \text{ arc sec} \] This means the stars are actually 3.125 arc seconds apart in the sky.
astronomical calculations
Astronomical calculations are essential for determining the true positions and motions of celestial objects. When using a telescope, you often need to calculate magnification, angular separation, and convert units. Magnification helps you understand the extent to which an image is enlarged. It is calculated by the formula: \[ \text{Magnification} = \frac{\text{Focal length of objective}}{\text{Focal length of eyepiece}} \] For example, if the objective's focal length is 0.8 meters and the eyepiece’s is 0.25 meters: \[ \text{Magnification} = \frac{0.8 \text{ m}}{0.25 \text{ m}} = 3.2 \] This calculation helps measure the actual angular separation between celestial objects more accurately.
telescope optics
Telescope optics involve understanding how lenses and mirrors work together to magnify distant celestial objects. The objective lens or mirror focuses light and creates an image. The eyepiece lens then magnifies this image. High-quality optical components ensure clear and sharp images. Key aspects include:
  • Objective Lens: It gathers and focuses light
  • Eyepiece: It further magnifies the focused light
  • Focal Length: Determines magnification power
Using these components effectively allows for precise observations of stars, planets, and other celestial bodies. Understanding telescope optics enhances your stargazing experience and supports more accurate astronomical measurements.

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Most popular questions from this chapter

Suppose we are observing an infrared source that is 500 pc away. It radiates like a \(50 \mathrm{K}\) blackbody and is 1 pc in extent. (a) What is the total energy per second per square meter reaching the Earth from this source? How does that compare with the total amount of solar radiation reaching the Earth per second per square meter. (b) Suppose we observe this source using a satellite with a \(1 \mathrm{m}\) diameter mirror, and we observe at a wavelength of \(100 \mu \mathrm{m} .\) What is the energy/s/Hz striking the telescope? (c) Suppose the telescope radiates like a blackbody at \(300 \mathrm{K}\), but with an efficiency of \(1 \% .\) (That is, the spectrum looks like that of a blackbody but with an intensity reduced by a factor of \(100 .\). What is the energy/Hz/s given off by the telescope at this wavelength? How does your answer compare with that in (b). (d) Redo part (c), assuming that we can cool the mirror to \(30 \mathrm{K}\) (still with a \(1 \%\) emission efficiency

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