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Resolving Power

Resolving Power. Astrophysics Lesson 4. Learning Objectives. Resolving power; Appreciation of diffraction pattern produced by circular aperture, Airy disc Resolving power of telescope, Rayleigh criterion, . Resolving Power .

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Resolving Power

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  1. Resolving Power Astrophysics Lesson 4

  2. Learning Objectives • Resolving power; Appreciation of diffraction pattern produced by circular aperture, Airy disc • Resolving power of telescope, Rayleigh criterion,

  3. Resolving Power • The resolving power of any optical instrument is an indication of how good it is at distinguishing two objects close to one another.  • For example, at a long distance two car headlights appear as a single blob of light. 

  4. Angles in radians or degrees • There are 2π radians in a circle = 360°. • 1 radian is approximately 57° (180/π). • Degrees are subdivided into arc-minutes (1° = 60 arc-minutes [60']) • and arc-seconds (1' = 60 arc seconds [60'']) • So 1°=60'=3,600''

  5. Angles in radians or degrees • Radians have the advantage that for small angles: • sin θ~ tan θ~θ • This makes trigonometrical functions easier.  However astronomers tend to use arc-seconds which are useful for describing very small areas of sky.

  6. Question • The Moon has a diameter of about 3500 km and is about 400,000 km from the Earth.  • What is the angle in radians that the Moon subtends to an observer on the Earth? • What is this in degrees?

  7. Answer • For small angles in radians, sin θ = tan θ = θ. • Angle of moon = 3500 km ÷ 400,000 km = 8.8 × 10-3 rad • 1 rad = 360/2π = 57.3° • Angle in degrees = 0.504° (= 30' 15'')

  8. Question 3 • Entirely coincidentally the angle subtended by the Sun is exactly the same as the angle subtended by the Moon (think Solar Eclipse!).  The distance between the Earth and the Sun is 150 x 106 km. • What is the diameter of the Sun?

  9. Answer 3 • For small angles in radians, sin q = tan q = q. • Angle of moon = 3500 km ÷ 400000 km = 8.8 x 10-3 rad • Diameter of the Sun = 150 x 106 km x 8.8 x 10-3 = 1.32 x 106 km

  10. Diffraction Effects • When light enters a telescope, it is passing through a gap.  It spreads out by the process of diffraction.  • You will remember that when light passes through a single slit, dark and bright fringes are made.  • The resulting pattern is called a Fraunhofer Diffraction pattern:

  11. Fraunhofer Diffraction Pattern

  12. Circular Diffraction • Fraunhofer diffraction also occurs with circular openings.  If we use a circular aperture we get an effect like this:

  13. The central bright spot is called an Airy Disc. • The physicist Lord Rayleigh studied the effect of overlapping of fringes and came up with the Rayleigh's Criterion.  The angular separation is given by the formula: • [θ - angular separation (rad); λ - wavelength (m); D - aperture width (m)]

  14. Calculation • Calculate the resolving power of a radio telescope using the diameter of the Earth. • Calculate the resolving power of your eye. • If headlights of car were about 5 km away, can we can tell that they are two separate lights?

  15. This is because the eye can resolve down to an angle about 3 × 10-4 radians. • To improve the resolution of a telescope, we need to have a large aperture and a short wavelength.

  16. Recap • Calculate the resolving power of a radio telescope using the diameter of the Earth. • Calculate the resolving power of your eye. • If headlights of car were about 5 km away, can we can tell that they are two separate lights?

  17. Problems of Ground Based Telescopes • The observation of objects in space is made difficult because the atmosphere is turbulent.  This results in the twinkling or scintillation of stars. 

  18. Telescope Locations • Light pollution from street lights does not help either.  • Dust in the atmosphere causes scattering of light. • Major observatories have moved as far away as possible from cities and are situated on high mountains.  • Um, like Hawaii, Canary Islands, Mexico. Nice.

  19. Or into space... • The best images of them all come from the Hubble Space Telescope (HST) which is in orbit above the Earth. • A Cassegrain instrument

  20. Mars images using HST

  21. Question What is the resolving power of a telescope of diameter 15 cm at a wavelength of 600 nm? Give your answer in arcseconds?

  22. Answer Use θ = λ/D = 600 × 10-9 / 0.15 = 4 × 10-6 rad 1 radian = ((3600 × 180)/π) arcseconds 4 × 10-6 rad = 4 × 10-6 ((3600 × 180)/π) = 0.825''

  23. In practice, although telescopes have much better resolution that the eye, this is limited by the atmosphere. Telescopes have large apertures to allow as much light to get in as possible.

  24. Objectives (Spare) • Astronomical telescope consisting of two converging lenses • Ray diagram to show the image formation in normal adjustment • Angular magnification in normal adjustment • M =      angle subtended by image at eye •          angle subtended by object at  unaided eye     • Focal lengths of the lenses • M = fo/fe • Focal point of concave mirror, cassegrain arrangement. • Draw ray diagrams to show path of rays through the telescope.

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