measuring angles and angular resolution
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Measuring Angles and Angular Resolution 1 Angles Angle is the - PDF document

Measuring Angles and Angular Resolution 1 Angles Angle is the ratio of two lengths: R: physical distance between observer and objects [km] S: physical distance along the arc between 2 objects Lengths are measured in same


  1. Measuring Angles and Angular Resolution 1

  2. Angles � Angle θ is the ratio of two lengths: � R: physical distance between observer and objects [km] � S: physical distance along the arc between 2 objects � Lengths are measured in same “units” (e.g., kilometers) � θ is “dimensionless” (no units), and measured in “radians” or “degrees” R S θ R 2

  3. “Angular Size” and “Resolution” � Astronomers usually measure sizes in terms of angles instead of lengths � because the distances are seldom well known θ S R 3

  4. Trigonometry + 2 2 R Y R Y S θ R S = physical length of the arc, measured in m Y = physical length of the vertical side [ m ] 4

  5. Trigonometric Definitions + 2 2 R Y R Y S θ R S θ ≡ R opposite side Y [ ] θ ≡ = tan adjacent side R opposite side Y 1 [ ] θ ≡ = = sin + hypotenuse 2 2 2 R Y R + 1 2 Y 5

  6. Angles: units of measure � 2 π ( ≈ 6.28) radians in a circle � 1 radian = 360 ˚ ÷ 2 π ≈ 57 ˚ � ⇒ ≈ 206,265 seconds of arc per radian � Angular degree ( ˚ ) is too large to be a useful angular measure of astronomical objects � 1º = 60 arc minutes � 1 arc minute = 60 arc seconds [arcsec] � 1º = 3600 arcsec � 1 arcsec ≈ (206,265) -1 ≈ 5 × 10 -6 radians = 5 µ radians 6

  7. Number of Degrees per Radian π 2 radians per circle ° 360 ° ≈ 1 radian = 57.296 π 2 ° ≈ 57 17'45" 7

  8. Trigonometry in Astronomy θ S Y R Usually R >> S (particularly in astronomy), so Y ≈ S S Y Y 1 θ ≡ ≈ ≈ ≈ R R + 2 2 2 R Y R + 1 2 Y [ ] [ ] θ ≈ θ ≈ θ tan sin 8

  9. Relationship of Trigonometric Functions for Small Angles Check it! 18 ˚ = 18 ˚ × (2 π radians per circle) ÷ (360 ˚ per circle) = 0.1 π radians ≈ 0.314 radians Calculated Results tan(18 ˚ ) ≈ 0.32 sin (18 ˚ ) ≈ 0.31 0.314 ≈ 0.32 ≈ 0.31 θ ≈ tan[ θ ] ≈ sin[ θ ] for | θ |< 0.1 π 9

  10. sin [ θ ] ≈ tan [ θ ] ≈ θ θ ≈ 0 for 1 sin( π x) tan( π x) π x 0.5 0 -0.5 -1 -0.5 -0.25 0 0.25 0.5 x Three curves nearly match for x ≤ 0.1 ⇒ π | x | < 0.1 π ≈ 0.314 radians 10

  11. Astronomical Angular “Yardsticks” � Easy yardstick: your hand held at arms’ length � fist subtends angle of ≈ 5 ˚ � spread between extended index finger and thumb ≈ 15˚ � Easy yardstick: the Moon � diameter of disk of Moon AND of Sun ≈ 0.5 ˚ = ½ ˚ ½ ˚ ≈ ½ · 1/60 radian ≈ 1/100 radian ≈ 30 arcmin = 1800 arcsec In the DRAWING: Point A: The sky appears blue due to scattering. The scattered light from the other ray is linearly polarized. Point B: When this person looks toward the sun the sky appears reddish because the most of the shorter wavelength light has already been scattered away. 11

  12. “Resolution” of Imaging System � Real systems cannot “resolve” objects that are closer together than some limiting angle � “Resolution” = “Ability to Resolve” � Reason: “Heisenberg Uncertainty Relation” � Fundamental limitation due to physics 12

  13. Image of Point Source 1. Source emits “spherical waves” 2. Lens “collects” only part of the sphere and “flips” its curvature λ D 3. “piece” of sphere converges to form image 13

  14. With Smaller Lens Lens “collects” a smaller part of sphere. Can’t locate the equivalent position (the “image”) as well Creates a “fuzzier” image 14

  15. Image of Two Point Sources Fuzzy Images “Overlap” and are difficult to distinguish (this is called “DIFFRACTION”) 15

  16. Image of Two Point Sources Apparent angular separation of the stars is ∆θ 16

  17. Resolution and Lens Diameter � Larger lens: � collects more of the spherical wave � better able to “localize” the point source � makes “smaller” images � smaller ∆θ between distinguished sources means BETTER resolution λ ∆ θ ≈ λ = wavelength of light D D = diameter of lens 17

  18. Equation for Angular Resolution λ λ = wavelength of light ∆ θ ≈ D = diameter of lens D � Better resolution with: � larger lenses � shorter wavelengths � Need HUGE “lenses” at radio wavelengths to get same angular resolution 18

  19. Resolution of Unaided Eye � Can distinguish shapes and shading of light of objects with angular sizes of a few arcminutes � Rule of Thumb: angular resolution of unaided eye is 1 arcminute 19

  20. Telescopes and magnification � Telescopes magnify distant scenes � Magnification = increase in angular size � (makes ∆θ appear larger) 20

  21. Simple Telescopes � Simple refractor telescope (as used by Galileo, Kepler, and their contemporaries) has two lenses � objective lens � collects light and forms intermediate image � “positive power” � Diameter D determines the resolution � eyepiece � acts as “magnifying glass” applied to image from objective lens � forms magnified image that appears to be infinitely far away 21

  22. Galilean Telescope f objective Ray incident “above” the optical axis emerges “above” the axis image is “upright” 22

  23. Galilean Telescope θ θ′ Ray entering at angle θ emerges at angle θ′ > θ Larger ray angle ⇒ angular magnification 23

  24. Keplerian Telescope f eyelens f objective Ray incident “above” the optical axis emerges “below” the axis image is “inverted” 24

  25. Keplerian Telescope θ′ θ Ray entering at angle θ emerges at angle θ′ where | θ′ | > θ Larger ray angle ⇒ angular magnification 25

  26. Telescopes and magnification � Ray trace for refractor telescope demonstrates how the increase in magnification is achieved � Seeing the Light, pp. 169-170, p. 422 � From similar triangles in ray trace, can show that f = − objective magnification f eyelens � f objective = focal length of objective lens � f eyelens = focal length of eyelens � magnification is negative ⇒ image is inverted 26

  27. Magnification: Requirements � To increase apparent angular size of Moon from “actual” to angular size of “fist” requires magnification of: ° 5 ° = × 10 0.5 � Typical Binocular Magnification � with binoculars, can easily see shapes/shading on Moon’s surface (angular sizes of 10's of arcseconds) � To see further detail you can use small telescope w/ magnification of 100-300 � can distinguish large craters w/ small telescope � angular sizes of a few arcseconds 27

  28. Ways to Specify Astronomical Distances � light year = distance light travels in 1 year 1 light year = 60 sec/min × 60 min/hr × 24 hrs/day × 365.25 days/year × (3 × 10 5 ) km/sec ≈ 9.5 × 10 12 km ≈ 5.9 × 10 12 miles ≈ 6 trillion miles 28

  29. Aside: parallax and distance � The only direct measure of distance astronomers have for objects beyond the solar system is parallax � Triangulation � Parallax: apparent motion of nearby stars (against a background of very distant stars) as Earth orbits the Sun � Requires taking images of the same star at two different times of the year Caution: NOT to scale Apparent Position of Foreground Star as seen from Location “B” A “Background” star Foreground star B (6 months later) Apparent Position of Foreground Star as seen from Location “A” Earth’s Orbit 29

  30. Parallax as Measure of Distance P Background star Image from “A” Image from “B” 6 months later � P is the “parallax” � typically measured in arcseconds 30

  31. Parallax as Measure of Distance � Apparent motion of 1 arcsec in 6 months defines the distance of 1 parsec ( parallax of 1 second ) � 1 parsec = 3.26 light years ≈ 3 × 10 13 km ≈ 20 × 10 12 miles = 20 trillion miles � D = P -1 � D is the distance (measured in pc) and P is parallax (in arcsec) 31

  32. Limitations to Magnification � Can you use a telescope (even a large one) to increase angular size of nearest star to match that of the Sun? � nearest star is α Cen (alpha Centauri) � Brightest star in constellation Centaurus � Diameter is similar to Sun’s 32

  33. α Centaurus � Near South Celestial Pole � Not visible from Southern Cross Rochester! α Centaurus 33

  34. Limitations to Magnification � Distance to α Cen is 1.3 pc � 1.3 pc ≈ 4.3 light years ≈ 1.5 × 10 13 km from Earth � Sun is 1.5 × 10 8 km from Earth � ⇒ would require angular magnification of 100,000 = 10 5 � ⇒ To obtain that magnification using telescope: f objective = 10 5 × f eyelens 34

  35. Limitations to Magnification � Can one magnify images by arbitrarily large factors? � Increasing magnification involves “spreading light out” over a larger imaging (detector) surface � necessitates ever-larger light-gathering power, larger telescopes � BUT: Remember diffraction � Wave nature of light, Heisenberg “uncertainty principle” � Diffraction is the unavoidable propensity of light to change direction of propagation, i.e., to “bend” � Cannot focus light from a point source to an arbitrarily small “spot” λ � Diffraction Limit of telescope ∆ θ ≈ D 35

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