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

R S

θ

R

Trigonometry “Angular Size” and “Resolution”

22 „ Astronomers usually sizes in terms R +Y of angles instead of lengths R because the distances are seldom well known S Y

θ θ S R R

S = physical length of the arc, measured in m Y = physical length of the vertical side [m]

Trigonometric Definitions Angles: units of measure R22+Y „ 2π (≈ 6.28) radians in a R 1 = 360˚ ÷ 2π≈57 ˚ S Y ⇒≈206,265 seconds of arc per radian

θ „ Angular (˚) is too large to be a useful R S angular measure of astronomical objects θ ≡ R 1º = 60 arc minutes opposite side Y 1 arc minute = 60 arc seconds [arcsec] tan[]θ ≡= adjacent side R 1º = 3600 arcsec -1 -6 opposite sideY 1 1 arcsec ≈ (206,265) ≈ 5 × 10 radians = 5 µradians sin[]θ ≡== RY22+ R 2 1+ Y 2

1 Number of Degrees per Radian in

Y 2π radians per circle θ S R 360° Usually R >> S (particularly in astronomy), so Y ≈ S 1 radian = ≈ 57.296° 2π SY Y 1 θ ≡≈≈ ≈ ≈ 57° 17'45" RR RY22+ R 2 1+ Y 2 θ ≈≈tan[θθ] sin[ ]

Relationship of Trigonometric sin[θ ] ≈ tan[θ ] ≈ θ for θ ≈ 0

Functions for Small Angles 1

sin(πx) Check it! tan(πx) 0.5 18˚ = 18˚ × (2π radians per circle) ÷ (360˚ per πx circle) = 0.1π radians ≈ 0.314 radians 0 Calculated Results -0.5 tan(18˚) ≈ 0.32 sin (18˚) ≈ 0.31 -1 0.314 ≈ 0.32 ≈ 0.31 -0.5 -0.25 0 0.25 0.5 x θ ≈ tan[θ ] ≈ sin[θ ] for |θ |<0.1π Three curves nearly match for x ≤ 0.1⇒ π|x|< 0.1π ≈ 0.314 radians

Astronomical Angular “Yardsticks” “Resolution” of Imaging System „ Easy yardstick: your hand held at arms’ length „ Real systems cannot “resolve” objects that fist subtends angle of ≈ 5˚ are closer together than some limiting spread between extended index finger and thumb ≈ 15˚ angle “Resolution” = “Ability to Resolve” „ Easy yardstick: the Moon „ Reason: “Heisenberg Uncertainty Relation” of disk of Moon AND of Sun ≈ 0.5˚ = ½˚ Fundamental limitation due to physics ½˚ ≈ ½ · 1/60 radian ≈ 1/100 radian ≈ 30 arcmin = 1800 arcsec

2 Image of Source With Smaller

1. Source emits “spherical waves” 2. Lens “collects” only part of the Lens “collects” a smaller part of sphere. and “flips” its curvature Can’t locate the equivalent (the “image”) as well Creates a “fuzzier” image

λ D

3. “piece” of sphere converges to form image

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

Apparent angular separation of the is ∆θ

Resolution and Lens Diameter Equation for

„ Larger lens: λ collects more of the spherical wave ∆θ ≈ λ = of better able to “localize” the D D = diameter of lens makes “smaller” images smaller ∆θ between distinguished sources means „ Better resolution with: BETTER resolution λ larger ∆≈θ λ = wavelength of light shorter D D = diameter of lens „ Need HUGE “lenses” at radio wavelengths to get same angular resolution

3 Resolution of Unaided and

„ Can distinguish shapes and shading of light „ Telescopes magnify distant scenes of objects with angular sizes of a few arcminutes „ Magnification = increase in angular size (makes ∆θ appear larger) „ Rule of Thumb: angular resolution of unaided eye is 1 arcminute

Simple Telescopes Galilean

„ Simple refractor telescope (as used by Galileo, Kepler, and their contemporaries) has two lenses lens „ collects light and forms intermediate image „ “positive ” „ Diameter D determines the resolution eyepiece

„ acts as “magnifying glass” applied to image from fobjective objective lens Ray incident “above” the optical axis „ forms magnified image that appears to be infinitely far emerges “above” the axis away image is “upright”

Galilean Telescope Keplerian Telescope

θ θ′

fobjective feyelens Ray entering at angle θ emerges at angle θ′ > θ Ray incident “above” the optical axis Larger ray angle ⇒ angular magnification emerges “below” the axis image is “inverted”

4 Keplerian Telescope 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 in ray trace, can show θ θ′ that f magnification =− objective feyelens

fobjective = of objective lens Ray entering at angle θ emerges at angle θ′ feyelens = focal length of eyelens where |θ′ |> θ

Larger ray angle ⇒ angular magnification „ magnification is negative ⇒ image is inverted

Ways to Specify Astronomical Magnification: Requirements Distances „ To increase apparent angular size of Moon from “actual” to angular size of “fist” requires magnification of: „ light year = distance light travels in 1 year 5° ° =×10 0.5 1 light year = 60 sec/min × 60 min/hr × 24 hrs/day × 365.25 days/year × (3 × 105) km/sec „ Typical Binocular Magnification ≈ 9.5 × 1012 km ≈ 5.9 × 1012 miles ≈ 6 trillion miles with , can easily see shapes/shading on Moon’s (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

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

“Background” star „ typically measured in arcseconds

Foreground star

B (6 months later) Apparent Position of Foreground Earth’s Orbit Star as seen from Location “A”

5 Limitations to Magnification Parallax as Measure of Distance „ Can you use a telescope (even a large one) to increase angular size of nearest star to match that of the Sun? „ Apparent motion of 1 arcsec in 6 months defines the distance of 1 parsec (parallax of 1 second) nearest star is α Cen (alpha Centauri) „ Brightest star in constellation Centaurus 1 parsec = 3.26 light years ≈ 3 × 1013 km ≈ 20 × 1012 miles = 20 trillion miles Diameter is similar to Sun’s

D = P-1 „ D is the distance (measured in pc) and P is parallax (in arcsec)

Limitations to Magnification α Centaurus „ Distance to α Cen is 1.3 pc 1.3 pc ≈ 4.3 light years ≈ 1.5×1013 km from Earth „ Near South Celestial „ Sun is 1.5 × 108 km from Earth Pole „ ⇒ would require angular magnification of Not visible from 5 Rochester! Southern Cross 100,000 = 10 „ ⇒ To obtain that magnification using telescope: 5 fobjective=10 × feyelens

α Centaurus

Limitations to Magnification Magnification: Limitations „ However, atmospheric effects typically „ Can one magnify images by arbitrarily large factors? dominate effects from diffraction „ Increasing magnification involves “spreading light most telescopes are limited by “seeing”: image out” over a larger imaging (detector) surface “smearing” due to atmospheric turbulence necessitates ever-larger light-gathering power, larger telescopes „ Rule of Thumb: „ BUT: Remember diffraction limiting resolution for visible light through the Wave nature of light, Heisenberg “” atmosphere is equivalent to that obtained by a Diffraction is the unavoidable propensity of light to change telescope with D ≈ 3.5" (≈ 90 mm) direction of propagation, i.e., to “bend” λ ∆≈θλ at = 500nm (Green light) Cannot light from a point source to an arbitrarily small D “spot” 500× 10−9 m „ Diffraction Limit of telescope λ =≈×=5.6 10−6 radians 5.6µ rad ∆≈θ 0.09 m D ≈≈1.2 arcsec 1/ 50 of eye's limit

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