Assignment #7 - With Solutions
Optics and Telescopes

Larry Bogan

Chapter #5 Chaisson and McMillan
Problem #1, 3, 4, 9

1. A CCD chip 1024x1024 pixels used with a certain telescope has a 10'x10'arc field of view. What angle of the sky corresponds to 1 pixel ? What would be the diameter of a typical seeing disk (2" arc radius) in pixels?

ANSWER -------------------

The CCD chip sees an area of the sky 10'x10' in angular size.
Each pixel then sees 10'/1024 = 0.0098'= 0.59" (since 1' = 60" arc)

A seeing disk of 4" diameter will be 4"/0.6" = 6.8 pixels

3. A 2 metre telescope can collect a given amount of light in 1 hour. Under the same conditions, how much time would be required for a 6 metre to perform the same task? A 12 metre telescope?

ANSWER -------------------

A telescope collects light in a rate proportional to the area of its primary mirror which is proportional to the diameter squared. Hence a 6 m telescope collects light (6/2)2 = 9 times faster than a 2 m telescope.
Time required to collect the same light is 60min/9 = 6.7 minutes

A 12 m telescope collects light (12/2)2 = 36 times fasters.
Time required for the 12m telescope = 60min/36 = 1.7 minutes = 100 seconds

4. A certain space-based telescope can achieve angular resolution of 0.05' for red light (wavelength 700 nm). What would be its resolution

ANSWER -------------------

The angular resolution is proportional to the wavelength of the electromagnetic radiation. (Same telescope)
angular resolution in radians = 1.2 x(wavelength)/(diameter)

(a) 700 nm = 0.70 micrometers so 3.5 micrometers is 5 times larger
hence resolution of 3.5 micrometers is 5 x 0.05' = 0.25'

(b) 140 nm is 1/5 of 700 nm so the resolution is 0.05' x 0.2 = 0.01'

9. Estimate the angular resolution of

  • (a) a radio interferometer with a 5000 km baseline, operating at a frequency of 5 GHz
  • (b) an infrared interferometer with a baseline of 50 m, operating at a wavelength of 1 micrometers.

    ANSWER -------------------

    Angular Resolution("arc) = 0.25 x (wavelength in micrometers)/(diameter in meters)

    (a)5 GHz = 5x109 Hz
    wavelength = (3x108 m/s)/ 5x109 Hz = 0.06 m
    = 60000 micrometers
    5000 km = 5x106 m
    So
    resolution = 0.25 (60000/5x106m) = 0.003" arc

    (b)Infrared Interferometer: wavelength = 1 micrometer, base = 50 m
    angular resolution = 0.25 (1/50) = 0.005" arc

    Other Exercises: (See Optics Page)

    1 - Determine the capabilities of the following two telescopes

    The telescopes use eyepieces with focal lengths of 7mm, 12mm, and 25mm.

    Make a table of the following properties

    ANSWER -------------------

    (a) The magnification = (focal length of the objective)/(focal length of the eyepiece)
    Telescope Magnifications
    7 mm12 mm25 mm
    A: dia = 125 mm
    fl = 600 mm    		865024
    B: dia = 333 mm
    fl = 1500 mm    		21412560

    (b) Field of view (degrees) = 57.3(object size)/(focal length of the objective) 

    A: 125 mm telescope:  FOV = 57.3 (20/600) = 1.9 degrees
B: 333 mm telescope:  FOV = 57.3 (20/1500) = 0.76 degrees

    (c) The light gathering power (LGP) compared with the eye of each telescope is given by the ratio of the of the area of the telescope objective to that of the pupil of the eye. That LGP is a brightness ratio and has an equivalence in magnitude difference. The eye can see the limit of 5.5 magnitude in the telescope but the telescope's LGP increases limit by the magnitude difference.

    LGP = [(diameter of objective)/(7 mm)]2
    Equivalent magnitude difference = 2.5 log10(B1/B2) = 2.5 log10(LGP) 

    A: 125 mm telescope: LGP = (125/7)2 = 319
		mag difference = 2.5 log10(319) = 6.3
		Limiting visible magnitude = 5.5 + 6.3 = 11.8
B: 333 mm telescope: LGP = (333/7)2 = 2263
		mag difference = 2.5 log10(2263) = 8.4
		Limiting visible magnitude = 5.5 + 8.4 = 13.9

    (d) The ideal angular resolution (in " arc) (AR) is given by 138/(diameter of objective in mm) 

    A: 125 mm telescope: AR = 138/125 = 1.1" arc
B: 333 mm telescope: AR = 138/333 = 0.4" arc

    (e) The brightness of an extended celestial object is proportional to the inverse of the focal ratio of the objective squared. Focal ratio (f#) = focal length/diameter

    Brightness, B ~ (D/fo)2 

    A: 125 mm telescope: f# = 600/125 = 4.8 
		B ~ (1/4.8)2 = 0.043
B: 333 mm telescope: f# = 1500/333 = 4.5
		B ~ (1/4.5)2 = 0.049
The two telescopes have almost the same focal ratios and hence yield almost the same 
brightness of image of the Moon. However, the image given by the larger telescope is 
brighter by 0.049/0.043 = 1.14

    2. - Find the shape of the orbit of a space probe sent to an asteroid with a circular orbit of radius 2.5 AU The probe will start at a perihelion near Earth orbit (1 AU from the Sun) and be at apihelion at 2.5 AU. Determine

    ANSWER -------------------

    (a) The orbit of the probe:
	rp = Perihelion = 1.0 AU
	ra = Aphelion = 2.5 AU
		Semi-major axis = (ra+rp)/2 = 3.5/2 = 1.75 AU
(b) The period of the orbit:
	P2 = a3   (Kepler's 3rd Law for the solar system)
	P = (1.75)3/2= 2.32 years
	Time of flight to the asteroid = 2.32/2 = 1.16 years
(c) The velocity of probe in its orbit at perihelion:
	vp = vc[(1+e)/(1-e)]1/2
	The circular velocity, vc= 2(pi)a/P  can be calculated
	But we would like the units in km/s 
		a = 1.75 AU = 1.75 x 1.5x108km/AU = 2.63x108km
		P = 2.32 years = 2.32 x (3.16x107 s/yr)=7.33x107s
	So vc = 6.28 (2.63x108km)/(7.33x107s)
			= 22.4 km/s
	The eccentricity, e, may be calculated from e = ra/a - 1
			e = (2.5/1.75) - 1 = 0.428
	vp = 22.4[(1+0.428)/(1-0.428)]1/2=22.4[(1.428)/(0.572)]1/2
		= 35.4 km/s
Compare this with the velocity of Earth in its orbit = 29.8 km/s
	Change in velocity of the probe to put it into its orbit to the asteroid
		= 35.4 - 29.8 = 5.6 km/s = 20,160 km/hr