# Chapter 8

Chapter 8 – Telescopes 1. Using diagrams, explain the principle of
a) An astronomical refractor.

The light from a celestial object it refracted by an Object Glass. This then travels down the tube and crosses at a focal point. The light then continues on into the Eyepiece and can be seen by the viewer.

b) A Newtonian reflec tor.
The light from a celestial object enters the tube and reaches the main mirror at the back. This mirror is a convex shape, so it reflects the light at an angle towards a focusing mirror, which is flat and at an angle to direct the light into an eyepiece and to the viewer’s eye.

c) What are the advantages and disadvantages of each type?
The advantages of a refracting telescope are: they are quite simple to handle; they do not go out of adjustment unless roughly treated; and they can survive quite a while.
The disadvantages are: because of the light spectrum, not all the different colour lights will be refracted the same meaning that some celestial objects may have circles of light around them; the image is inverse, when the image is actually upside down; and they can be quite expensive.

The advantages of a reflecting telescope are: they are cheaper than a refracting telescope; and they do not have the problem of circles of light surrounding celestial objects.
The disadvantages are: the mirrors inside the telescope need re-coating with either silver or aluminium to continue to have maximum reflection; they are easily damaged if roughly handled; and there is a small mirror in the middle of the tube, which means it scatters the light and causes some loss in contrast.

2. Give the magnifications produced in the following cases:
a) A 6-inch (15cm) f/8 reflector with a ½ inch (1.27cm) eyepiece.

Focal ratio = focal length/lens diameter.
Focal length = focal ratio x lens diameter
FL = 15 x 6
FL= 90cm

M=F/f                                M=?
M=90/1.3                          F=90
M=70.9                              f=1.3
M=  x71

b) A 3 inch (7.5cm) f/10 refractor with a ½ inch (1.27cm) eyepiece.
Fr=Fl/Ld                            Fr= 10
Fl=Fr x Ld                          Fl= ?
Fl=10 x 7.5                         Ld= 7.5
Fl=75cm

M=F/f                                M= ?
M=75/1.3                           F= 75
M=57.7                               f= 1.3
M=   x58

c) An 8 inch (20cm) f/8 reflector with ¼ inch (0.6cm) eyepiece.
Fr=Fl/Ld                            Fr= 8
Fl=Fr x Ld                          Fl= ?
Fl=8 x 20                           Ld= 20
Fl=160cm

M=F/f                                M= ?
M=160/0.6                       F= 160
M=266.7                            f= 0.6
M=   x267

d)A 4 inch (10cm) f/12 refractor with a 1 inch (2.5cm) eyepiece.
Fr=Fl/Ld                            Fr= 12
Fl=Fr x Ld                          Fl= ?
Fl=12 x 10                           Ld= 10
Fl=120cm

M=F/f                                M= ?
M=120/2.5                        F= 120
M=48                                  f= 2.5
M=   x48

e)A 3 inch (7.5cm) f/12 refractor with an 1⁄8 inch (0.3cm) eyepiece.
Fr=Fl/Ld                            Fr= 12
Fl=Fr x Ld                          Fl= ?
Fl=12 x 7.5                         Ld= 7.5
Fl=90cm

M=F/f                                M= ?
M=90/0.3                         F= 90
M=300                              f= 0.3
M=   x300

3. You have a 6 inch (15cm) reflector with a focal length of 48 inches (122cm). If you were choosing three eyepieces to use with this telescope, what would be their focal lengths? Give your reasons.

The focal ratio is 122/15 = 8.13 = f/8. So I would chose a 1 inch eyepiece, which would give a magnification of x48, a ½ inch eyepiece that gives a x96 magnification and a ¼ inch eyepiece to give a x192 magnification. To go any bigger with that size reflector will diminish the amount of light causing large magnificated celestial objects to appear dim or not appear at all.

4. Two telescopes, X and Y, are both optically good. X is a refractor of 3 inches (7.5cm) aperture, with a focal length of 48 inches (122cm); it has an altazimuth tripod mounting. Y is a reflector of the Newtonian type, aperture 6 inches (15cm), and has an equatorial mount on a tripod. Each telescope is provided with an eyepiece of focal length ½ inch (1.27cm).
a) What magnification will be obtained with each telescope?

X-    M=F/f                                M= ?
M=122/1.3                                 F= 122
M=93.8                                      f= 1.3
M=   x94

Y-    M=F/f                                M= ?
M=122/1.3                                F= 122
M=93.8                                      f= 1.3
M=   x94

b) Which telescope will show the fainter stars – and why?
Telescope Y will show the fainter stars because it has a bigger aperture, which means it can catch more light. Also, whilst reflecting, there is no light loss; unlike with refracting.

c) If you had a choice between these telescopes for your own use, which would you choose, and why?
I would choose telescope X. Both telescopes have the same magnification, and even though there other one has a larger aperture, it will have some reduced light due to the flat mirror. Telescope X has refracting lenses which means I don’t have to make any adjustments and I do not need to treat the lenses. Also I would prefer to have a altazimuth mounting than a equatorial one.

5. a) If you had a choice between 7 x 50 binoculars and a altazimuth 2 inch (5cm) refractor, which would you prefer, and why?
I would prefer the 2 inch refractor because it has a mounting tripod, meaning there is no disturbance from shaking hands; which is a problem with a pair of binoculars.

b) What are disadvantages of binoculars with a magnification of greater than about x12?
Without mounting, there is a risk of having very shaky hands and with something with such a high magnification, but a small field of vision, shaky hands will mean I will see nearly nothing.

6. Calculate the percentage reduction of the light received by a Newtonian reflector with a 10cm mirror, due to a 2cm obstruction in the tube caused by the flat.
πr²                       πr²
π5²                       π2²
π25                       π4
78.54cm²            12.57cm²

12.57/78.54 x 100 = 16%

We lose about 16% of the original light collected.