Data Taken by. Exploring Telescopes

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1 Astro 30/Harpell Date: Instructor s signature Data Taken by (your name) LabPartners: Exploring Telescopes There are a few factors to consider when shopping for a telescope, or simply comparing different types. The primary consideration is the size of the Objective (lens or Mirror). In general, the larger the objective, the greater the amount of the light the telescope can gather (and hence the shorter the exposure time), the greater the resolving power (and hence the more detail that can be seen), and the greater the effective maximum magnifying ability of the telescope. Magnification Magnification depends on both the focal length of the telescope and the focal length of the eyepiece according to the relation: M = ftel/feye Thus, telescopes with the same focal lengths will have different magnifications with different eyepieces. For example, a telescope with a focal length of 100 centimeters and an eyepiece with focal length of ten millimeters will have a magnification of M = 100/1.0 = 100 Note that eyepiece focal lengths are typically given in millimeters. Also notice that both focal lengths must be in the same units (i.e. centimeters). As you can see, the smaller

2 the eyepiece, the greater the magnification it can provide. The focal length of a telescope depends on the curvature its lens or mirror. The F Ratio It is standard in the telescope-making world to list both the diameter of a telescope and its f ratio. While this sounds confusing, it is really quite simple. The f ratio is simply the ratio of the focal length of the telescope to the diameter of its objective. If the telescope described above (with a focal length of 100 cm) had a mirror 10 cm in diameter, then its f ratio would be: F ratio = 100/10 = 10 F ratios are commonly written is a special format. For this particular telescope, its f ratio would be written as f/10. Commonly, the diameter and the f ratio are listed together--for example a 10 cm objective, and f/10. These two bits of information allow you to easily calculate the focal length (10 x the diameter = 100 cm). Telescopes with small f ratios are considered to be fast while telescopes with large f ratios are not so fast, less fast, or downright slow The designation of fast vs. slow has to do with imaging and exposure times. The shorter the necessary exposure time, the faster the telescope. A 10 cm, f/10 would provide high magnification but would cut down on the light available per pixel. A 10 cm f/5 would provide less magnification, but would provide twice as much light per pixel and would thus be twice as fast, or require one half of the exposure time. Resolving Power and Magnification Returning to the issue of magnification, recall that the magnification for a given telescope depends on the focal length of the eyepiece being used. For the telescope described above, changing from a 10 mm eyepiece, to a 5 mm eyepiece would change the magnification from 100 to 200. With this in mind, why do we usually use longer focal length eyepieces (i.e. around 20 to 50 mm), as opposed to much shorter eyepieces, which would provide unlimited magnification? The answer has to do with the amount of detail that can be seen. Basically, if you magnify a slightly fuzzy image, you will get a larger, very fuzzy image. Technically, this is described as the resolving power of the telescope: α = 1.22λ/D where α is the smallest angular separation between two objects that can be resolved, λ is the wavelength of the light being viewed, and D is the diameter of the Telescope s objective. If we assume green light, then a handy formula is: α = 14/D where D is in centimeters, and α is in arc seconds. Thus a 10-centimeter objective would provide resolving power of 1.4 arc seconds. If the telescope has a resolving power of 1.4 arc seconds, it would make no sense to magnify an image so large that angular separations smaller than 1.4 arc sections are visible. Such small details would be blurred out in the magnified image. Effective Maximum Magnification Problems arise when the atmosphere comes into play, as it always does on Earth. Typically, light from a distant source is refracted by the moving atmosphere and creates a seeing disk on the retina of the viewer, or on the detector. This seeing disk is typically

3 larger than one arc second, meaning that a 14 cm telescope would achieve the same theoretical resolution as a 14-meter telescope because of atmospheric limitations. That s not the whole story, however, since the effective magnification of the telescope is also related to the size of the objective (in general, the more light the clearer the image) by the relation: Mmax = 10D where D is in centimeters. (note: this is an approximate relationship which depends on many factors such as the quality of the optical system, observing conditions, etc). Thus the telescope described above would have a Maximum effective magnification of 100. Smallest Useful Eyepiece Since you don t want to magnify objects more than the maximum effective magnification, there is a limit to the size of the eyepiece you can use. Remember, the smaller the focal length of the eyepiece, the greater the magnification. Also, the ratio of the focal lengh of the telescope to the focal length of the eyepiece equals the magnification. Thus, for the example telescope, the ratio of the focal length of the telescope to that of the eyepiece equals 100, so: f tel /f eye = 100. In other words 1000mm/f eye = 100 Or f eye = 10 mm Part I Telescope Properties Now it s your turn. There will be various telescopes located around the roomyou re your report briefly describe the telescopes in addition to filling out the table below. To help you in your description, fill in the missing information in the tables below. Basic Telescope Information (Table 1) Telescope Name Telescope Type Objective Diameter Focal Length f-ratio (f-number) Example reflecting 100 mm 1000 mm 10 LX200GPS Schmidt-Cassegrain 356 mm 3556 mm RCX400 Ritchey-Chrétien 305 mm 2438 mm Orion 120 Refractor 120 mm 1000 mm Telescope Properties (Table 2) Telescope Name Theoretical resolving power (arc seconds) Magnification for 10 and 40 mm eyepiece 10 mm 40 mm Maximum possible magnification Example LX200GPS RCX400 Orion 120 Smallest useful eyepiece (mm)

4 Part 2 Angular Measurements of the Moon and Jupiter 1. What is the angular size of the moon tonight? (You ll need to look this number up or find out from your instructor.) What is the distance to the moon in km tonight? Angular size of the moon Distance to the moon (d) 2. Once you are outside at a telescope, write down the name of the telescope and the focal length of the eyepiece. You will do this for two telescopes so that you have two different magnifications. 3. Can you see the whole moon in the telescope? Hint: choose the telescope and eyepiece with the smallest magnification. If so, estimate the percentage or fraction of the field of view that the moon takes up. If you cannot see the entire moon, estimate the percentage of the moon that is visible. Based on your estimate and on your answer for question 1, what is the field of view size for this eyepiece and telescope combination? For example, if the moon has an angular diameter of 0.5 degrees takes up ½ the diameter of the field of view, then the field of view is 1.0 degrees. If entire moon is visible, portion of the field of view it takes up Telescope 1 Telescope 2 If part of moon is visible, portion of moon visible Telescope 1 Telescope 2 Field of View: Telescope 1 Telescope 2 4. Commandeer one of the telescopes. Switch out the current eyepiece for a microguide eyepiece. The microguide is a 9mm focal length eyepiece. The eyepiece has a red LED that can be turned on to illuminate a grid that should look like this: Tonight we will be using the markings that go across the center of the microguide

5 (Labeled 1 in the diagram above) as a ruler. The angular size of each division depends on the focal length of the telescope you are using. Calculate the distance in arc seconds between scale divisions (SD) using the following equation: SD = f where f is the focal length of your telescope. SD 5. Using your moon map, identify two craters and measure their diameters in scale divisions (number of markings on the microguide eyepiece). Fill the information in Table 3 below. 6. Calculate the angular size of the craters in arc seconds by multiplying your answer to part 5 by your answer to part 4. Fill the information in table 3 below. 7. Calculate the approximate size of each crater in km using this equation: π s = d θ where d is the distance to the moon in km and θ is the crater diameter in arc seconds. Fill in the information in table 3 below. 8. Locate the Copernicus Crater and the Eratosthenes Crater. Repeat the process used in steps 5-7, this time measuring the distance between the centers of the two craters Crater Name Crater measurements (Table 3) Crater Diameter in Crater Diameter number of divisions in arc seconds (θ) Crater Diameter in km (s) Distance between Copernicus and Eratosthenes Distance in number of divisions Distance in arc seconds (θ) Distance in km (s) 9. Point the telescope at Jupiter. (if visible) Measure the angular size of Jupiter and the angular distance from Jupiter to the closest visible moon and the farthest visible moon. Jupiter System Measurements (Table 4) Measurement Number of divisions arc seconds Jupiter Diameter Distance from Jupiter to closest moon

6 Distance from Jupiter to farthest moon