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Astronomy 102, Fall 2003

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Field of View of the Telescope

Goal: To determine the angular diameter of the field of view of the Celestron-8 telescope when using the 25mm and the 10mm eyepieces.

Due Date: Competency in these skills is expected by the end of your second week of using the telescope. This lab must be completed by the February 14 lab grading date.

Summary: In order to become more accustomed to the telescope, a simple exercise to determine the angular diameter of the field of view of the telescope is useful. The field of view is that portion of the sky visible through the telescope. The field of view is different for each eyepiece. Higher magnification results in a smaller field of view. Knowing the field of view of the telescope for each eyepiece gives you a reference to estimate the size of the celestial objects you are observing. It is also very helpful to know when you are star hopping. For small objects like planets, or for more precise measurements, the method of transit times is to be preferred. As you will see, we will use the method of transit times to measure the angular diameter of the field of view of the telescope.

Introduction

Since the celestial sphere (the sky) appears to rotate in one day, the stars drift through the field of a stationary telescope. Since the rotation of the sky is well known (360o in 24 hours), we can use that fact to measure the field of view of the telescope by simply timing how long it takes for a star to drift through the field of view.

Example:

Suppose for example that a star takes 2 minutes to drift across the diameter of the field of view. Then we can readily compute the field of view (FOV):

FOV = 2min *(1 hour/ 60 min) * (360o/24 hour) = 0.50o OR 30' (minutes of arc)

This is approximately the apparent diameter of the Moon. In this telescope/eyepiece combination, the Moon would barely fit in the field of view.

The above example is strictly valid only if we use a star located on the celestial equator (declination =0o). Stars located away from the celestial equator also make one turn in the sky in 24 hours, but their apparent motion takes place on smaller circles and their angular speed across the sky is smaller. An extreme example is the pole star, Polaris. It is located very close to the North Celestial Pole and if you point your telescope at it with the drive turned off, you will see that it hardly moves at all. This is exactly the same situation as we have on Earth. Someone standing at the North pole would make one turn in one day, but wouldn't be moving at all! At the same time, people on the Equator are the ones who are moving the fastest due do the Earth's rotation. We can correct for this effect in our field of view measurement by introducing the cosine of the declination (latitude in our Earth example) of the star you are using. The field of view is correctly given by:

FOV= (transit time in hours) * cos (declination) * (360o/24 hours)

Note the following conversions for a star located on the celestial equator:

transit time angle
24 hours 360o
1 hour 15o
1 minute 15'
1 second 15"

Recall that 1o (degree)= 60' (minutes of arc) =3600" (seconds of arc)

Procedure

When you are all done, compute the average transit times and the field of view in degrees and in arc minutes, rounded off to 0.01o/1' , for each eyepiece and for the finderscope. Write these numbers clearly so you can refer to them easily for the rest of the semester. Outside of the lab, comment on the accuracy of your measurements and possible sources of error.



Last modified: 2003-January-7, by Robert A. Knop Jr.

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