1)What is the light gahtering power of a 8 inch diameter telescope compared to t
ID: 296599 • Letter: 1
Question
1)What is the light gahtering power of a 8 inch diameter telescope compared to the human eye?
2) How much more times the light gathering power does a telescope with a 11 inch diameter objective have than a telescope with a 3 inch diameter.
3) What is the theoretical angular resolution of a 5 inch diameter telescope measured in seconds of arc? Round your answer to the nearest hundredth.
Exmaple: 0.1452 is submitted as 0.15
4)What is the magnification of an image when using a 10 inch diameter telescope with a focal length of 900 mm when using an eyepiece with a 15 mm focal length?
5)What is the maximum practicl magnification of a telescope with a 10 inch diameter objective and a focal length of 1200 mm?
Telescope A has a diameter of 6 inches and a focal length of 1200mm.
Telescope B has a diameter of 10 inches and a focal length of 800mm.
Which telescope could distinguish two stars separated by 0.7"?
A)
B only
B)
neither A nor B
C)
both A and B
D)
A only
A)
B only
B)
neither A nor B
Explanation / Answer
1) A = ? ( ??/2 ) 2
The pupil of a human eye is approximately 7 mm.
Telescope diameter = 8 inch =203 mm
light gahtering power of a 8 inch diameter telescope compared to the human eye = 2032/ 72 =841 times
2) the light gathering power of a telescope with a 11 inch diameter objective than a telescope with a 3 inch diameter = 112 / 32 =13.44 times
3) For visible light, the minimum separation of two objects which can just be resolved is: ? = 4.56/ ?? , where ? is the angular resolution of two objects (in arcseconds), and D is the diameter of the telescope's primary mirror (in inches)
? = 4.56/ 5 = 0.912 = 0.91
4) The magnifying power of a telescope is determined by the focal length of both the telescope and the eyepiece used.
MAGNIFICATION OF TELESCOPE = TELESCOPE FOCAL LENGTH / EYEPIECE FOCAL LENGTH
=900/15 = 60
5) for a person to see two objects that have a separation of less than 2 minutes of arc, a telescope needs to magnify the separation to one the eye can resolve, which is 2 minutes of arc, or 120 arc-seconds.
So then we have: Mmax x PR(Resolving Power) = 120
Mmax x 120/DO = 120
Mmax = DO = 10 inch
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