The rays are refracted from boundary of objective lens to make an image at f0 in normal adjustment as shown in figure. These rays again refracted from boundary of eye-lens and eye-ring ab is formed. Eye-ring ab is image of object lens AB formed at eyepiece. It is the best position of eye when employing telescope as maximum amount of light enters object lens from outside thus creating the wide field of view. At the distance closer to eye lens than eye-ring, no further improvement in view is attained. Using lens equation for above figure, one can got the value of distance v from eye-ring as
1/v + 1/(f1 + f2) = 1/f2
On rearranging terms of the equation, we get
v = f2/f1(f1 + f2)............Eq.1
Objective diameter: eye-ring diameter = Height of object/ Height of image
= Distance of object/Distance of image
AB/ab = u/v = [(f1 + f2)/(f2/f1)(f1 + f2)] = f1/f2.......Eq.2
But the angular magnification of telescope is provided by relation as
M = f1/f2
This signifies that for angular magnification to be large, f1 should be much greater than f2
So, angular magnification for the telescope in normal adjustment can also be stated as
M = Diameter of objective/diameter of eye ring.......Eq.3
Astronomical Telescope with Image Formed at Near Point:
When telescope is not in normal adjustment the eye required an accommodation to focus image to a numerical distance D (the least distance of distinct vision).
The image formed by objective lens fall between focus of the eye lens and the eye lens so that the final image is virtual as shown in figure given above.
Angular magnification is given by M = α'/α
Or in another form, it can be written as
M = (h/u)/(h/f1) = f1/u.......Eq.4
But value of u can be attained using the Eq.
1/u + 1/v = 1/f
1/u + 1/-D = 1/f2
u = f2D/f2 + D.......Eq.5
Substituting the value of u from Eq.5 in Eq.4, we get the expression for magnification as
M = (f1/f2)((f2 + D)/D)
M = f1/f2(1 + f2/D)
But as you know
f1/f2 = M(angular magnification)
This signifies that angular magnification can also be expressed as
M = Diameter of objective/Diameter of eye ring.
Unlike the astronomical telescope utilized to observe distant objects or objects at infinity, like stars and moon, terrestrial telescope is utilized to view distant objects on land. This signifies that it is significant for final image to be erect. It thus comprises of the erecting lens placed between object lens and eye-lens. Erecting lens is placed such that center of curvature of its faces coincides with focus of the object lens. This is done so that angular magnification of the instrument is retained and also the final image remains erect.
Despite the importance of the erecting lens, its disadvantages are:
i) Erecting lens decreases intensity of light through eye-piece.
ii) Instrument is now longer by 4f, where f is focal length of erecting lens.
The Reflector Telescope:
The astronomical telescope with lens as its objective is known as a refractor telescope while the reflector telescope has the large curved mirror as its objective. It comprises of parabolic mirror of large size as its objective. Due to its large size, mirror collects large amount of light from distant planets and brought to focus to be photographed. Advantages of using this type of telescope are:
i) It has the large angular magnification. Expression
Where, f1 is the focal length of objective.
As it is hard to manufacture the lens with very large focal length, there is limitation to magnification. This can be attained with the mirror; it is preferable to use the very large mirror as objective.
ii) Telescope is free from chromatic aberration caused by the objective lens.
iii) There will be no loss of light because of absorption and reflection of light as on surface of a lens.
iv) Large diameter of objective increases resolving power of telescope.
Resolving power of the telescope (two distant objects just seem separated) is provided by
P = D/1.22λ = 1/Θ
Here Θ is smallest angle that is subtended at the telescope by two distant objects that can just be seen separated. Other symbols utilized here are: D is diameter of objective lens and λ is the mean wavelength of light from object.
It can be seen that Θ is inversely proportional to resolving power of the telescope. If value of Θ is smaller, then resolving power is greater. Value of Θ doesn't depend on focal length of objective but only depends on diameter.
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