1. Biconvex lens of power 0.1 dioptre.
2.
3. Magnification:
The magnifying power of a telescope is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective.
i.e. m = \(\frac{β}{\alpha}\) …….(1)[from figure]
But from ∆CIM, tanα = \(\frac{IM}{IC}\), α = \(\frac{IM}{IC}\)
(For small values tan α ≈ α)
from ∆C^{1} IM, tanβ = \(\frac{IM}{IC^1}\) ,β = \(\frac{IM}{IC^1}\)
substituting α and β in eq (1) we get
But IC = f_{o} (the focal length objective lens)
and IC^{1} = f_{e} (the focal length eyepiece lens.)
∴ \(m =\frac{f_0}{f_e}\)
In this case the length of the telescope tube is (f_{o} + f_{e}).
Case 1 :
When the image formed by the objective is within the focal length of the eyepiece, Then the final image is formed at the least distant of distinct vision. In this case, magnifying power.
m = \(\frac{f_0}{f_e}\Big[1+\frac{f_e}{D}\Big]\)
4.
- As magnifying power is negative, the final image in an astronomical telescope is inverted.
- To have large magnifying power, f_{0} must be as large as possible and f_{e} must be as small as possible.
- As intermediate image is between the two lens, cross wire (ora measuring device) can be used.
- In normal setting of telescope, the final image is at infinity.