Telescope & Magnifying Power

Fig-1 shows a telescope in the normal adjustment i.e. when the final image is formed at infinity and the eye is most relaxed. In this situation , 

• the magnifying power of the  telescope will be numerically equal to  ratio of the focal length of the objective to the focal length of the eye piece i.e.   f₀ / f_e  .  Sign-wise, it will be negative as the final image is inverted with respect to the object.
• Separation between the objective and the eyepiece  ( also called length of the instrument) will be equal to the sum of the focal lengths of the two lenses i.e. objective and eyepiece.
• The  intermediate image is real and inverted.
• The overall magnifying power will be more in magnitude if the focal of the objective  is increased and focal length of the eyepiece is decreased.
•   If focal length of the objective is increased, the size of the intermediate image  grows in same proportion.

In the second fig, the telescope is so adjusted that the final image is positioned at the near point with respect to the eyepiece. In this situation 

• Magnifying power  will be of magnitude  ( f₀ / f_e ) (1+ f_e / D). This value of the magnifying power is more than that of in normal adjustment.
• Length of the instrument  is now  f_{0 +   (} f_e D )/ (f_e +D)  because the intermediate image  will now be at distance  f_e D / (f_e +D)  from the eyepiece. Note that this length is  shorter than the length in case of the normal adjustment.