Saved Bookmarks
| 1. |
A terrestrial telescope has an objective of focal length 180cm and an eyepiece of focal length 5.0cm. The erecting lens has focal length of 3.5cm. What is the separation betwee the objective and the eyepiece?What is the magnifying power of the telescope? |
|
Answer» Solution :The image formed by the objective FALLS at 2F of the erecting lens which forms an ERECT image of the same size. The separation between the objective and the EYEPIECE = `f_0+f_e+4xxfocal length of the erecting lens `180+5+4xx3.5` = 199cm. Magnifying power of the TELESCOPE `f_o/f_e` = 180/5 = 36 |
|