question_answer

Which two main considerations are kept in mind while designing the 'objective? of an astronomical telescope? Obtain an expression for the angular magnifying power and the length of the tube of an astronomical telescope in its 'normal adjustment' position.                    

Answer:

                Two main considerations with an astronomical telescope are: (i) its light gathering power and (ii) its resolving power. The light gathering power depends on the area of the objective. With larger diameters, fainter objects can be observed. The resolving power, or the ability to observe two objects distinctly, which are very nearly in the same direction, also depends on the diameter of the objective. When the final image is formed at infinity: Normal adjustment. As shown in Fig. when a parallel beam of light is incident on the objective, it forms a real, inverted and diminished image A?B? in its focal plane. The eyepiece is so adjusted that the image A?B; exactly lies at its focus. Therefore, the final image is formed at infinity, and is highly magnified and inverted with respect to the object. Magnifying power in normal adjustment. It is defined as the ratio of the angle subtended at the eye by the final image as seen through the telescope to the angle subtended at the eye by the object seen directly, when both the image and the object lie at infinity. As the object is very far off, the angle subtended by it at the eye is practically equal to the angle \[\alpha \]subtended by it at the objective. Thus \[\angle A'OB'=\alpha \] And let \[\angle A'EB'=\beta \] \[\therefore \] Magnifying power, \[m=\frac{\beta }{\alpha }=\frac{\tan \beta }{\tan \alpha }\,\,[\because \,\alpha ,\,\beta \,are\,small\,angle]\] \[=\frac{A'B'/B'E}{A'B'/OB'}=\frac{OB'}{B'E}\] Applying new Cartesian sign convention, \[OB'=+{{f}_{0}}=\,Dis\tan ce\,of\,A'B'\,from\]\[the\,objective\,along\,the\,incident\,light\] \[B'E=-{{f}_{e}}=\,Dis\tan ce\,of\,A'B'\,from\,\]\[the\,eyepiece\,against\,the\,incident\,light\] \[\therefore \] \[m=-\frac{{{f}_{0}}}{{{f}_{e}}}\] Clearly for large magnifying power, \[{{f}_{0}}>>{{f}_{e}}.\] The negative sign for m indicates that the image is real and inverted. The length of the telescope in normal adjustment\[={{f}_{0}}+{{f}_{e}}\].