Question

the magnifying power of an astronomical telescope for normal adjustment is (a) -fo/fe (b) -fo/fe (c) - fe/fo (d) -fo+fe

Answer 1

answer : (a ) and (b) [ because both options are same e.g., -$\frac{f_0}{f_e}$ ]

explanation : you already know,

magnification of astronomical telescope, M = angle made by the final image at eye/ angle made the object at eye

see figure ,

M = $\frac{tan\beta}{tan\alpha}$

= $\frac{AB/BE}{AB/OB}$

= $\frac{OB}{BE}$

= $\frac{f_0}{-f_e}$

= -$\frac{f_0}{f_e}$

hence, magnifying power of astronomical telescope is -$\frac{f_0}{f_e}$