Question
In an astronomical telescope in normal adjustment a straight black line of length $$L$$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $$I.$$ The magnification of the telescope is
A.
$$\frac{L}{I} + 1$$
B.
$$\frac{L}{I} - 1$$
C.
$$\frac{{L + 1}}{{L - 1}}$$
D.
$$\frac{L}{I}$$
Answer :
$$\frac{L}{I}$$
Solution :
We know, magnification of telescope, we have
$$\eqalign{ & M = \frac{{{f_0}}}{{{f_e}}}, \cr & {\text{Here,}}\,\,\frac{{{f_e}}}{{{f_e} + u}} = \frac{{ - I}}{L} \cr & \Rightarrow \frac{{{f_e}}}{{{f_e} - \left( {{f_o} + {f_e}} \right)}} = \frac{{ - I}}{L} \cr & \Rightarrow \frac{{{f_e}}}{{{f_o}}} = \frac{I}{L} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,M = \frac{L}{I} \cr} $$
We know, magnification of telescope, we have
$$\eqalign{ & M = \frac{{{f_0}}}{{{f_e}}}, \cr & {\text{Here,}}\,\,\frac{{{f_e}}}{{{f_e} + u}} = \frac{{ - I}}{L} \cr & \Rightarrow \frac{{{f_e}}}{{{f_e} - \left( {{f_o} + {f_e}} \right)}} = \frac{{ - I}}{L} \cr & \Rightarrow \frac{{{f_e}}}{{{f_o}}} = \frac{I}{L} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,M = \frac{L}{I} \cr} $$
