• # question_answer In an astronomical telescope, the focal length of objective lens and eyepiece are 150cm and 6cm respectively. In case when final image is formed at least distance of clear vision $(D=25cm)$. The magnifying power is: A)  29                               B)  30 C)  31                               D)  32

The magnification is given by $m=\frac{{{F}_{O}}}{{{F}_{e}}}\left( 1+\frac{Fe}{D} \right)=\frac{150}{6}\,\left( 1+\frac{6}{25} \right)=31$