The focal length of the objective and the eyepiece of a telescope are 50 cm and 5 cm respectively. If the telescope is focussed for distinct vision on a scale distant 2 m from its objective, then its magnifying power will be :

The focal length of the objective, $$ f_o = 50 $$ cm, and the focal length of the eyepiece, $$ f_e = 5 $$ cm. The telescope is focused for distinct vision on a scale 2 m (200 cm) from the objective. The least distance of distinct vision, $$ D $$, is 25 cm. First, find the image distance for the objective lens. The object distance for the objective is $$ u_o = -200 $$ cm (negative because the object is on the side from which light is incident). Using the lens formula: $$$ \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} $$$ Substitute the values: $$$ \frac{1}{v_o} - \frac{1}{-200} = \frac{1}{50} $$$ $$$ \frac{1}{v_o} + \frac{1}{200} = \frac{1}{50} $$$ $$$ \frac{1}{v_o} = \frac{1}{50} - \frac{1}{200} $$$ $$$ \frac{1}{50} = \frac{4}{200}, \quad \frac{1}{200} = \frac{1}{200} $$$ $$$ \frac{1}{v_o} = \frac{4}{200} - \frac{1}{200} = \frac{3}{200} $$$ $$$ v_o = \frac{200}{3} \approx 66.67 \text{ cm} $$$ The magnification by the objective lens is given by the ratio $$ m_o = \frac{v_o}{u_o} $$: $$$ m_o = \frac{200/3}{-200} = -\frac{1}{3} $$$ For the eyepiece, the final image is formed at the least distance of distinct vision, so the image distance $$ v_e = -25 $$ cm (negative because the image is virtual and on the same side as the incident light). The angular magnification of the eyepiece when the final image is at $$ D $$ is: $$$ M_e = 1 + \frac{D}{f_e} $$$ Substitute $$ D = 25 $$ cm and $$ f_e = 5 $$ cm: $$$ M_e = 1 + \frac{25}{5} = 1 + 5 = 6 $$$ The overall magnifying power of the telescope is the product of the linear magnification of the objective and the angular magnification of the eyepiece: $$$ M = m_o \times M_e $$$ Substitute the values: $$$ M = \left(-\frac{1}{3}\right) \times 6 = -2 $$$ The negative sign indicates that the final image is inverted. Hence, the correct answer is Option D.