The focal lengths of objective lens and eye lens of a Galilean Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition the Magnifying Power of the Galilean Telescope should be:

In a Galilean telescope, the objective lens is converging with focal length $$f_o = 30$$ cm, and the eye lens is diverging with focal length $$f_e = -3$$ cm. The final image is formed at the least distance of distinct vision, $$D = 25$$ cm, on the same side as the object for the eye lens, so $$v_e = -25$$ cm.

Using the thin lens formula for the eye lens: $$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$$, we get $$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{-3}$$. Solving, $$\frac{1}{u_e} = \frac{1}{3} - \frac{1}{25} = \frac{25 - 3}{75} = \frac{22}{75}$$, so $$u_e = \frac{75}{22}$$ cm. The positive value of $$u_e$$ confirms the intermediate image is virtual and on the same side as the final image, which is characteristic of a Galilean telescope.

The magnifying power of a Galilean telescope when the final image is at the least distance of distinct vision is given by $$M = \frac{f_o}{|f_e|}\left(1 + \frac{|f_e|}{D}\right) = \frac{30}{3}\left(1 + \frac{3}{25}\right) = 10 \times \frac{28}{25} = \frac{280}{25} = +11.2$$. The positive sign indicates an erect image, consistent with the Galilean telescope.