A telescope has focal length of objective and eye-piece as 200 cm...

In relaxed eye state of telescope or when final image is formed at infinity, the magnification of telescope is given by $|M|=\frac{f_o}{f_e}$ where $$\begin{aligned} f_o & =\text{ focal length of objective }=200 \mathrm{ cm} \\ f_e & =\text{ focal length of eye-piece }=5 \mathrm{ cm}\end{aligned}$$ Hence, $|M|=\frac{200}{5}=40 |M|=\frac{200}{5}=40$ Note : The aperture of the objective lens is kept large so that more and more rays coming from the heavenly body may enter the telescope and a bright image is formed by the objective lens. The aperture of the eye lens is kept comparatively small so that all the rays may enter the eye.