Physics - Geometrical Optics Question with Solution | TestHub

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 $\begin{array}{rl}{f}_o& =\text{ focal length of objective }=200\mathrm{cm}\\ f_e& =\text{ focal length of eye-piece }=5\mathrm{cm}\end{array}$ 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.