An astronomical telescope has a magnifying power $10$. The focal length of the eyepiece is $20 \, cm$. The focal length of the objective is.....$cm$.

The focal length of the objective lens of a telescope is $30 \text{ cm}$ and that of its eye lens is $3 \text{ cm}$. It is focused on a scale at a distance $2 \text{ m}$ from it. The distance of the objective lens from the eye lens to see the clear image is: (in $\text{ cm}$)

The focal lengths of the objective and eyepiece of an astronomical telescope are $60\, cm$ and $5\, cm$ respectively. When the final image is formed at the least distance of distinct vision $(25\, cm)$, the magnifying power and the length of the telescope will be, respectively:

An astronomical telescope has objective and eye-piece lenses of powers $0.5\, D$ and $20\, D$ respectively. Its magnifying power will be:

In an astronomical telescope in normal adjustment,a straight black line of length $L$ is drawn on the inside part of the objective lens. The eyepiece forms a real image of this line. The length of this image is $I$. The magnification of the telescope is:

The focal length of the objective and the eyepiece of a telescope are $100\, cm$ and $5\, cm$ respectively. If the final image is formed at the least distance of distinct vision,the magnification of the telescope is: