Question

The focal length of eye lens and object lens of a telescope is 4 mm and 4 cm respectively. If final image of an far object is at $\infty$. Then the magnifying power and length of the tube are:

• A. 10, 4.4 cm
• B. 4, 44 cm
• C. 44, 10 cm
• D. 10, 44 cm

Answer 1

The correct option is A 10, 4.4 cm

Magnification is the amount that a telescope enlarges its subject. Its equal to the telescopes focal length divided by the eyepieces focal length. As a rule of thumb, a telescopes maximum useful magnification is 50 times its aperture in inches (or twice its aperture in millimeters).
That is, $M=fo/fe$
In this case, the focal length of eye lens and object lens of a telescope is 4 mm = 0.4 cm and 4 cm respectively.
So, Magnification $M=fo/fe=4/0.4=10.$
Focal length of the eyepiece is the distance from the center of the eyepiece lens to the point at which light passing through the lens is brought to a focus.
Focal length of the objective is the distance from the center of the objective lens (or mirror) to the point at which incoming light is brought to a focus.
The length of the tube is given as sum of the focal lengths of the eye lens and the object lens.
So, Length of the tube $=fo+fe=4+0.4=4.4cm.$
Hence, the magnifying power and length of the tube are: 10, 4.4 cm.