Question

The focal length of the eye lens and object lens of a telescope is $4\;{\text{mm}}$ and $4\;{\text{cm}}$. If the final image of a far object is at $\infty $. Then the magnifying power and length of the tube are
A) $10,4.4\;{\text{cm}}$
B) $4,4.4\;{\text{cm}}$
C) $44,10\;{\text{cm}}$
D) $10,44\;{\text{cm}}$

Answer 1

Hint: The above problem can be solved by using the working principle of the telescope. The magnifying power of the telescope is the ability of the telescope to enlarge the size of the image of the distant object. It depends on the focal lengths of the eyepiece lens and object lens.

Complete step by step answer:
Given: The focal length of the eye lens is ${f_e} = 4\;{\text{mm}} = 4\;{\text{mm}} \times \dfrac{{1\;{\text{cm}}}}{{10\;{\text{mm}}}} = 0.4\;{\text{cm}}$.
The focal length of the object lens is ${f_o} = 4\;{\text{cm}}$.
The expression to calculate the magnifying power of telescope is given as:
$m = \dfrac{{{f_o}}}{{{f_e}}}......\left( 1 \right)$
Substitute $4\;{\text{cm}}$ for ${f_o}$ and $0.4\;{\text{cm}}$ for ${f_e}$ in the expression (1) to calculate the magnifying power of the telescope.
$m = \dfrac{{4\;{\text{cm}}}}{{0.4\;{\text{cm}}}}$
$m = 10$
The expression to calculate the length of the tube of the telescope is given as:
$L = {f_o} + {f_e}......\left( 2 \right)$
Substitute $4\;{\text{cm}}$ for ${f_o}$ and $0.4\;{\text{cm}}$ for ${f_e}$ in the expression (2) to calculate the length of the tube of the telescope.
$L = 4\;{\text{cm}} + 0.4\;{\text{cm}}$
$L = 4.4\;{\text{cm}}$

Thus, the magnifying power of the telescope is 10, the length of the tube of the telescope is $4.4\;{\text{cm}}$ and the option (A) is the correct answer.

Additional Information:
The telescope consists of two converging lenses placed along the same axis. These two lenses are connected through the tube. The one lens is called an eye lens and placed near the human eye. The other lens faces the distant object and is called the objective lens. The focal length and aperture of the objective lens is more than the eye lens.

Note: Be careful in substituting the value of focal length of eye and objective lens. The magnifying power is the ratio of the focal length of the objective lens to focal length of the eye lens. The length of the tube of the telescope is equal to the sum of focal lengths of the eye and objective lens.