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More # 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 areA) $10,4.4\;{\text{cm}}$ B) $4,4.4\;{\text{cm}}$ C) $44,10\;{\text{cm}}$ D) $10,44\;{\text{cm}}$ Verified
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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.

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.