Question

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

Answer 1

Use the formula for the magnification of the telescope. This formula gives the relation between the magnification of the telescope and focal lengths of the eyepiece lens and objective lens of the telescope. The length of the tube is the sum of the focal lengths of the eyepiece lens and objective lens.

Formula used:
The magnification $m$ of a telescope is given by
$m = \dfrac{{{f_o}}}{{{f_e}}}$ …… (1)
Here, ${f_o}$ is the focal length of the objective lens and ${f_e}$ is focal length of the eyepiece lens.

Complete step by step answer:
We have given that the focal length of the eyepiece lens is $4\,{\text{mm}}$ and the focal length of the objective lens is $4\,{\text{cm}}$.
${f_e} = 4\,{\text{mm}}$
$\Rightarrow{f_o} = 4\,{\text{cm}}$
Convert the focal length of the eyepiece lens from millimeter to centimeter.
${f_e} = \left( {4\,{\text{mm}}} \right)\left( {\dfrac{{{{10}^{ - 1}}\,{\text{cm}}}}{{1\,{\text{mm}}}}} \right)$
$\Rightarrow {f_e} = 0.4\,{\text{cm}}$
Hence, the focal length of the eyepiece lens is $0.4\,{\text{cm}}$.
Let us first calculate the magnification of the telescope.Substitute $4\,{\text{cm}}$ for ${f_o}$ and $0.4\,{\text{cm}}$ for ${f_e}$ in equation (1).
$m = \dfrac{{4\,{\text{cm}}}}{{0.4\,{\text{cm}}}}$
$\Rightarrow m = 10$
Hence, the magnification of the telescope is 10.

Let us now calculate the length of the tube of the telescope.The length of the tube of the telescope is the sum of the distance from centre of the eyepiece lens to the focal point of the eyepiece lens and distance from the same focal point to centre of the objective lens.In other words, we can define the length of a tube of a telescope as the sum of focal lengths of the objective lens and eyepiece lens.
$L = {f_e} + {f_o}$
Substitute $4\,{\text{cm}}$ for ${f_o}$ and $0.4\,{\text{cm}}$ for ${f_e}$ in equation (1).
$L = \left( {0.4\,{\text{cm}}} \right) + \left( {4\,{\text{cm}}} \right)$
$\therefore L = 4.4\,{\text{cm}}$
Hence, the length of the tube is $4.4\,{\text{cm}}$.
Therefore, the magnification of the telescope and length of the tube are 10 and $4.4\,{\text{cm}}$ respectively.

Hence, the correct option is A.

Note: The students should not forget to convert the unit of focal length of the eyepiece lens from millimeter to centimeter (i.e. to the CGS system of units). If this unit conversion is not done then the final answers for the magnification and length of the tube of the telescope will be incorrect as length of the tube is in centimeter.