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An astronomical telescope has an eyepiece of a focal-length of $5cm$. If the angular magnification in normal adjustment is 10 when the final image is at least distance of distinct vision ($25cm$) from the eyepiece, then angular magnification will be:
A) $10$
B) $12$
C) $50$
D) $60$

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Last updated date: 05th Mar 2024
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Hint: In order to solve this question you have to know the formula for the angular magnification for both the normal adjustment and when the final image is at least distance of distinct vision from the eyepiece. Also, remember all the concepts related to an astronomical telescope.

Formula used:
$m' = - \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)$
here ${f_e}$ is the focal length of the eyepiece of the given astronomical telescope
\[{f_o}\] is the focal length of the given astronomical telescope
$D$ is the least distance of distinct vision of the human eye

Complete step by step solution:
All the information given in the question are:
An astronomical telescope has an eyepiece of focal length, ${f_e} = 5cm$
Angular magnification in normal adjustments, $m = 10cm$
Final image is at least distance of distinct vision, $D = 25cm$
Firstly apply the formula for magnification for normal adjustment that is,
$m = - \dfrac{{{f_o}}}{{{f_e}}}$
On putting all the given values,
$10 = \dfrac{{{f_o}}}{5}$
On further solving, we have
$ \Rightarrow {f_o} = 50cm$
Now, when final image is at least distance of distinct vision from eyepiece, then angular magnification $m'$ is given by
$m' = - \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)$
Here ${f_e}$ is the focal length of the eyepiece of the given astronomical telescope
\[{f_o}\] is the focal length of the given astronomical telescope
$D$ is the least distance of distinct vision of the human eye
$ \Rightarrow m' = 10\left( {1 + \dfrac{5}{{25}}} \right)$
On further solving, we get the angular magnification as
$ \Rightarrow m' = 12$

Therefore, the correct option is (B).

Note: A astronomical telescope is an optical instrument used for observing far away distance. It has an object with an eyepiece with a short focal length and a large focal length to observe distant objects or celestial bodies. The ability of a telescope to magnify a distant object is known as its magnifying power. Mathematically, it is equal to the ratio of the focal length of the objective to the focal length of the eyepiece.