Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The angular magnification of an astronomical telescope of a distant object is $5$. If the distance between the objective and eyepiece is $36cm$ and the final image is formed at infinity, the focal length of its objective and eyepiece will be:(A) ${f_o} = 30cm$ and ${f_e} = 6cm$ (B) ${f_o} = 0.72$ and ${f_e} = 3cm$ (C) ${f_o} = 30cm$ and ${f_e} = 10cm$ (D) ${f_0} = 50cm$ and ${f_e} = 10cm$

Last updated date: 24th Jul 2024
Total views: 63.9k
Views today: 0.63k
Verified
63.9k+ views
Hint: Use the angular magnification equation which will be in terms of objective focal length and eyepiece focal length. This will give a direct relation between the focal length of the eyepiece and the focal length of the objective of the telescope. The distance between the lens is equal to the focal lengths of the objective and the eyepiece. Using these relations to arrive at the answer.

Complete step by step solution:
Let the distance between the objective lens and the eyepiece be $L$ . Let the focal length of the objective and the eyepiece be ${f_o}$ and ${f_e}$ respectively. Let the angular magnification be represented as $M$ .
Angular magnification of an astronomical telescope is given by
$M = \dfrac{{{f_o}}}{{{f_e}}}$
Substituting the given values of ${f_o}$ and ${f_e}$ in $M$ , we get
$5 = \dfrac{{{f_o}}}{{{f_e}}}$
$\Rightarrow 5{f_e} = {f_o}$
Also, $L = {f_o} + {f_e}$
By substituting the given value of $L$ , we get
$36 = {f_o} + {f_e}$
By substituting the already derived relation for ${f_o}$ in the above equation, we get
$36 = 5{f_e} + {f_e}$
$\therefore {f_e} = \dfrac{{36}}{6}$
From the above equation, we get ${f_e} = 6cm$ .
$\therefore {f_o} = 5{f_e}$
By substituting the calculated value for ${f_e}$ ,
${f_o} = 5 \times 6 = 30cm$

So, the calculated values are: ${f_o} = 30cm$ and the calculated value for ${f_e} = 6cm$.
$\therefore$ option (A) is the correct option.

Note:
In an astronomical telescope, the diminished image of the objective is made to fall on the focus of the eyepiece. That is why the distance between the lenses is equal to the focal lengths of the lenses. The eyepiece gives an enlarged, magnified image of this virtual object( image of an objective).