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Question: The angular magnification of an astronomical telescope of a distant object is \[5\]. If the distance...

The angular magnification of an astronomical telescope of a distant object is 55. If the distance between the objective and eyepiece is 36cm36cm and the final image is formed at infinity, the focal length of its objective and eyepiece will be:
(A) fo=30cm{f_o} = 30cm and fe=6cm{f_e} = 6cm
(B) fo=0.72{f_o} = 0.72 and fe=3cm{f_e} = 3cm
(C) fo=30cm{f_o} = 30cm and fe=10cm{f_e} = 10cm
(D) f0=50cm{f_0} = 50cm and fe=10cm{f_e} = 10cm

Explanation

Solution

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 LL . Let the focal length of the objective and the eyepiece be fo{f_o} and fe{f_e} respectively. Let the angular magnification be represented as MM .
Angular magnification of an astronomical telescope is given by
M=fofeM = \dfrac{{{f_o}}}{{{f_e}}}
Substituting the given values of fo{f_o} and fe{f_e} in MM , we get
5=fofe5 = \dfrac{{{f_o}}}{{{f_e}}}
5fe=fo\Rightarrow 5{f_e} = {f_o}
Also, L=fo+feL = {f_o} + {f_e}
By substituting the given value of LL , we get
36=fo+fe36 = {f_o} + {f_e}
By substituting the already derived relation for fo{f_o} in the above equation, we get
36=5fe+fe36 = 5{f_e} + {f_e}
fe=366\therefore {f_e} = \dfrac{{36}}{6}
From the above equation, we get fe=6cm{f_e} = 6cm .
fo=5fe\therefore {f_o} = 5{f_e}
By substituting the calculated value for fe{f_e} ,
fo=5×6=30cm{f_o} = 5 \times 6 = 30cm

So, the calculated values are: fo=30cm{f_o} = 30cm and the calculated value for fe=6cm{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).