Question
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 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) fo=30cm and fe=6cm
(B) fo=0.72 and fe=3cm
(C) fo=30cm and fe=10cm
(D) f0=50cm and fe=10cm
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 L . Let the focal length of the objective and the eyepiece be fo and fe respectively. Let the angular magnification be represented as M .
Angular magnification of an astronomical telescope is given by
M=fefo
Substituting the given values of fo and fe in M , we get
5=fefo
⇒5fe=fo
Also, L=fo+fe
By substituting the given value of L , we get
36=fo+fe
By substituting the already derived relation for fo in the above equation, we get
36=5fe+fe
∴fe=636
From the above equation, we get fe=6cm .
∴fo=5fe
By substituting the calculated value for fe ,
fo=5×6=30cm
So, the calculated values are: fo=30cm and the calculated value for fe=6cm.
∴ 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).