JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is \(30\,cm\). The focal length of the objective, when the angular magnification of the telescope is \(2\) , will be \(.....cm\)
- A \(20\)
- B \(30\)
- C \(10\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
\(f _{0}+ f _{ e }=30\) \(m =\frac{ f _{0}}{ f _{ e }}\) \(2=\frac{ f _{0}}{ f _{ e }} \Rightarrow f _{0}=2 f _{ e }\) So \(f _{0}+\frac{ f _{0}}{2}=30\) \(f _{0}=20\,cm\)
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